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IGCSE Chemistry · ⁨IGCSE 化学⁩

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IGCSE化学(0620)は、物質の状態と原子構造から始まり、化学量論、電気化学、熱力学、反応速度、酸と塩基、周期表、金属類、環境化学、有機化学を経て、実験技術と分析で終わる。

化学量論が成績を左右する。 独立した問題として出題されるだけでなく、電気化学、熱力学、反応速度の内部でも登場するため、モルに関する学習時間は複数倍の回报を得られる。

他の確実な失点原因は方程式である。未平衡、状態記号の欠落、あるいはイオン方程式を完全な分子方程式で記述すること。6コンポーネントの問題集があり、基礎(Core)と拡張(Extended)に分かれるので、受験するクラスを確認してから復習を始めよ。

  • 1

    States of matter

    Watch lesson · ⁨レッスンを視聴⁩
    1.1

    The three states of matter

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State the distinguishing properties of solids, liquids and gases
    2 Describe the structures of solids, liquids and gases in terms of particle separation, arrangement and motion
    3 Describe changes of state in terms of melting, boiling, evaporating, freezing and condensing 5 Explain changes of state in terms of kinetic particle theory, including the interpretation of heating and cooling curves
    4 Describe the effects of temperature and pressure on the volume of a gas 6 Explain, in terms of kinetic particle theory, the effects of temperature and pressure on the volume of a gas
    日本語
    コア サプリメント
    1 固体、液体、気体の区別される性質を述べる
    2 粒子間距離、配列、運動の観点から、固体、液体、気体の構造を記述する
    3 融解、沸騰、蒸発、凝固、凝縮の観点から、状態変化を記述する 5 運動論的粒子理論の観点から状態変化を説明し、加熱曲線および冷却曲線の解釈を行う
    4 温度および圧力が気体の体積に与える効果を記述する 6 運動論的粒子理論の観点から、温度および圧力が気体の体積に与える効果を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Solid, liquid, gas: particle motion

    Everything around you is made of tiny particles 粒子 — these can be atoms 原子, molecules 分子, or ions 离子. The kinetic particle theory 粒子动理论 says these particles are always moving. How close the particles are, how they are arranged, and how they move decides whether matter is a solid 固体, a liquid 液体, or a gas 气体.

    Properties you can observe

    You do not need a microscope to tell the three states apart. They behave in different ways:

    • A solid has a fixed shape and a fixed volume 体积. It does not flow and you cannot compress 压缩 it (squeeze it smaller).
    • A liquid has a fixed volume but no fixed shape. It flows and takes the shape of its container. It is almost impossible to compress.
    • A gas has no fixed shape and no fixed volume. It flows and spreads out to fill the whole container. A gas is easy to compress.

    The table below sums up these properties. Density 密度 means how much mass 质量 is packed into a given volume.

    Property Solid Liquid Gas
    Shape fixed takes the shape of the container fills the whole container
    Volume fixed fixed fills the whole container
    Can it be compressed? no almost none yes, easily
    Does it flow? no yes yes
    Density high high low

    The particle picture

    The kinetic particle theory explains these properties by looking at three things: the separation 间距 of the particles (how far apart they are), their arrangement 排列 (the pattern), and their motion 运动 (how they move).

    Solid Liquid Gas
    Separation touching, very close close together far apart
    Arrangement regular 规则 pattern random, no pattern random, no pattern
    Motion vibrate 振动 about fixed positions move and slide past each other move quickly in all directions

    Strong forces of attraction 引力 hold the particles together. In a solid these forces are strong enough to hold every particle in place, so a solid keeps its shape. In a liquid the forces are weaker, so particles can move around. In a gas the particles move so fast that the forces hardly act at all, so the gas spreads out.

    日本語
    Solid, liquid, gas: particle motion

    Everything around you is made of tiny particles 粒子 — these can be atoms 原子, molecules 分子, or ions 离子. The kinetic particle theory 粒子动理论 says these particles are always moving. How close the particles are, how they are arranged, and how they move decides whether matter is a solid 固体, a liquid 液体, or a gas 气体.

    Properties you can observe

    You do not need a microscope to tell the three states apart. They behave in different ways:

    • A solid has a fixed shape and a fixed volume 体积. It does not flow and you cannot compress 压缩 it (squeeze it smaller).
    • A liquid has a fixed volume but no fixed shape. It flows and takes the shape of its container. It is almost impossible to compress.
    • A gas has no fixed shape and no fixed volume. It flows and spreads out to fill the whole container. A gas is easy to compress.

    The table below sums up these properties. Density 密度 means how much mass 质量 is packed into a given volume.

    Property Solid Liquid Gas
    Shape fixed takes the shape of the container fills the whole container
    Volume fixed fixed fills the whole container
    Can it be compressed? no almost none yes, easily
    Does it flow? no yes yes
    Density high high low

    The particle picture

    The kinetic particle theory explains these properties by looking at three things: the separation 间距 of the particles (how far apart they are), their arrangement 排列 (the pattern), and their motion 运动 (how they move).

    Solid Liquid Gas
    Separation touching, very close close together far apart
    Arrangement regular 规则 pattern random, no pattern random, no pattern
    Motion vibrate 振动 about fixed positions move and slide past each other move quickly in all directions

    Strong forces of attraction 引力 hold the particles together. In a solid these forces are strong enough to hold every particle in place, so a solid keeps its shape. In a liquid the forces are weaker, so particles can move around. In a gas the particles move so fast that the forces hardly act at all, so the gas spreads out.

    Particle diagrams of a solid, a liquid and a gas: the solid is a regular packed grid, the liquid is close but random, the gas is sparse and spread out
    Particles are packed and regular in a solid, close and random in a liquid, and far apart in a gas
    Explore · ⁨探索⁩

    States of matter · ⁨物質の状態⁩

    Heat the box and watch the particles break free: a vibrating solid melts to a flowing liquid, then spreads out as a gas. Same particles — just more energy. · ⁨箱を加熱して粒子が解放される様子を見てください:振動する固体が溶けて流れ出る液体になり、さらに広がって気体になります。同じ粒子ですが、エネルギーが大きくなっただけです。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    particles/ˈpɑːtɪklz/ 粒子
    atoms/ˈætəmz/ 原子
    molecules/ˈmɒlɪkjuːlz/ 分子
    ions/ˈaɪɒnz/ イオン
    kinetic particle theory/kɪˈnetɪk ˈpɑːtɪkl ˈθɪəri/ 運動論的粒子説
    solid/ˈsɒlɪd/ 固体
    liquid/ˈlɪkwɪd/ 液体
    gas/ɡæs/ 気体
    volume/ˈvɒljuːm/ 体積
    compress/kəmˈpres/ 圧縮する
    density/ˈdensɪti/ 密度
    mass/mæs/ 質量
    separation/ˌsepəˈreɪʃn/ 分離
    arrangement/əˈreɪndʒmənt/ arrangement
    motion/ˈməʊʃn/ 運動
    regular/ˈreɡjʊlə/ 規則的
    vibrate/vaɪˈbreɪt/ 振動する
    forces of attraction/ˈfɔːsɪz ɒv əˈtrækʃn/ 引力
    1.1

    Changes of state

    English

    When you heat or cool a substance, it can change from one state to another. You must know the name of each change.

    Change What happens Name
    solid → liquid melting 熔化 melting
    liquid → solid freezing 凝固 freezing
    liquid → gas (at the surface, below the boiling point) evaporating 蒸发 evaporation
    liquid → gas (all through the liquid) boiling 沸腾 boiling
    gas → liquid condensing 凝结 condensation

    A pure solid melts at one fixed temperature, the melting point 熔点. A pure liquid boils at one fixed temperature, the boiling point 沸点. The same substance freezes at its melting point and condenses at its boiling point.

    A few substances change straight from solid to gas without melting first. This is called sublimation 升华. In the photo below, warmed solid iodine in a beaker turns directly into a purple gas; the gas then cools on the round flask of ice above and turns back into a solid.

    Explaining changes of state with the particle theory

    Each change of state is really a change in the energy 能量 of the particles.

    • When you heat a solid, the particles gain energy and vibrate faster. At the melting point the particles have enough energy to break away from their fixed places and slide around — the solid melts.
    • When you heat a liquid, the particles move faster. At the boiling point they have enough energy to fully escape the forces of attraction and become a gas.
    • Cooling does the opposite. The particles lose energy, move more slowly, and the forces of attraction pull them back together.

    During a change of state the energy goes into breaking the forces of attraction, not into making the particles move faster. This is why the temperature stays the same while a substance is melting or boiling.

    Heating and cooling curves

    A heating curve 加热曲线 is a graph of temperature against time as you heat a substance steadily. A cooling curve 冷却曲线 is the same graph as the substance cools.

    On a heating curve there are two flat (level) parts:

    • The first flat part is at the melting point. Here solid and liquid are both present. The heat energy breaks the forces holding the solid together, so the temperature does not rise.
    • The second flat part is at the boiling point. Here liquid and gas are both present, and the temperature again stays constant.

    A cooling curve is the reverse. It has a flat part at the boiling point (the gas condenses) and a flat part at the melting point (the liquid freezes). As the particles slow down, thermal energy 热能 is released to the surroundings.

    日本語
    Ice cubes melting in a glass
    Ice melting to water: a change of state as the solid warms.

    When you heat or cool a substance, it can change from one state to another. You must know the name of each change.

    Change What happens Name
    solid → liquid melting 熔化 melting
    liquid → solid freezing 凝固 freezing
    liquid → gas (at the surface, below the boiling point) evaporating 蒸发 evaporation
    liquid → gas (all through the liquid) boiling 沸腾 boiling
    gas → liquid condensing 凝结 condensation

    A pure solid melts at one fixed temperature, the melting point 熔点. A pure liquid boils at one fixed temperature, the boiling point 沸点. The same substance freezes at its melting point and condenses at its boiling point.

    Solid, liquid and gas linked by arrows: melting and boiling go one way, freezing and condensing go back
    Heating: solid → liquid → gas; cooling reverses each change

    A few substances change straight from solid to gas without melting first. This is called sublimation 升华. In the photo below, warmed solid iodine in a beaker turns directly into a purple gas; the gas then cools on the round flask of ice above and turns back into a solid.

    A beaker of solid iodine on a hotplate filled with bright purple iodine gas, with a flask of ice resting on top where the gas turns back to a solid
    Warmed solid iodine turns straight into a purple gas (sublimation)
    A block of dry ice giving off thick white fog as it turns straight into gas
    The everyday example: dry ice is solid carbon dioxide, and it sublimes too — it never leaves a puddle

    Explaining changes of state with the particle theory

    Each change of state is really a change in the energy 能量 of the particles.

    • When you heat a solid, the particles gain energy and vibrate faster. At the melting point the particles have enough energy to break away from their fixed places and slide around — the solid melts.
    • When you heat a liquid, the particles move faster. At the boiling point they have enough energy to fully escape the forces of attraction and become a gas.
    • Cooling does the opposite. The particles lose energy, move more slowly, and the forces of attraction pull them back together.

    During a change of state the energy goes into breaking the forces of attraction, not into making the particles move faster. This is why the temperature stays the same while a substance is melting or boiling.

    Heating and cooling curves

    A heating curve 加热曲线 is a graph of temperature against time as you heat a substance steadily. A cooling curve 冷却曲线 is the same graph as the substance cools.

    On a heating curve there are two flat (level) parts:

    • The first flat part is at the melting point. Here solid and liquid are both present. The heat energy breaks the forces holding the solid together, so the temperature does not rise.
    • The second flat part is at the boiling point. Here liquid and gas are both present, and the temperature again stays constant.

    A cooling curve is the reverse. It has a flat part at the boiling point (the gas condenses) and a flat part at the melting point (the liquid freezes). As the particles slow down, thermal energy 热能 is released to the surroundings.

    A heating curve of temperature against time, rising then flat at the melting point, rising then flat at the boiling point, then rising again
    Temperature stays constant at the melting and boiling points while forces of attraction are broken
    Explore · ⁨探索⁩

    Heating a substance through its states · ⁨物質の状態を通じた加熱⁩

    Step up the temperature. Adding heat gives the particles more energy until they break free — and the temperature pauses at each change of state while the energy does that work. · ⁨温度を上げます。熱を加えると粒子にエネルギーが加わり、解放されるまで動きます—そして、エネルギーがその作業をする間に、各状態変化で温度は一時的に停滞します。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    melting/ˈmeltɪŋ/ 融解
    freezing/ˈfriːzɪŋ/ 凝固
    evaporating/ɪˈvæpəreɪtɪŋ/ 蒸発
    boiling/ˈbɔɪlɪŋ/ 沸点
    condensing/kənˈdensɪŋ/ 凝縮
    melting point/ˈmeltɪŋ pɔɪnt/ 融点
    boiling point/ˈbɔɪlɪŋ pɔɪnt/ 沸点
    sublimation/ˌsʌblɪˈmeɪʃn/ ⟦number⟧ sublimation ⟦number⟧ → <昇華>
    energy/ˈenədʒi/ エネルギー
    heating curve/ˈhiːtɪŋ kɜːv/ 加熱曲線
    cooling curve/ˈkuːlɪŋ kɜːv/ 冷却曲線
    thermal energy/ˈθɜːml ˈenədʒi/ 熱エネルギーに変換する
    1.1

    Gases: temperature, pressure and volume

    English

    A gas pushes on the walls of its container. This push, spread over the area of the wall, is the pressure 压强 of the gas. Pressure comes from the gas particles hitting the walls.

    Effect of temperature

    If you heat a fixed mass of gas while keeping the pressure the same, its volume increases.

    Using the particle theory: heating gives the particles more kinetic energy 动能, so they move faster. They hit the walls harder and more often. To keep the pressure the same, the gas must take up more space, so the volume gets bigger.

    If instead the volume is fixed (a sealed, rigid container), heating the gas makes the pressure rise, because the faster particles collide 碰撞 with the walls harder and more often.

    Effect of pressure

    If you increase the pressure on a fixed mass of gas while keeping the temperature the same, its volume decreases. Squeezing the gas into a smaller space means the particles hit the walls more often, which is what a higher pressure means.

    日本語

    A gas pushes on the walls of its container. This push, spread over the area of the wall, is the pressure 压强 of the gas. Pressure comes from the gas particles hitting the walls.

    Effect of temperature

    If you heat a fixed mass of gas while keeping the pressure the same, its volume increases.

    A heated gas: faster particles take up more space at constant pressure
    A heated gas: faster particles spread out and take up more space

    Using the particle theory: heating gives the particles more kinetic energy 动能, so they move faster. They hit the walls harder and more often. To keep the pressure the same, the gas must take up more space, so the volume gets bigger.

    If instead the volume is fixed (a sealed, rigid container), heating the gas makes the pressure rise, because the faster particles collide 碰撞 with the walls harder and more often.

    Effect of pressure

    If you increase the pressure on a fixed mass of gas while keeping the temperature the same, its volume decreases. Squeezing the gas into a smaller space means the particles hit the walls more often, which is what a higher pressure means.

    At the same temperature, squeezing a gas into a smaller volume packs the particles closer so they hit the walls more often, which is a higher pressure
    Squeezing a gas into a smaller volume makes the particles hit the walls more often, so the pressure rises
    Explore · ⁨探索⁩

    Gases, pressure & volume · ⁨気体、圧力 & 体積⁩

    p = k / V

    Squeeze the volume and the pressure rises (Boyle's law). · ⁨体積を圧縮すると圧力が上昇します(ボイルの法則)。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    pressure/ˈpreʃə/ 圧力
    kinetic energy/kɪˈnetɪk ˈenədʒi/ 運動エネルギー
    collide/kəˈlaɪd/ 衝突
    1.2

    Diffusion

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe and explain diffusion in terms of kinetic particle theory 2 Describe and explain the effect of relative molecular mass on the rate of diffusion of gases
    日本語
    コア サプリメント
    1 運動論的粒子理論の観点から、拡散を記述・説明する 2 相対分子量が気体の拡散速度に与える効果を記述・説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Diffusion: random motion, one-way flow

    Diffusion 扩散 is the spreading of particles from a region where they are crowded to a region where they are spread out — that is, from high concentration 浓度 to low concentration. It happens because particles are always moving in random directions.

    Diffusion explains why you can smell food from across a room: the smell particles move and mix with the air until they reach your nose. Diffusion happens in gases and in liquids, but not in solids, because solid particles cannot move from place to place.

    Rate of diffusion and molecular mass

    Lighter gas particles move faster than heavier ones at the same temperature. So a gas with a smaller relative molecular mass 相对分子质量 (a smaller mass for each molecule) has a faster rate 速率 of diffusion.

    A classic experiment shows this. Cotton wool soaked in ammonia 氨气 ($\text{NH}_3$) is put at one end of a long glass tube. Cotton wool soaked in hydrogen chloride 氯化氢 ($\text{HCl}$) is put at the other end. Both gases diffuse along the tube and meet to form a white ring of ammonium chloride 氯化铵 ($\text{NH}_4\text{Cl}$).

    $$\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl}$$

    Ammonia has $M_r = 17$ and hydrogen chloride has $M_r = 36.5$. Ammonia is lighter, so it diffuses faster and travels further along the tube. The white ring forms nearer the hydrogen chloride end.

    Worked example. Bromine ($M_r = 160$) and ammonia ($M_r = 17$) are released at the same moment from opposite ends of a long tube. Which travels further before they meet? Compare the relative molecular masses. Ammonia's $M_r$ is far smaller, so its molecules move faster at the same temperature and it diffuses more quickly. Ammonia therefore travels the greater distance, and the two gases meet nearer the bromine end. The rule is always "lighter means faster" - argue from $M_r$, never from how big the formula looks on paper.

    日本語
    Diffusion: random motion, one-way flow

    Diffusion 扩散 is the spreading of particles from a region where they are crowded to a region where they are spread out — that is, from high concentration 浓度 to low concentration. It happens because particles are always moving in random directions.

    Particles crowded on one side of a box spread out over time until they fill the box evenly, moving from high to low concentration
    Particles diffuse from high to low concentration until they are spread out evenly

    Diffusion explains why you can smell food from across a room: the smell particles move and mix with the air until they reach your nose. Diffusion happens in gases and in liquids, but not in solids, because solid particles cannot move from place to place.

    Rate of diffusion and molecular mass

    Lighter gas particles move faster than heavier ones at the same temperature. So a gas with a smaller relative molecular mass 相对分子质量 (a smaller mass for each molecule) has a faster rate 速率 of diffusion.

    A classic experiment shows this. Cotton wool soaked in ammonia 氨气 ($\text{NH}_3$) is put at one end of a long glass tube. Cotton wool soaked in hydrogen chloride 氯化氢 ($\text{HCl}$) is put at the other end. Both gases diffuse along the tube and meet to form a white ring of ammonium chloride 氯化铵 ($\text{NH}_4\text{Cl}$).

    $$\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl}$$

    Ammonia has $M_r = 17$ and hydrogen chloride has $M_r = 36.5$. Ammonia is lighter, so it diffuses faster and travels further along the tube. The white ring forms nearer the hydrogen chloride end.

    A long glass tube with ammonia diffusing in from the left and hydrogen chloride from the right; they meet to form a white ring closer to the hydrogen chloride end
    Ammonia ($M_r=17$) is lighter, so it diffuses faster and further; the white ring forms nearer the HCl end

    Worked example. Bromine ($M_r = 160$) and ammonia ($M_r = 17$) are released at the same moment from opposite ends of a long tube. Which travels further before they meet? Compare the relative molecular masses. Ammonia's $M_r$ is far smaller, so its molecules move faster at the same temperature and it diffuses more quickly. Ammonia therefore travels the greater distance, and the two gases meet nearer the bromine end. The rule is always "lighter means faster" - argue from $M_r$, never from how big the formula looks on paper.

    Explore · ⁨探索⁩

    Diffusion · ⁨拡散⁩

    Release the cloud and the particles spread on their own from the crowded corner until they fill the box evenly — diffusion, sped up by heat. · ⁨雲を解放すると、粒子は混み合った隅から自分自身の力で広がり、均等に箱を満たします—これは熱によって加速された拡散です。⁩

    Explore · ⁨探索⁩

    Diffusion · ⁨拡散⁩

    Set the concentration on each side. Particles spread from high to low concentration until evenly mixed. · ⁨両側の濃度を設定せよ。粒子は高い方から低い方へ濃度差がなくなるまで拡散する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    diffusion/dɪˈfjuːʒn/ 拡散
    concentration/ˌkɒnsənˈtreɪʃn/ 濃度
    relative molecular mass/ˈrelətɪv məˈlekjʊlə mæs/ 相対分子量
    rate/reɪt/ 速度
    ammonia/æˈməʊnɪə/ アンモニア
    hydrogen chloride/ˈhaɪdrədʒn ˈklɔːraɪd/ 塩化水素
    ammonium chloride/æˈməʊnɪəm ˈklɔːraɪd/ 塩化アンモニウム
    Watch lesson · ⁨レッスンを視聴⁩
    1.2

    Exam tips

    • During melting and boiling the temperature stays constant (the flat parts of a heating curve), because the energy breaks the forces between particles instead of making them move faster.
    • Evaporation happens only at the surface and at any temperature; boiling happens throughout the liquid at one fixed temperature.
    • Explain gas behaviour with the particle theory: heating a gas at fixed volume raises the pressure because the particles hit the walls harder and more often — the particles themselves do not get bigger.
    • Diffusion is faster for lighter particles (smaller $M_r$) and cannot happen in solids. In the ammonia/hydrogen chloride tube the lighter ammonia travels further, so the white ring forms nearer the HCl end.
  • 2

    Atoms, elements and compounds

    Watch lesson · ⁨レッスンを視聴⁩
    2.1

    Elements, compounds and mixtures

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the differences between elements, compounds and mixtures
    日本語
    コア サプリメント
    1 元素、化合物、および混合物の違いを説明せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    All substances are made from about 100 simple building blocks. Knowing how they are joined lets you sort every substance into one of three groups.

    • An element 元素 is a substance made of only one type of atom 原子. You cannot break it into anything simpler by a chemical reaction. Examples: copper, oxygen, carbon.
    • A compound 化合物 is two or more elements chemically joined (bonded) together. The atoms are joined in a fixed ratio. Examples: water, carbon dioxide. A compound has different properties from the elements in it.
    • A mixture 混合物 is two or more substances that are just mixed, not chemically joined. The parts keep their own properties and can be separated by physical methods. Example: air.

    The key difference: in a compound the elements are bonded and can only be separated by chemical reactions; in a mixture they are not bonded and are easy to separate.

    日本語
    A pure vanadium crystal bar and cube
    A pure element (vanadium): elements are the simplest substances.

    All substances are made from about 100 simple building blocks. Knowing how they are joined lets you sort every substance into one of three groups.

    • An element 元素 is a substance made of only one type of atom 原子. You cannot break it into anything simpler by a chemical reaction. Examples: copper, oxygen, carbon.
    • A compound 化合物 is two or more elements chemically joined (bonded) together. The atoms are joined in a fixed ratio. Examples: water, carbon dioxide. A compound has different properties from the elements in it.
    • A mixture 混合物 is two or more substances that are just mixed, not chemically joined. The parts keep their own properties and can be separated by physical methods. Example: air.

    The key difference: in a compound the elements are bonded and can only be separated by chemical reactions; in a mixture they are not bonded and are easy to separate.

    Three boxes of particles: an element with one type of atom, a compound with two atom types bonded in a fixed ratio, and a mixture of unbonded atoms
    An element has one type of atom; a compound has different atoms bonded in a fixed ratio; a mixture is not bonded
    Explore · ⁨探索⁩

    Elements, compounds and mixtures lab · ⁨元素、化合物、混合物の実験⁩

    Classify everyday particle examples by composition. · ⁨日常の粒子の例を組成に基づいて分類します。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    element/ˈelɪmənt/ 元素
    atom/ˈætəm/ 原子
    compound/ˈkɒmpaʊnd/ 化合物
    mixture/ˈmɪkstʃə/ 混合物
    2.2

    Atomic structure

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the structure of the atom as a central nucleus containing neutrons and protons surrounded by electrons in shells
    2 State the relative charges and relative masses of a proton, a neutron and an electron
    3 Define proton number/atomic number as the number of protons in the nucleus of an atom
    4 Define mass number/nucleon number as the total number of protons and neutrons in the nucleus of an atom
    5 Determine the electronic configuration of elements and their ions with proton number 1 to 20, e.g. 2,8,3
    6 State that: (a) Group VIII noble gases have a full outer electron shell (b) the number of outer shell electrons is equal to the group number in Groups I to VII (c) the number of occupied electron shells is equal to the period number
    日本語
    コア サプリメント
    1 原子の構造として、中心に中性子と陽子を含む原子核があり、その周囲を電子が殻において回っていることを説明せよ
    2 陽子、中性子、および電子の相対電荷および相対質量を述べよ
    3 原子番号を、原子の原子核内の陽子の数として定義せよ
    4 質量数を、原子の原子核内の陽子と中性子の総数として定義せよ
    5 原子番号1〜20の元素およびそのイオンの電子配置を求めよ。例: 2,8,3
    6 以下を述べよ: (a) 第VIII族の希ガスは最外殻電子が満たされている (b) 最外殻電子数は第I〜VII族において族番号に等しい (c) 占有している電子殻の数は周期番号に等しい

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Inside the atom

    Every atom has a small, dense centre called the nucleus 原子核. Around it, electrons 电子 move in shells 电子层 (energy levels). The nucleus contains two kinds of particle: protons 质子 and neutrons 中子.

    Each particle has a relative mass and a relative charge 电荷. You must learn these values:

    Particle Relative mass Relative charge
    proton 1 $+1$
    neutron 1 $0$
    electron $\frac{1}{1840}$ (almost 0) $-1$

    An atom has no overall charge because it has equal numbers of protons ($+1$ each) and electrons ($-1$ each).

    Proton number and mass number

    Two numbers describe an atom:

    • The proton number 质子数 (also called the atomic number 原子序数) is the number of protons in the nucleus. It tells you which element the atom is.
    • The mass number 质量数 (also called the nucleon number 核子数) is the total number of protons and neutrons in the nucleus.

    So the number of neutrons $=$ mass number $-$ proton number.

    Electronic configuration

    The electrons fill the shells from the inside out. The first shell holds up to 2 electrons; the next shells hold up to 8 each (for the first 20 elements). The electronic configuration 电子排布 lists how many electrons are in each shell, starting from the inside.

    For example, an atom with 13 electrons has the configuration $2,8,3$. Sodium (proton number 11) is $2,8,1$. Calcium (proton number 20) is $2,8,8,2$.

    The configuration links to the Periodic Table 周期表:

    • A Group 族 number (Groups I to VII) equals the number of electrons in the outer shell. So $2,8,1$ is in Group I.
    • A Period 周期 number equals the number of shells that hold electrons. So $2,8,1$ has three shells, so it is in Period 3.
    • The noble gases 稀有气体 in Group VIII (or 0) have a full outer shell, which makes them very unreactive.
    日本語

    Inside the atom

    Every atom has a small, dense centre called the nucleus 原子核. Around it, electrons 电子 move in shells 电子层 (energy levels). The nucleus contains two kinds of particle: protons 质子 and neutrons 中子.

    A lithium atom: an orange nucleus of protons and neutrons, with two electron shells holding 2 then 1 electrons
    An atom has a tiny nucleus of protons and neutrons, with electrons in shells around it

    Each particle has a relative mass and a relative charge 电荷. You must learn these values:

    Particle Relative mass Relative charge
    proton 1 $+1$
    neutron 1 $0$
    electron $\frac{1}{1840}$ (almost 0) $-1$

    An atom has no overall charge because it has equal numbers of protons ($+1$ each) and electrons ($-1$ each).

    Proton number and mass number

    Two numbers describe an atom:

    • The proton number 质子数 (also called the atomic number 原子序数) is the number of protons in the nucleus. It tells you which element the atom is.
    • The mass number 质量数 (also called the nucleon number 核子数) is the total number of protons and neutrons in the nucleus.

    So the number of neutrons $=$ mass number $-$ proton number.

    Electronic configuration

    The electrons fill the shells from the inside out. The first shell holds up to 2 electrons; the next shells hold up to 8 each (for the first 20 elements). The electronic configuration 电子排布 lists how many electrons are in each shell, starting from the inside.

    For example, an atom with 13 electrons has the configuration $2,8,3$. Sodium (proton number 11) is $2,8,1$. Calcium (proton number 20) is $2,8,8,2$.

    The configuration links to the Periodic Table 周期表:

    • A Group 族 number (Groups I to VII) equals the number of electrons in the outer shell. So $2,8,1$ is in Group I.
    • A Period 周期 number equals the number of shells that hold electrons. So $2,8,1$ has three shells, so it is in Period 3.
    • The noble gases 稀有气体 in Group VIII (or 0) have a full outer shell, which makes them very unreactive.
    A sodium atom with electrons drawn in three shells as 2, 8, 1; the one outer electron means Group I and the three shells in use mean Period 3
    Sodium's configuration 2,8,1 links to the Periodic Table: one outer electron means Group I, three shells means Period 3
    Explore · ⁨探索⁩

    Electron shells · ⁨電子殻⁩

    Change the atomic number and watch the shells fill (2, 8, 8) — the electron arrangement of the first 20 elements. · ⁨原子番号を変化させ、殻が埋まっていく様子(2, 8, 8)を見よ——初めから20番までの元素の電子配置である。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    nucleus/ˈnjuːklɪəs/ 核
    electrons/ɪˈlektrɒnz/ 電子
    shells/ʃelz/ 殻の数
    protons/ˈprəʊtɒnz/ 陽子
    neutrons/ˈnjuːtrɒnz/ 中性子
    charge/tʃɑːdʒ/ 電荷
    proton number/ˈprəʊtɒn ˈnʌmbə/ 原子番号
    atomic number/əˈtɒmɪk ˈnʌmbə/ 原子番号
    mass number/mæs ˈnʌmbə/ 質量数
    nucleon number/ˈnjuːklɪən ˈnʌmbə/ 核子数
    electronic configuration/ɪlekˈtrɒnɪk kənˌfɪɡjəˈreɪʃn/ 電子配置
    Periodic Table/ˌpɪərɪˈɒdɪk ˈteɪbl/ 元素周期表
    Group/ɡruːp/ グループ
    Period/ˈpɪərɪəd/ 期間
    noble gases/ˈnəʊbl ˈɡæsɪz/ 希ガス
    Watch lesson · ⁨レッスンを視聴⁩
    2.3

    Isotopes

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Define isotopes as different atoms of the same element that have the same number of protons but different numbers of neutrons 3 State that isotopes of the same element have the same chemical properties because they have the same number of electrons and therefore the same electronic configuration
    2 Interpret and use symbols for atoms, e.g. $^{12}_{6}\text{C}$, and ions, e.g. $^{35}_{17}\text{Cl}^-$ 4 Calculate the relative atomic mass of an element from the relative masses and abundances of its isotopes
    日本語
    コア サプリメント
    1 同位体を、同じ元素の異なる原子で、陽子数が同じだが中性子数が異なるものとして定義せよ 3 同一元素の同位体が電子数が同じため、同じ電子配置を持ち、したがって同じ化学的性質を持つことを述べよ
    2 原子の記号、例:$^{12}_{6}\text{C}$、およびイオンの記号、例:$^{35}_{17}\text{Cl}^-$ を解釈して使用せよ 4 同位体の相対質量および存在比から元素の相対原子質量を計算せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Isotopes 同位素 are atoms of the same element that have the same number of protons but different numbers of neutrons. Because the proton number is the same, they are the same element. Because the neutron number is different, they have different mass numbers.

    You write an atom with its mass number on top and proton number below, like $^{12}_{6}\text{C}$. For an ion you add the charge, like $^{35}_{17}\text{Cl}^{-}$.

    Isotopes of an element have the same chemical properties. This is because chemical reactions only involve electrons, and isotopes have the same number of electrons and the same electronic configuration. (The extra neutrons change only the mass.)

    Calculating relative atomic mass

    The relative atomic mass 相对原子质量 ($A_r$) of an element is the average mass of its atoms, taking into account how common each isotope is. The abundance 丰度 is the percentage of each isotope.

    $$A_r = \frac{\sum (\text{isotope mass} \times \text{abundance})}{100}$$

    For example, chlorine is 75% $^{35}\text{Cl}$ and 25% $^{37}\text{Cl}$:

    $$A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = 35.5$$

    Worked example. Boron has two isotopes, $^{10}\text{B}$ and $^{11}\text{B}$, and its $A_r$ is 10.8. Find the percentage of each. Let the abundance of $^{11}\text{B}$ be $x$, so the abundance of $^{10}\text{B}$ is $(100 - x)$. Then

    $$\frac{11x + 10(100 - x)}{100} = 10.8$$

    so $11x + 1000 - 10x = 1080$, giving $x = 80$. Boron is 80% $^{11}\text{B}$ and 20% $^{10}\text{B}$. Check it against common sense: 10.8 lies closer to 11, so the heavier isotope must be the more common one. If your answer gives the majority to the isotope further from $A_r$, you have them the wrong way round.

    日本語

    Isotopes 同位素 are atoms of the same element that have the same number of protons but different numbers of neutrons. Because the proton number is the same, they are the same element. Because the neutron number is different, they have different mass numbers.

    You write an atom with its mass number on top and proton number below, like $^{12}_{6}\text{C}$. For an ion you add the charge, like $^{35}_{17}\text{Cl}^{-}$.

    Isotopes of an element have the same chemical properties. This is because chemical reactions only involve electrons, and isotopes have the same number of electrons and the same electronic configuration. (The extra neutrons change only the mass.)

    Calculating relative atomic mass

    The relative atomic mass 相对原子质量 ($A_r$) of an element is the average mass of its atoms, taking into account how common each isotope is. The abundance 丰度 is the percentage of each isotope.

    $$A_r = \frac{\sum (\text{isotope mass} \times \text{abundance})}{100}$$

    For example, chlorine is 75% $^{35}\text{Cl}$ and 25% $^{37}\text{Cl}$:

    $$A_r = \frac{(35 \times 75) + (37 \times 25)}{100} = 35.5$$
    Chlorine's two isotopes drawn as circles sized by abundance, 75 percent Cl-35 and 25 percent Cl-37, so the weighted-average relative atomic mass 35.5 sits closer to 35
    Relative atomic mass is a weighted average: chlorine is 75 percent Cl-35 and 25 percent Cl-37, so A_r is 35.5

    Worked example. Boron has two isotopes, $^{10}\text{B}$ and $^{11}\text{B}$, and its $A_r$ is 10.8. Find the percentage of each. Let the abundance of $^{11}\text{B}$ be $x$, so the abundance of $^{10}\text{B}$ is $(100 - x)$. Then

    $$\frac{11x + 10(100 - x)}{100} = 10.8$$

    so $11x + 1000 - 10x = 1080$, giving $x = 80$. Boron is 80% $^{11}\text{B}$ and 20% $^{10}\text{B}$. Check it against common sense: 10.8 lies closer to 11, so the heavier isotope must be the more common one. If your answer gives the majority to the isotope further from $A_r$, you have them the wrong way round.

    Explore · ⁨探索⁩

    Protons, neutrons & isotopes · ⁨陽子、中性子 & 同位体⁩

    Change the neutrons to make isotopes (same element, different mass number); change the electrons to make ions. · ⁨中性子を変化させると同位体(同じ元素、質量数が異なる)になり、電子を変化させるとイオンになります。⁩

    Explore · ⁨探索⁩

    Isotope lab · ⁨同位体の実験⁩

    Classify isotope facts by proton, neutron and mass number. · ⁨陽子、中性子、質量数に基づいて同位体の事実を分類する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    isotopes/ˈaɪsətəʊps/ 同位体
    relative atomic mass/ˈrelətɪv əˈtɒmɪk mæs/ 相対原子質量
    abundance/əˈbʌndəns/ 存在比
    Watch lesson · ⁨レッスンを視聴⁩
    2.4

    Ions and ionic bonds

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the formation of positive ions, known as cations, and negative ions, known as anions 5 Describe the giant lattice structure of ionic compounds as a regular arrangement of alternating positive and negative ions
    2 State that an ionic bond is a strong electrostatic attraction between oppositely charged ions 6 Describe the formation of ionic bonds between ions of metallic and non-metallic elements, including the use of dot-and-cross diagrams
    3 Describe the formation of ionic bonds between elements from Group I and Group VII, including the use of dot-and-cross diagrams 7 Explain in terms of structure and bonding the properties of ionic compounds: (a) high melting points and boiling points (b) good electrical conductivity when aqueous or molten and poor when solid
    4 Describe the properties of ionic compounds: (a) high melting points and boiling points (b) good electrical conductivity when aqueous or molten and poor when solid
    日本語
    コア サプリメント
    1 カチオンと呼ばれる正イオンと、アニオンと呼ばれる負イオンの形成を説明する 5 正のイオンと負のイオンが交互に規則正しく配列した、イオン化合物の巨大格子構造を説明する
    2 イオン結合は、反対の電荷を持つイオン間の強い静電的引力であることを述べる 6 金属元素と非金属元素のイオン間におけるイオン結合の形成を、点線図を用いて説明する
    3 Group IおよびGroup VIIの元素間のイオン結合の形成を、点線図を用いて説明する 7 イオン化合物の性質を構造および結合の観点から説明する:(a) 高い融点および沸点 (b) 水溶液または溶融状態では電気伝導性が良く、固体では悪い
    4 イオン性化合物の性質を説明せよ: (a) 高い融点および沸点 (b) 水溶液または熔融状態では高い電気伝導性、固体では低い電気伝導性

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Ionic bonding: electron transfer

    An ion 离子 is an atom (or group of atoms) that has lost or gained electrons, so it has an electric charge.

    • A metal atom loses electrons to form a positive ion, called a cation 阳离子.
    • A non-metal atom gains electrons to form a negative ion, called an anion 阴离子.

    Atoms do this to get a full outer shell, like a noble gas.

    How an ionic bond forms

    An ionic bond 离子键 is a strong electrostatic attraction 静电引力 between oppositely charged ions (a $+$ ion and a $-$ ion pull together).

    Ionic bonds form between a metal 金属 and a non-metal 非金属. Take sodium chloride, $\text{NaCl}$. Sodium ($2,8,1$) gives its one outer electron to chlorine ($2,8,7$). Now sodium is $\text{Na}^{+}$ ($2,8$) and chlorine is $\text{Cl}^{-}$ ($2,8,8$). Both have full outer shells, and the opposite charges attract.

    You can show this with a dot-and-cross diagram: draw each atom's outer-shell electrons as dots for one element and crosses for the other, then show the electron moving from the metal to the non-metal.

    The structure and properties of ionic compounds

    An ionic compound 离子化合物 is not made of separate molecules 分子. The ions pack together into a giant lattice 晶格 — a regular pattern of huge numbers of alternating 交替 positive and negative ions.

    This structure explains the properties:

    Property Reason
    high melting point 熔点 and boiling point 沸点 the strong electrostatic attraction between ions needs a lot of energy to break
    poor electrical conductivity 导电性 when solid the ions are fixed in place and cannot move
    good conductor when molten 熔融 or aqueous 水溶液 the ions are now free to move and carry charge
    日本語
    Ionic bonding: electron transfer

    An ion 离子 is an atom (or group of atoms) that has lost or gained electrons, so it has an electric charge.

    • A metal atom loses electrons to form a positive ion, called a cation 阳离子.
    • A non-metal atom gains electrons to form a negative ion, called an anion 阴离子.

    Atoms do this to get a full outer shell, like a noble gas.

    How an ionic bond forms

    An ionic bond 离子键 is a strong electrostatic attraction 静电引力 between oppositely charged ions (a $+$ ion and a $-$ ion pull together).

    Ionic bonds form between a metal 金属 and a non-metal 非金属. Take sodium chloride, $\text{NaCl}$. Sodium ($2,8,1$) gives its one outer electron to chlorine ($2,8,7$). Now sodium is $\text{Na}^{+}$ ($2,8$) and chlorine is $\text{Cl}^{-}$ ($2,8,8$). Both have full outer shells, and the opposite charges attract.

    You can show this with a dot-and-cross diagram: draw each atom's outer-shell electrons as dots for one element and crosses for the other, then show the electron moving from the metal to the non-metal.

    Sodium's single outer electron transfers to chlorine; sodium becomes Na+ (2,8) and chlorine becomes Cl- (2,8,8)
    Sodium gives its outer electron to chlorine; both reach full outer shells and the opposite charges attract (the blue electron came from sodium)

    The structure and properties of ionic compounds

    An ionic compound 离子化合物 is not made of separate molecules 分子. The ions pack together into a giant lattice 晶格 — a regular pattern of huge numbers of alternating 交替 positive and negative ions.

    A grid of alternating small blue Na+ ions and larger orange Cl- ions, repeating in a regular pattern
    Ions pack into a giant lattice: a regular, repeating pattern of alternating positive and negative ions
    A clear halite crystal broken into flat-faced cubes with square corners
    A halite crystal: sodium chloride's cubic shape is the ionic lattice inside it, made big enough to see

    This structure explains the properties:

    Property Reason
    high melting point 熔点 and boiling point 沸点 the strong electrostatic attraction between ions needs a lot of energy to break
    poor electrical conductivity 导电性 when solid the ions are fixed in place and cannot move
    good conductor when molten 熔融 or aqueous 水溶液 the ions are now free to move and carry charge
    Explore · ⁨探索⁩

    Forming an ionic bond (NaCl) · ⁨イオン結合の形成(NaCl)⁩

    Step through it. A metal hands its outer electron to a non-metal; the oppositely charged ions then attract and pack into a giant lattice. · ⁨手順を追って確認します。金属が最外殻の電子を非金属に渡すと、反対の電荷を持つイオン同士が引き合い、巨大な格子状構造を形成します。⁩

    Explore · ⁨探索⁩

    Ionic bonding (electron transfer) · ⁨イオン結合(電子の移動)⁩

    Step through it: the metal hands its outer electron(s) to the non-metal, both reach full shells, and the resulting + and − ions attract — an ionic bond. · ⁨手順を追って確認します:金属が最外殻の電子を非金属に渡し、両方とも最外殻が満タンになり、生じた+と−のイオン同士が引き合って、イオン結合が形成されます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    ion/ˈaɪɒn/ イオン
    cation/ˈkætaɪən/ カチオン
    anion/ˈænaɪən/ アニオン
    ionic bond/aɪˈɒnɪk bɒnd/ イオン結合
    electrostatic attraction/ɪˌlektrəʊˈstætɪk əˈtrækʃn/ 静電気的引力
    metal/ˈmetl/ 金属
    non-metal/nɒn ˈmetl/ 非金属
    ionic compound/aɪˈɒnɪk ˈkɒmpaʊnd/ イオン化合物
    molecules/ˈmɒlɪkjuːlz/ 分子
    lattice/ˈlætɪs/ 格子
    alternating/ˈɔːltəneɪtɪŋ/ 交流
    melting point/ˈmeltɪŋ pɔɪnt/ 融点
    boiling point/ˈbɔɪlɪŋ pɔɪnt/ 沸点
    electrical conductivity/ɪˈlektrɪkl kɒndəkˈtɪvɪti/ 電気伝導性
    molten/ˈməʊltn/ 溶融状態(_molten_)
    aqueous/ˈeɪkwɪəs/ 水溶液(_aqueous_)
    Watch lesson · ⁨レッスンを視聴⁩
    2.5

    Simple molecules and covalent bonds

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State that a covalent bond is formed when a pair of electrons is shared between two atoms leading to noble gas electronic configurations
    2 Describe the formation of covalent bonds in simple molecules, including $\text{H}_2$, $\text{Cl}_2$, $\text{H}_2\text{O}$, $\text{CH}_4$, $\text{NH}_3$ and $\text{HCl}$. Use dot-and-cross diagrams to show the electronic configurations in these and similar molecules 4 Describe the formation of covalent bonds in simple molecules, including $\text{CH}_3\text{OH}$, $\text{C}_2\text{H}_4$, $\text{O}_2$, $\text{CO}_2$ and $\text{N}_2$. Use dot-and-cross diagrams to show the electronic configurations in these and similar molecules
    3 Describe in terms of structure and bonding the properties of simple molecular compounds: (a) low melting points and boiling points (b) poor electrical conductivity 5 Explain in terms of structure and bonding the properties of simple molecular compounds: (a) low melting points and boiling points in terms of weak intermolecular forces (specific types of intermolecular forces are not required) (b) poor electrical conductivity
    日本語
    コア サプリメント
    1 共有結合とは、2つの原子間で電子対が共有され、希ガスの電子配置が得られることで形成されるものであることを述べよ
    2 単純分子における共有結合の形成を、$\text{H}_2$, $\text{Cl}_2$, $\text{H}_2\text{O}$, $\text{CH}_4$, $\text{NH}_3$ および $\text{HCl}$ を含めて説明する。これらの分子および類似する分子における電子配置を示すために点線図を使用する 4 単純分子における共有結合の形成を、$\text{CH}_3\text{OH}$, $\text{C}_2\text{H}_4$, $\text{O}_2$, $\text{CO}_2$ および $\text{N}_2$ を含めて説明する。これらの分子および類似する分子における電子配置を示すために点線図を使用する
    3 単純分子化合物の性質を構造および結合の観点から説明する:(a) 低い融点および沸点 (b) 電気伝導性が悪い 5 単純分子化合物の性質を構造および結合の観点から説明する:(a) 弱い分子間力(特定の種類の分子間力は不要)による低い融点および沸点 (b) 電気伝導性が悪い

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Covalent bonding: a shared pair

    A covalent bond 共价键 forms when two atoms share a pair of electrons. By sharing, each atom gets a full outer shell (a noble gas configuration). Covalent bonds form between non-metal atoms.

    Some molecules to know:

    • $\text{H}_2$ — two hydrogen atoms share one pair of electrons (a single bond).
    • $\text{Cl}_2$, $\text{HCl}$ — one shared pair each.
    • $\text{H}_2\text{O}$ — oxygen shares one pair with each of two hydrogen atoms.
    • $\text{NH}_3$ — nitrogen shares a pair with each of three hydrogen atoms.
    • $\text{CH}_4$ — carbon shares a pair with each of four hydrogen atoms.
    • $\text{O}_2$ and $\text{CO}_2$ have double bonds (two shared pairs); $\text{N}_2$ has a triple bond (three shared pairs); $\text{C}_2\text{H}_4$ and $\text{CH}_3\text{OH}$ also use shared pairs.

    In a dot-and-cross diagram for a molecule, you draw the outer electrons of each atom and show which pairs are shared in the overlap between the atoms.

    Properties of simple molecular compounds

    These substances are made of small, separate molecules.

    • They have low melting points and boiling points. The covalent bonds inside each molecule are strong, but the intermolecular forces 分子间作用力 (the forces between one molecule and the next) are weak, so little energy is needed to separate the molecules.
    • They are poor conductors of electricity, because the molecules have no overall charge and no free electrons or ions to carry charge.
    日本語
    Covalent bonding: a shared pair
    A ball-and-stick model of a water molecule
    A model of a water molecule: atoms share electrons in covalent bonds.

    A covalent bond 共价键 forms when two atoms share a pair of electrons. By sharing, each atom gets a full outer shell (a noble gas configuration). Covalent bonds form between non-metal atoms.

    Some molecules to know:

    • $\text{H}_2$ — two hydrogen atoms share one pair of electrons (a single bond).
    • $\text{Cl}_2$, $\text{HCl}$ — one shared pair each.
    • $\text{H}_2\text{O}$ — oxygen shares one pair with each of two hydrogen atoms.
    • $\text{NH}_3$ — nitrogen shares a pair with each of three hydrogen atoms.
    • $\text{CH}_4$ — carbon shares a pair with each of four hydrogen atoms.
    • $\text{O}_2$ and $\text{CO}_2$ have double bonds (two shared pairs); $\text{N}_2$ has a triple bond (three shared pairs); $\text{C}_2\text{H}_4$ and $\text{CH}_3\text{OH}$ also use shared pairs.

    In a dot-and-cross diagram for a molecule, you draw the outer electrons of each atom and show which pairs are shared in the overlap between the atoms.

    Dot-and-cross diagrams for hydrogen (one shared pair), water (two shared pairs and lone pairs) and methane (four shared pairs)
    In a covalent bond, atoms share pairs of electrons so each reaches a full outer shell

    Properties of simple molecular compounds

    These substances are made of small, separate molecules.

    • They have low melting points and boiling points. The covalent bonds inside each molecule are strong, but the intermolecular forces 分子间作用力 (the forces between one molecule and the next) are weak, so little energy is needed to separate the molecules.
    • They are poor conductors of electricity, because the molecules have no overall charge and no free electrons or ions to carry charge.
    Explore · ⁨探索⁩

    Why a molecule has its shape · ⁨分子が特定の形状を持つ理由⁩

    Shared electron pairs repel one another and spread out as far apart as they can. Four bonding pairs and no lone pairs gives a tetrahedral shape, like methane (CH4). · ⁨共有電子対同士は反発し合い、可能な限り離れて配置される。4つの結合電子対で非共有電子対がない場合、メタン (CH4) のように四面体構造となる。⁩

    Explore · ⁨探索⁩

    Covalent bonding (sharing) · ⁨共有結合(共有)⁩

    Step through it: two non-metal atoms overlap and share a pair of electrons — counting for both — so each reaches a full outer shell. That shared pair is the covalent bond; O₂ shares two pairs (a double bond). · ⁨手順を追う:2つの非金属原子が重なり合い、電子対を共有して双方の価電子殻を満たす。この共有電子対が共有結合であり、O₂ は2組の電子対(二重結合)を共有している。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    covalent bond/ˈkəʊvələnt bɒnd/ 共有結合
    intermolecular forces/ˌɪntəməˈlekjʊlə ˈfɔːsɪz/ 分子間力
    Watch lesson · ⁨レッスンを視聴⁩
    2.6

    Giant covalent structures

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the giant covalent structures of graphite and diamond 3 Describe the giant covalent structure of silicon(IV) oxide, $\text{SiO}_2$
    2 Relate the structures and bonding of graphite and diamond to their uses, limited to: (a) graphite as a lubricant and as an electrode (b) diamond in cutting tools 4 Describe the similarity in properties between diamond and silicon(IV) oxide, related to their structures
    日本語
    コア サプリメント
    1 グラファイトおよびダイヤモンドの巨大共有結合構造を説明する 3 シリコン(IV)酸化物の巨大共有結合構造、$\text{SiO}_2$ を説明する
    2 グラファイトおよびダイヤモンドの構造および結合と用途との関連性を記述する。用途は以下に限定される:(a) グラファイト潤滑剤および電極として (b) ダイヤモンド切断工具として 4 ダイヤモンドとシリコン(IV)酸化物の性質の類似点を、それらの構造に関連して説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Some covalent substances are not small molecules. Instead, millions of atoms are joined by covalent bonds into one giant covalent structure 巨型共价结构. The two you must know are both forms of carbon.

    Diamond 金刚石: each carbon atom is bonded to four other carbon atoms. This makes a very strong, rigid 3-D network. Diamond is extremely hard, so it is used in cutting tools 切割工具.

    Graphite 石墨: each carbon atom is bonded to only three others, forming flat layers 层. There are weak forces between the layers, so the layers can slide over each other — this makes graphite a good lubricant 润滑剂. The fourth outer electron of each carbon is free; these delocalised electrons 离域电子 can move, so graphite conducts electricity and is used as an electrode 电极.

    Silicon(IV) oxide 二氧化硅 ($\text{SiO}_2$) has a giant covalent structure like diamond, so it is also very hard and has a very high melting point.

    日本語

    Some covalent substances are not small molecules. Instead, millions of atoms are joined by covalent bonds into one giant covalent structure 巨型共价结构. The two you must know are both forms of carbon.

    Diamond 金刚石: each carbon atom is bonded to four other carbon atoms. This makes a very strong, rigid 3-D network. Diamond is extremely hard, so it is used in cutting tools 切割工具.

    Graphite 石墨: each carbon atom is bonded to only three others, forming flat layers 层. There are weak forces between the layers, so the layers can slide over each other — this makes graphite a good lubricant 润滑剂. The fourth outer electron of each carbon is free; these delocalised electrons 离域电子 can move, so graphite conducts electricity and is used as an electrode 电极.

    Diamond with each carbon bonded to four others in a network, beside graphite's flat hexagonal layers held by weak forces
    Diamond bonds each carbon to four others (hard); graphite forms flat layers with weak forces between them (slippery)
    A specimen of natural graphite: a dull grey-black mineral splitting into thin flakes
    Natural graphite: those weak forces between the layers are why the specimen flakes, and why it marks paper

    Silicon(IV) oxide 二氧化硅 ($\text{SiO}_2$) has a giant covalent structure like diamond, so it is also very hard and has a very high melting point.

    Explore · ⁨探索⁩

    Giant covalent lab · ⁨巨大共有結合の実験室⁩

    Compare giant covalent structures by bonding and properties. · ⁨結合と性質によって巨大共有結合構造を比較せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    giant covalent structure/ˈdʒaɪənt ˈkəʊvələnt ˈstrʌktʃə/ 巨大共有結合構造
    diamond/ˈdaɪəmənd/ ダイヤモンド
    cutting tools/ˈkʌtɪŋ tuːlz/ 切削工具
    graphite/ˈɡræfaɪt/ グラファイト
    layers/ˈleɪəz/ レイヤー
    lubricant/ˈluːbrɪkənt/ 潤滑剤
    delocalised electrons/dɪˈlɒkəlaɪzd ɪˈlektrɒnz/ 非局在電子
    electrode/ɪˈlektrəʊd/ 電極
    silicon(IV) oxide/ˈsɪlɪkən ˈɒksaɪd/ 四酸化ケイ素
    2.7

    Metallic bonding

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe metallic bonding as the electrostatic attraction between the positive ions in a giant metallic lattice and a ‘sea’ of delocalised electrons
    2 Explain in terms of structure and bonding the properties of metals: (a) good electrical conductivity (b) malleability and ductility
    日本語
    コア サプリメント
    1 金属結合を、巨大な金属格子中の正イオンと「海」のような非局在電子との間の静電気的引力として説明せよ
    2 構造および結合の観点から、金属の性質を説明せよ: (a) 高い電気伝導性 (b) 展性および延性

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A metal is a giant structure of positive ions surrounded by a 'sea' of delocalised electrons that are free to move through the whole metal. Metallic bonding 金属键 is the strong electrostatic attraction between these positive ions and the sea of electrons.

    This explains two key properties of metals:

    • Good electrical conductivity: the delocalised electrons are free to move and carry charge through the metal.
    • Malleability 展性 (can be hammered into sheets) and ductility 延性 (can be pulled into wires): the layers of positive ions can slide over each other without breaking the metallic bond, so the metal changes shape instead of shattering.
    日本語

    A metal is a giant structure of positive ions surrounded by a 'sea' of delocalised electrons that are free to move through the whole metal. Metallic bonding 金属键 is the strong electrostatic attraction between these positive ions and the sea of electrons.

    A regular lattice of positive metal ions with free electrons scattered in the gaps between them
    A metal is positive ions in a 'sea' of delocalised electrons that are free to move and carry charge

    This explains two key properties of metals:

    • Good electrical conductivity: the delocalised electrons are free to move and carry charge through the metal.
    • Malleability 展性 (can be hammered into sheets) and ductility 延性 (can be pulled into wires): the layers of positive ions can slide over each other without breaking the metallic bond, so the metal changes shape instead of shattering.
    Explore · ⁨探索⁩

    Inside a metal — and why it behaves that way · ⁨金属内部の構造と、その物理的性質の理由⁩

    Step through it. Positive ions sit in a shared sea of delocalised electrons — that one picture explains conduction and why metals bend instead of snapping. · ⁨手順を追う。正イオンが非局在電子の海の中に座っている — この図は伝導と、なぜ金属が折れることなく曲げられるかを説明する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    metallic bonding/məˈtælɪk ˈbɒndɪŋ/ 金属結合
    malleability/ˌmæləˈbɪlɪti/ 延性
    ductility/dʌkˈtɪlɪti/ 延性
    Watch lesson · ⁨レッスンを視聴⁩
    2.7

    Exam tips

    • Number of neutrons = mass number − proton number. The proton number is what decides which element an atom is.
    • Isotopes have the same chemical properties because reactions involve only electrons, and isotopes share the same electron configuration; only the mass differs.
    • Match structure to properties: ionic = giant lattice (high melting point; conducts only when molten or aqueous); simple molecular = weak forces between molecules (low melting point; no conduction); giant covalent = very hard, very high melting point.
    • Graphite conducts and is slippery (one free electron per carbon; layers slide); diamond does neither (all four electrons bonded, so hard and non-conducting). Metals conduct and bend thanks to the sea of delocalised electrons.
    • For relative atomic mass, take the weighted average: $A_r = \dfrac{\sum(\text{isotope mass} \times \text{abundance})}{100}$.
  • 3

    Stoichiometry

    Watch lesson · ⁨レッスンを視聴⁩
    English

    Stoichiometry 化学计量 is the study of the amounts of substances in a reaction — how much reacts and how much is made. This topic is mostly about counting atoms and doing calculations.

    日本語

    Stoichiometry 化学计量 is the study of the amounts of substances in a reaction — how much reacts and how much is made. This topic is mostly about counting atoms and doing calculations.

    3.1

    Chemical formulae

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State the formulae of the elements and compounds named in the subject content
    2 Define the molecular formula of a compound as the number and type of different atoms in one molecule 5 Define the empirical formula of a compound as the simplest whole number ratio of the different atoms or ions in a compound
    3 Deduce the formula of a simple compound from the relative numbers of atoms present in a model or a diagrammatic representation 6 Deduce the formula of an ionic compound from the relative numbers of the ions present in a model or a diagrammatic representation or from the charges on the ions
    4 Construct word equations and symbol equations to show how reactants form products, including state symbols 7 Construct symbol equations with state symbols, including ionic equations
    8 Deduce the symbol equation with state symbols for a chemical reaction, given relevant information
    日本語
    コア サプリメント
    1 学習内容で名付けられた元素および化合物の化学式を述べよ
    2 化合物の分子式を、1つの分子に含まれる異なる原子の数および種類として定義する 5 化合物の実験式を、化合物内の異なる原子またはイオンの最も単純な整数比として定義する
    3 モデルまたは模式図に存在する原子の相対数から、単純化合物の化学式を導き出す 6 モデルまたは模式図に存在するイオンの相対数、あるいはイオンの電荷から、イオン化合物の化学式を導き出す
    4 反応物から生成物がどのように形成されるかを示す語式方程式および記号方程式を構築する(状態記号を含む) 7 状態記号を含む記号方程式を構築する(イオン方程式を含む)
    8 必要な情報を与えられた場合、化学反応に対する状態記号付きの記号方程式を導き出す

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A formula 化学式 shows which atoms 原子 are in a substance, and how many of each. There are two kinds of formula you must know.

    • The molecular formula 分子式 is the actual number of each atom in one molecule 分子. For glucose it is $\text{C}_6\text{H}_{12}\text{O}_6$.
    • The empirical formula 实验式 is the simplest whole-number ratio 比例 of the atoms or ions 离子 in a compound 化合物. For glucose it is $\text{CH}_2\text{O}$.

    Working out a formula

    If you are given a model or diagram, just count the atoms of each element and write them as a formula.

    For an ionic compound 离子化合物 you can work out the formula from the charges on the ions. The total positive charge must balance the total negative charge, because the compound has no overall charge. Some common ions:

    Positive ions Negative ions
    $\text{Na}^{+}$, $\text{K}^{+}$, $\text{H}^{+}$ $\text{Cl}^{-}$, $\text{OH}^{-}$, $\text{NO}_3^{-}$
    $\text{Mg}^{2+}$, $\text{Ca}^{2+}$, $\text{Cu}^{2+}$ $\text{O}^{2-}$, $\text{SO}_4^{2-}$, $\text{CO}_3^{2-}$
    $\text{Al}^{3+}$ $\text{N}^{3-}$

    For example, $\text{Na}^{+}$ and $\text{O}^{2-}$: you need two $\text{Na}^{+}$ to balance one $\text{O}^{2-}$, so the formula is $\text{Na}_2\text{O}$.

    日本語

    A formula 化学式 shows which atoms 原子 are in a substance, and how many of each. There are two kinds of formula you must know.

    • The molecular formula 分子式 is the actual number of each atom in one molecule 分子. For glucose it is $\text{C}_6\text{H}_{12}\text{O}_6$.
    • The empirical formula 实验式 is the simplest whole-number ratio 比例 of the atoms or ions 离子 in a compound 化合物. For glucose it is $\text{CH}_2\text{O}$.

    Working out a formula

    If you are given a model or diagram, just count the atoms of each element and write them as a formula.

    For an ionic compound 离子化合物 you can work out the formula from the charges on the ions. The total positive charge must balance the total negative charge, because the compound has no overall charge. Some common ions:

    Positive ions Negative ions
    $\text{Na}^{+}$, $\text{K}^{+}$, $\text{H}^{+}$ $\text{Cl}^{-}$, $\text{OH}^{-}$, $\text{NO}_3^{-}$
    $\text{Mg}^{2+}$, $\text{Ca}^{2+}$, $\text{Cu}^{2+}$ $\text{O}^{2-}$, $\text{SO}_4^{2-}$, $\text{CO}_3^{2-}$
    $\text{Al}^{3+}$ $\text{N}^{3-}$

    For example, $\text{Na}^{+}$ and $\text{O}^{2-}$: you need two $\text{Na}^{+}$ to balance one $\text{O}^{2-}$, so the formula is $\text{Na}_2\text{O}$.

    Aluminium 3+ and oxygen 2- ions with their charge numbers crossing over to become the subscripts in Al2O3
    For an ionic compound, the ion charges cross over to give the formula (here $\text{Al}_2\text{O}_3$)
    Explore · ⁨探索⁩

    Formula writing route · ⁨化学式の書き方⁩

    Follow charges or valencies to a neutral chemical formula. · ⁨電荷または価数に従って、電荷の釣り合った化学式を導く。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    stoichiometry/ˌstəʊɪkɪˈɒmətri/ 化学量論
    formula/ˈfɔːmjʊlə/ 公式
    atoms/ˈætəmz/ 原子
    molecular formula/məˈlekjʊlə ˈfɔːmjʊlə/ 分子式
    molecule/ˈmɒlɪkjuːl/ 分子
    empirical formula/emˈpɪrɪkl ˈfɔːmjʊlə/ 経験式
    ratio/ˈreɪʃɪəʊ/ 比
    ions/ˈaɪɒnz/ イオン
    compound/ˈkɒmpaʊnd/ 化合物
    ionic compound/aɪˈɒnɪk ˈkɒmpaʊnd/ イオン化合物
    Watch lesson · ⁨レッスンを視聴⁩
    3.1

    Writing equations

    English

    An equation shows how reactants 反应物 (the starting substances) change into products 生成物 (the substances made).

    • A word equation 文字方程式 uses names: magnesium + oxygen → magnesium oxide.
    • A symbol equation 化学方程式 uses formulae and must be balanced 配平 — the same number of each atom on both sides.
    $$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$

    You add state symbols 状态符号 in brackets to show the physical state: $(s)$ solid, $(l)$ liquid, $(g)$ gas, and $(aq)$ for aqueous 水溶液 (dissolved in water).

    $$\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$$

    Ionic equations

    An ionic equation 离子方程式 shows only the ions that actually change. Ions that are the same on both sides are spectator ions 旁观离子 and are left out. For example, when an acid reacts with an alkali:

    $$\text{H}^{+}(aq) + \text{OH}^{-}(aq) \rightarrow \text{H}_2\text{O}(l)$$
    日本語

    An equation shows how reactants 反应物 (the starting substances) change into products 生成物 (the substances made).

    • A word equation 文字方程式 uses names: magnesium + oxygen → magnesium oxide.
    • A symbol equation 化学方程式 uses formulae and must be balanced 配平 — the same number of each atom on both sides.
    $$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$
    Two magnesium atoms plus an oxygen molecule react to form two units of magnesium oxide, with two Mg and two O on each side
    A balanced symbol equation has the same number of each atom on both sides

    You add state symbols 状态符号 in brackets to show the physical state: $(s)$ solid, $(l)$ liquid, $(g)$ gas, and $(aq)$ for aqueous 水溶液 (dissolved in water).

    $$\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$$

    Ionic equations

    An ionic equation 离子方程式 shows only the ions that actually change. Ions that are the same on both sides are spectator ions 旁观离子 and are left out. For example, when an acid reacts with an alkali:

    $$\text{H}^{+}(aq) + \text{OH}^{-}(aq) \rightarrow \text{H}_2\text{O}(l)$$
    Explore · ⁨探索⁩

    Balance the equation · ⁨方程式をバランスさせる⁩

    Step the coefficients until every element has the same number of atoms on both sides. · ⁨係数を調整して、両側のすべての元素について原子数が一致するようにする。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    reactants/rɪˈæktənts/ 反応物である
    products/ˈprɒdʌkts/ 生成物
    word equation/wɜːd ɪˈkweɪʒn/ 文字式
    symbol equation/ˈsɪmbl ɪˈkweɪʒn/ 記号式
    balanced/ˈbælənst/ バランス取れた
    state symbols/steɪt ˈsɪmblz/ 状態記号
    aqueous/ˈeɪkwɪəs/ 水溶液(_aqueous_)
    ionic equation/aɪˈɒnɪk ɪˈkweɪʒn/ イオン方程式
    spectator ions/spekˈteɪtə ˈaɪɒnz/ 観測イオン
    3.2

    Relative masses

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe relative atomic mass, $A_r$, as the average mass of the isotopes of an element compared to 1/12th of the mass of an atom of $^{12}\text{C}$
    2 Define relative molecular mass, $M_r$, as the sum of the relative atomic masses. Relative formula mass, $M_r$, will be used for ionic compounds
    3 Calculate reacting masses in simple proportions. Calculations will not involve the mole concept
    日本語
    コア サプリメント
    1 相対原子質量、$A_r$ を、ある元素の同位体の平均質量を、$^{12}\text{C}$ の原子の質量の 1/12 と比較したものとして定義する
    2 相対分子量、$M_r$ を、相対原子質量の総和として定義する。イオン化合物に対しては相対式量、$M_r$ が用いられる
    3 単純な比率で反応物の質量を計算する。計算にはモル概念は含まれない

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The relative atomic mass 相对原子质量 ($A_r$) of an element 元素 is the average mass 质量 of its atoms compared to $\tfrac{1}{12}$ of the mass of one $^{12}\text{C}$ atom.

    The relative molecular mass 相对分子质量 ($M_r$) is the sum of the relative atomic masses of all the atoms in the molecule. For ionic compounds we use the relative formula mass 相对式量, found the same way from the formula.

    $$M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18$$

    Reacting masses by simple proportion

    You can sometimes find a reacting mass without the mole. If 24 g of magnesium makes 40 g of magnesium oxide, then 12 g of magnesium (half as much) makes 20 g of magnesium oxide.

    日本語
    A laboratory analytical balance
    A laboratory balance measures mass — the basis of mole calculations.

    The relative atomic mass 相对原子质量 ($A_r$) of an element 元素 is the average mass 质量 of its atoms compared to $\tfrac{1}{12}$ of the mass of one $^{12}\text{C}$ atom.

    The relative molecular mass 相对分子质量 ($M_r$) is the sum of the relative atomic masses of all the atoms in the molecule. For ionic compounds we use the relative formula mass 相对式量, found the same way from the formula.

    $$M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18$$

    Reacting masses by simple proportion

    You can sometimes find a reacting mass without the mole. If 24 g of magnesium makes 40 g of magnesium oxide, then 12 g of magnesium (half as much) makes 20 g of magnesium oxide.

    Explore · ⁨探索⁩

    Formula mass lab · ⁨式量実験⁩

    mass = moles x Mr · ⁨質量 = モル数 × 式量(Mr)⁩

    Change number of formula units and see total mass scale. · ⁨式単位数を変えて総質量スケールを確認する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    relative atomic mass/ˈrelətɪv əˈtɒmɪk mæs/ 相対原子質量
    element/ˈelɪmənt/ 元素
    mass/mæs/ 質量
    relative molecular mass/ˈrelətɪv məˈlekjʊlə mæs/ 相対分子量
    relative formula mass/ˈrelətɪv ˈfɔːmjʊlə mæs/ 相対式質量
    Watch lesson · ⁨レッスンを視聴⁩
    3.3

    The mole

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    2 State that the mole, mol, is the unit of amount of substance and that one mole contains $6.02 \times 10^{23}$ particles, e.g. atoms, ions, molecules; this number is the Avogadro constant
    3 Use the relationship $\text{amount of substance (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$ to calculate: (a) amount of substance (b) mass (c) molar mass (d) relative atomic mass or relative molecular/formula mass (e) number of particles, using the value of the Avogadro constant
    4 Use the molar gas volume, taken as $24\text{ dm}^3$ at room temperature and pressure, r.t.p., in calculations involving gases
    1 State that concentration can be measured in $\text{g/dm}^3$ or $\text{mol/dm}^3$ 5 Calculate stoichiometric reacting masses, limiting reactants, volumes of gases at r.t.p., volumes of solutions and concentrations of solutions expressed in $\text{g/dm}^3$ and $\text{mol/dm}^3$, including conversion between $\text{cm}^3$ and $\text{dm}^3$
    6 Use experimental data from a titration to calculate the moles of solute, or the concentration or volume of a solution
    7 Calculate empirical formulae and molecular formulae, given appropriate data
    8 Calculate percentage yield, percentage composition by mass and percentage purity, given appropriate data
    日本語
    コア サプリメント
    2 モル、mol は物質の量の単位であり、1モルには $6.02 \times 10^{23}$ 個の粒子(例:原子、イオン、分子)が含まれることを述べる。この数はアボガドロ定数である
    3 関係式 $\text{amount of substance (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$ を用いて次のものを計算する:(a) 物質の量 (b) 質量 (c) 摩尔質量 (d) 相対原子質量または相対分子量/式量 (e) 粒子数(アボガドロ定数の値を用いる)
    4 気体に関する計算において、室温・常圧下(r.t.p.)でのモル体積を $24\text{ dm}^3$ とみなして用いる
    1 濃度を $\text{g/dm}^3$ または $\text{mol/dm}^3$ で測定できることを述べる 5 化学量論的な反応物の質量、限界反応物、r.t.p. における気体の体積、溶液の体積、および $\text{g/dm}^3$ および $\text{mol/dm}^3$ で表された溶液の濃度を計算する。$\text{cm}^3$ と $\text{dm}^3$ の間の換算も含む
    6 滴定の実験データを用いて、溶質のモル数、または溶液の濃度や体積を計算する
    7 適切なデータが与えられた場合、実験式および分子式を計算する
    8 適切なデータに基づき、収率、質量に対する組成の割合、および純度の百分率を計算する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The mole 摩尔 (symbol mol) is the unit for the amount of substance 物质的量. One mole of any substance contains $6.02 \times 10^{23}$ particles 粒子 (atoms, ions or molecules). This number is the Avogadro constant 阿伏伽德罗常数.

    The molar mass 摩尔质量 is the mass of one mole, in grams per mole (g/mol). Its number is the same as the $A_r$ or $M_r$. The key relationship is:

    $$\text{amount (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$$

    Worked example. How many moles are in 36 g of water? Molar mass of water $= 18$ g/mol.

    $$n = \frac{36}{18} = 2 \text{ mol}$$

    To find the number of particles, multiply the moles by the Avogadro constant.

    Volumes of gases

    At room temperature and pressure (r.t.p.), one mole of any gas takes up the same volume 体积. This molar gas volume 摩尔气体体积 is $24 \text{ dm}^3$.

    $$\text{volume of gas (dm}^3) = \text{amount (mol)} \times 24$$

    Worked example. What volume does $0.25$ mol of carbon dioxide occupy at r.t.p.?

    $$\text{volume} = 0.25 \times 24 = 6 \text{ dm}^3$$

    Concentration of solutions

    The concentration 浓度 of a solution 溶液 can be given in $\text{g/dm}^3$ or in $\text{mol/dm}^3$.

    $$\text{concentration (mol/dm}^3) = \frac{\text{amount of solute (mol)}}{\text{volume (dm}^3)}$$

    Remember to convert volume: $1 \text{ dm}^3 = 1000 \text{ cm}^3$, so divide a volume in $\text{cm}^3$ by 1000.

    日本語

    The mole 摩尔 (symbol mol) is the unit for the amount of substance 物质的量. One mole of any substance contains $6.02 \times 10^{23}$ particles 粒子 (atoms, ions or molecules). This number is the Avogadro constant 阿伏伽德罗常数.

    The molar mass 摩尔质量 is the mass of one mole, in grams per mole (g/mol). Its number is the same as the $A_r$ or $M_r$. The key relationship is:

    $$\text{amount (mol)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$$

    Worked example. How many moles are in 36 g of water? Molar mass of water $= 18$ g/mol.

    $$n = \frac{36}{18} = 2 \text{ mol}$$

    To find the number of particles, multiply the moles by the Avogadro constant.

    Volumes of gases

    At room temperature and pressure (r.t.p.), one mole of any gas takes up the same volume 体积. This molar gas volume 摩尔气体体积 is $24 \text{ dm}^3$.

    Three balloons of the same size holding one mole of hydrogen, oxygen and carbon dioxide: different masses but the same 24 cubic decimetres
    One mole of any gas fills 24 cubic decimetres at room temperature and pressure
    $$\text{volume of gas (dm}^3) = \text{amount (mol)} \times 24$$

    Worked example. What volume does $0.25$ mol of carbon dioxide occupy at r.t.p.?

    $$\text{volume} = 0.25 \times 24 = 6 \text{ dm}^3$$

    Concentration of solutions

    The concentration 浓度 of a solution 溶液 can be given in $\text{g/dm}^3$ or in $\text{mol/dm}^3$.

    $$\text{concentration (mol/dm}^3) = \frac{\text{amount of solute (mol)}}{\text{volume (dm}^3)}$$

    Remember to convert volume: $1 \text{ dm}^3 = 1000 \text{ cm}^3$, so divide a volume in $\text{cm}^3$ by 1000.

    A mole map with moles in the centre, linked to mass, gas volume, concentration and number of particles by their conversion factors
    Moles sit at the centre: multiply going outwards, divide coming back (concentration is moles ÷ volume)
    Explore · ⁨探索⁩

    Mole particle lab · ⁨モル粒子実験⁩

    particles = n x Avogadro constant · ⁨粒子数 = n × アボガドロ定数⁩

    Change moles and see particle number scale with Avogadro constant. · ⁨モル数を変えて、アボガドロ定数に対応する粒子数スケールを確認する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    mole/məʊl/ モル
    amount of substance/əˈmaʊnt ɒv ˈsʌbstəns/ 物質量
    particles/ˈpɑːtɪklz/ 粒子
    Avogadro constant/ˌævəˈɡædrəʊ ˈkɒnstənt/ アボガドロ定数
    molar mass/ˈməʊlə mæs/ モル質量
    volume/ˈvɒljuːm/ 体積
    molar gas volume/ˈməʊlə ɡæs ˈvɒljuːm/ モル体積
    concentration/ˌkɒnsənˈtreɪʃn/ 濃度
    solution/səˈluːʃn/ 溶液
    3.3

    Doing reaction calculations

    English

    Most calculations follow the same steps: change the known amount into moles, use the balanced equation to find the moles of what you want, then change back into mass, volume or concentration.

    Worked example. What mass of magnesium oxide forms when 12 g of magnesium burns completely? $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. ($A_r$: Mg $= 24$, O $= 16$.)

    Moles of Mg $= 12 / 24 = 0.5$ mol. The ratio Mg : MgO is $1 : 1$, so $0.5$ mol of MgO forms. Its molar mass is $24 + 16 = 40$ g/mol, so the mass is $0.5 \times 40 = 20$ g — the same answer as the simple-proportion method above.

    Limiting reactant

    When two reactants are mixed, often one runs out first. This is the limiting reactant 限量反应物. It decides how much product can form; any other reactant is in excess (left over).

    Worked example. 8 g of hydrogen reacts with 8 g of oxygen: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$. Which is the limiting reactant, and what mass of water forms? ($A_r$: H $= 1$, O $= 16$.)

    Moles of $\text{H}_2 = 8 / 2 = 4$ mol; moles of $\text{O}_2 = 8 / 32 = 0.25$ mol. The equation needs $2\text{H}_2$ for every one $\text{O}_2$, so $0.25$ mol of $\text{O}_2$ reacts with only $0.5$ mol of $\text{H}_2$. There is far more hydrogen than that, so oxygen is the limiting reactant and the hydrogen is in excess. The oxygen makes $2 \times 0.25 = 0.5$ mol of water, so the mass is $0.5 \times 18 = 9$ g. Always work in moles, not masses — the reactant with the smaller mass is not always the one that runs out.

    Titration calculation

    In a titration 滴定 you measure the volume of one solution that reacts with another. From the volume and concentration you find the moles of one solute 溶质, then use the equation ratio to find the moles, concentration or volume of the other.

    Worked example. $25.0 \text{ cm}^3$ of sodium hydroxide solution is exactly neutralised by $20.0 \text{ cm}^3$ of $0.10 \text{ mol/dm}^3$ hydrochloric acid. $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$. Find the concentration of the sodium hydroxide.

    Moles of HCl $= 0.10 \times \dfrac{20.0}{1000} = 0.0020$ mol. The ratio is $1 : 1$, so there are $0.0020$ mol of NaOH in $25.0 \text{ cm}^3$:

    $$\text{concentration} = \frac{0.0020}{25.0 / 1000} = 0.08 \text{ mol/dm}^3$$

    Empirical and molecular formulae from data

    To find an empirical formula from masses or percentages:

    1. Divide each element's mass (or %) by its $A_r$.
    2. Divide all the answers by the smallest one to get the simplest ratio.

    To find the molecular formula, compare the empirical formula mass with the real $M_r$ and multiply up.

    Worked example. A compound contains $40\%$ calcium, $12\%$ carbon and $48\%$ oxygen by mass. Find its empirical formula. ($A_r$: Ca $= 40$, C $= 12$, O $= 16$.)

    Divide each percentage by its $A_r$: Ca $= 40/40 = 1$, C $= 12/12 = 1$, O $= 48/16 = 3$. The ratio is $1 : 1 : 3$, so the empirical formula is $\text{CaCO}_3$.

    Percentages

    • Percentage yield 产率 compares how much product you actually got with the most you could get: $\dfrac{\text{actual}}{\text{theoretical}} \times 100$.
    • Percentage composition by mass 质量分数 of an element $= \dfrac{\text{mass of the element}}{M_r} \times 100$.
    • Percentage purity 纯度 $= \dfrac{\text{mass of pure substance}}{\text{mass of impure sample}} \times 100$.

    Worked example. Find the percentage by mass of nitrogen in ammonium nitrate, $\text{NH}_4\text{NO}_3$. ($A_r$: N $= 14$, H $= 1$, O $= 16$.)

    $M_r = (2 \times 14) + (4 \times 1) + (3 \times 16) = 80$. The mass of nitrogen is $2 \times 14 = 28$, so

    $$\% \text{ nitrogen} = \frac{28}{80} \times 100 = 35\%$$

    Worked example. A reaction could make at most 5.0 g of product, but only 4.0 g is actually collected. Find the percentage yield.

    $$\% \text{ yield} = \frac{4.0}{5.0} \times 100 = 80\%$$

    Some product is always lost — left behind in the apparatus, or used up in side reactions — so the yield is almost never $100\%$.

    Worked example. A 50 g sample of impure calcium carbonate contains 45 g of pure calcium carbonate. Find its percentage purity.

    $$\% \text{ purity} = \frac{45}{50} \times 100 = 90\%$$
    日本語
    A titration being carried out in a laboratory
    A titration measures exactly how much acid reacts with a base.

    Most calculations follow the same steps: change the known amount into moles, use the balanced equation to find the moles of what you want, then change back into mass, volume or concentration.

    Worked example. What mass of magnesium oxide forms when 12 g of magnesium burns completely? $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$. ($A_r$: Mg $= 24$, O $= 16$.)

    Moles of Mg $= 12 / 24 = 0.5$ mol. The ratio Mg : MgO is $1 : 1$, so $0.5$ mol of MgO forms. Its molar mass is $24 + 16 = 40$ g/mol, so the mass is $0.5 \times 40 = 20$ g — the same answer as the simple-proportion method above.

    Limiting reactant

    When two reactants are mixed, often one runs out first. This is the limiting reactant 限量反应物. It decides how much product can form; any other reactant is in excess (left over).

    Four A particles and two B particles react to make two products, leaving two A particles unreacted
    B runs out first, so it is the limiting reactant and sets how much product forms; the spare A is left over in excess

    Worked example. 8 g of hydrogen reacts with 8 g of oxygen: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$. Which is the limiting reactant, and what mass of water forms? ($A_r$: H $= 1$, O $= 16$.)

    Moles of $\text{H}_2 = 8 / 2 = 4$ mol; moles of $\text{O}_2 = 8 / 32 = 0.25$ mol. The equation needs $2\text{H}_2$ for every one $\text{O}_2$, so $0.25$ mol of $\text{O}_2$ reacts with only $0.5$ mol of $\text{H}_2$. There is far more hydrogen than that, so oxygen is the limiting reactant and the hydrogen is in excess. The oxygen makes $2 \times 0.25 = 0.5$ mol of water, so the mass is $0.5 \times 18 = 9$ g. Always work in moles, not masses — the reactant with the smaller mass is not always the one that runs out.

    Titration calculation

    In a titration 滴定 you measure the volume of one solution that reacts with another. From the volume and concentration you find the moles of one solute 溶质, then use the equation ratio to find the moles, concentration or volume of the other.

    Worked example. $25.0 \text{ cm}^3$ of sodium hydroxide solution is exactly neutralised by $20.0 \text{ cm}^3$ of $0.10 \text{ mol/dm}^3$ hydrochloric acid. $\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$. Find the concentration of the sodium hydroxide.

    Moles of HCl $= 0.10 \times \dfrac{20.0}{1000} = 0.0020$ mol. The ratio is $1 : 1$, so there are $0.0020$ mol of NaOH in $25.0 \text{ cm}^3$:

    $$\text{concentration} = \frac{0.0020}{25.0 / 1000} = 0.08 \text{ mol/dm}^3$$
    The titration calculation as a flow: the known acid gives its moles, the equation's 1 to 1 ratio gives the moles of sodium hydroxide, and dividing by its volume gives the concentration
    The titration calculation: known solution to moles, ratio, then the unknown concentration

    Empirical and molecular formulae from data

    To find an empirical formula from masses or percentages:

    1. Divide each element's mass (or %) by its $A_r$.
    2. Divide all the answers by the smallest one to get the simplest ratio.

    To find the molecular formula, compare the empirical formula mass with the real $M_r$ and multiply up.

    Worked example. A compound contains $40\%$ calcium, $12\%$ carbon and $48\%$ oxygen by mass. Find its empirical formula. ($A_r$: Ca $= 40$, C $= 12$, O $= 16$.)

    Divide each percentage by its $A_r$: Ca $= 40/40 = 1$, C $= 12/12 = 1$, O $= 48/16 = 3$. The ratio is $1 : 1 : 3$, so the empirical formula is $\text{CaCO}_3$.

    Percentages

    • Percentage yield 产率 compares how much product you actually got with the most you could get: $\dfrac{\text{actual}}{\text{theoretical}} \times 100$.
    • Percentage composition by mass 质量分数 of an element $= \dfrac{\text{mass of the element}}{M_r} \times 100$.
    • Percentage purity 纯度 $= \dfrac{\text{mass of pure substance}}{\text{mass of impure sample}} \times 100$.

    Worked example. Find the percentage by mass of nitrogen in ammonium nitrate, $\text{NH}_4\text{NO}_3$. ($A_r$: N $= 14$, H $= 1$, O $= 16$.)

    $M_r = (2 \times 14) + (4 \times 1) + (3 \times 16) = 80$. The mass of nitrogen is $2 \times 14 = 28$, so

    $$\% \text{ nitrogen} = \frac{28}{80} \times 100 = 35\%$$

    Worked example. A reaction could make at most 5.0 g of product, but only 4.0 g is actually collected. Find the percentage yield.

    $$\% \text{ yield} = \frac{4.0}{5.0} \times 100 = 80\%$$

    Some product is always lost — left behind in the apparatus, or used up in side reactions — so the yield is almost never $100\%$.

    Worked example. A 50 g sample of impure calcium carbonate contains 45 g of pure calcium carbonate. Find its percentage purity.

    $$\% \text{ purity} = \frac{45}{50} \times 100 = 90\%$$
    Explore · ⁨探索⁩

    Reaction calculation route · ⁨反応計算の流れ⁩

    Follow a known mass through a balanced equation to the answer. · ⁨既知の質量から balanced equation を通じて答えに至るまで追跡する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    limiting reactant/ˈlɪmɪtɪŋ rɪˈæktənt/ limiting reactant(限定試薬)
    titration/taɪˈtreɪʃn/ 滴定
    solute/ˈsɒljuːt/ 溶質
    percentage yield/pəˈsentɪdʒ jiːld/ 収率
    percentage composition by mass/pəˈsentɪdʒ ˌkɒmpəˈzɪʃn baɪ mæs/ 質量百分組成
    percentage purity/pəˈsentɪdʒ ˈpjʊərɪti/ 百分率純度
    3.3

    Exam tips

    • Always turn masses, gas volumes and concentrations into moles first, use the balanced equation's ratio, then convert back. Never compare masses directly across an equation.
    • Key formulas: $n = \dfrac{\text{mass}}{M_r}$, gas volume $= n \times 24\ \text{dm}^3$ at r.t.p., and concentration $= \dfrac{n}{\text{volume in dm}^3}$. Change cm³ to dm³ by dividing by 1000.
    • To balance a symbol equation you may only change the big numbers in front of a formula, never the small subscripts inside it.
    • The limiting reactant is the one that runs out (fewest moles once you allow for the ratio); it sets how much product forms. The other reactant is in excess.
    • Percentage yield = actual ÷ theoretical × 100; percentage purity = mass of pure substance ÷ mass of sample × 100. Both are always 100% or less.
  • 4

    Electrochemistry

    Watch lesson · ⁨レッスンを視聴⁩
    4.1

    Electrolysis

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Define electrolysis as the decomposition of an ionic compound, when molten or in aqueous solution, by the passage of an electric current 8 Describe the transfer of charge during electrolysis to include: (a) the movement of electrons in the external circuit (b) the loss or gain of electrons at the electrodes (c) the movement of ions in the electrolyte
    2 Identify in simple electrolytic cells: (a) the anode as the positive electrode (b) the cathode as the negative electrode (c) the electrolyte as the molten or aqueous substance that undergoes electrolysis
    3 Identify the products formed at the electrodes and describe the observations made during the electrolysis of: (a) molten lead(II) bromide (b) concentrated aqueous sodium chloride (c) dilute sulfuric acid using inert electrodes made of platinum or carbon/graphite 9 Identify the products formed at the electrodes and describe the observations made during the electrolysis of aqueous copper(II) sulfate using inert carbon/graphite electrodes and when using copper electrodes
    4 State that metals or hydrogen are formed at the cathode and that non-metals (other than hydrogen) are formed at the anode
    5 Predict the identity of the products at each electrode for the electrolysis of a binary compound in the molten state 10 Predict the identity of the products at each electrode for the electrolysis of a halide compound in dilute or concentrated aqueous solution
    11 Construct ionic half-equations for reactions at the anode (to show oxidation) and at the cathode (to show reduction)
    6 State that metal objects are electroplated to improve their appearance and resistance to corrosion
    7 Describe how metals are electroplated
    日本語
    コア サプリメント
    1 電解を、イオン性化合物が溶融状態または水溶液中で電気流を通過させることで分解される現象として定義する 8 電解中の電荷移動について説明せよ:(a) 外部回路における電子の移動 (b) 電極での電子の損失または獲得 (c) 電解質内のイオンの移動
    2 単純な電解セルにおいて次のものを識別せよ:(a) アノードは正極である (b) カソードは負極である (c) 電解質は電解を受ける溶融物質または水溶液である
    3 白金または炭素/グラファイト製の不活性電極を用いて、次のものの電解中に電極で生成される生成物と観察される現象を識別して説明せよ:(a) 溶融臭化鉛(II) (b) 濃塩化ナトリウム水溶液 (c) 希硫酸 9 不活性炭素/グラファイト電極を使用した場合と、銅電極を使用する場合の両方について、硫酸銅(II)水溶液の電解中に電極で生成される生成物と観察される現象を識別して説明せよ
    4 カソードでは金属または水素が生成され、アノードでは非金属(水素を除く)が生成されることを述べる
    5 溶融状態の二元化合物を電解した際の各電極で生成される生成物の種類を予測する 10 希水溶液または濃水溶液におけるハロゲン化物化合物を電解した際の各電極で生成される生成物の種類を予測する
    11 アノード(酸化を示す)およびカソード(還元を示す)での反応に対するイオン半方程式を立てる
    6⟧ 金属製品の外観向上および腐食抵抗強化のために電気メッキが行われることを述べる
    7 金属のめっき法を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Electrolysis: ions discharge at the electrodes

    Electrolysis 电解 is the breaking down (decomposition 分解) of an ionic compound 离子化合物 — when it is molten 熔融 or in aqueous 水溶液 solution — by passing an electric current 电流 through it.

    It only works when the substance is molten or dissolved, because then the ions 离子 are free to move and carry the charge. A solid ionic compound cannot be electrolysed because its ions are locked in place.

    The electrolytic cell

    The set-up is called an electrolytic cell 电解池. Two electrodes 电极 (solid conductors) dip into the electrolyte 电解质 — the molten or aqueous substance being broken down.

    • The anode 阳极 is the positive ($+$) electrode.
    • The cathode 阴极 is the negative ($-$) electrode.

    The electrodes are often inert 惰性 (they do not react), such as platinum 铂 or carbon/graphite 石墨.

    What is formed at each electrode

    There is a simple rule:

    • Metals 金属 or hydrogen 氢气 are formed at the cathode.
    • Non-metals 非金属 (other than hydrogen) are formed at the anode.

    For aqueous solutions, the product can depend on whether the solution is dilute 稀 or concentrated 浓. Here are the three Core examples:

    Electrolyte At the cathode ($-$) At the anode ($+$)
    molten lead(II) bromide, $\text{PbBr}_2$ lead 铅 (silvery liquid) bromine 溴 (red-brown vapour)
    concentrated aqueous sodium chloride hydrogen (bubbles of gas) chlorine 氯气 (pale green gas)
    dilute sulfuric acid 硫酸 hydrogen (bubbles of gas) oxygen 氧气 (bubbles of gas)

    Worked example. Predict the products of electrolysing molten zinc chloride, $\text{ZnCl}_2$. Because it is molten, only zinc ions and chloride ions are present - there is no water to complicate things. Zinc is a metal, so it forms at the cathode: $\text{Zn}^{2+}$ ions gain electrons and silvery zinc appears. Chlorine is a non-metal, so it forms at the anode: pale green chlorine gas bubbles off. Always check the state first. A molten compound simply gives you its own two elements, while an aqueous one brings water into the competition - which is why concentrated sodium chloride solution gives hydrogen at the cathode rather than sodium.

    How the charge moves

    During electrolysis:

    • Electrons 电子 move through the wires (the external circuit 外电路) from the power supply.
    • At the cathode, positive ions gain electrons. Gaining electrons is reduction 还原.
    • At the anode, negative ions lose electrons. Losing electrons is oxidation 氧化.
    • Inside the electrolyte, the ions move: positive ions go to the cathode and negative ions go to the anode.

    You can write a half-equation 半反应式 for each electrode. For molten lead(II) bromide:

    $$\text{Pb}^{2+} + 2e^{-} \rightarrow \text{Pb} \quad (\text{cathode, reduction})$$
    $$2\text{Br}^{-} \rightarrow \text{Br}_2 + 2e^{-} \quad (\text{anode, oxidation})$$

    Electrolysis of copper(II) sulfate

    The product at the anode depends on the electrode:

    • With inert carbon electrodes: copper 铜 forms at the cathode and oxygen forms at the anode. The blue colour of the solution slowly fades as copper is removed.
    • With copper electrodes: copper forms at the cathode, while the anode itself dissolves into the solution. The blue colour stays the same. This is used to purify copper.

    Electroplating

    Electroplating 电镀 means covering a metal object with a thin layer of another metal. This improves its appearance and its resistance to corrosion 腐蚀 (rusting and wearing away).

    To electroplate an object:

    • the object to be coated is made the cathode;
    • the plating metal is made the anode;
    • the electrolyte is a solution containing ions of the plating metal.
    日本語
    Electrolysis: ions discharge at the electrodes

    Electrolysis 电解 is the breaking down (decomposition 分解) of an ionic compound 离子化合物 — when it is molten 熔融 or in aqueous 水溶液 solution — by passing an electric current 电流 through it.

    A water electrolysis unit
    An electrolysis cell splits a compound using an electric current.

    It only works when the substance is molten or dissolved, because then the ions 离子 are free to move and carry the charge. A solid ionic compound cannot be electrolysed because its ions are locked in place.

    A small electrolysis demonstration: two upturned graduated tubes stand in a beaker of water, wired to a 9-volt battery, with gas collected at the top of each tube and about twice as much gas in one tube as the other
    Electrolysis of water: gas bubbles off at each electrode and collects in the tubes above, with about twice as much hydrogen as oxygen

    The electrolytic cell

    The set-up is called an electrolytic cell 电解池. Two electrodes 电极 (solid conductors) dip into the electrolyte 电解质 — the molten or aqueous substance being broken down.

    • The anode 阳极 is the positive ($+$) electrode.
    • The cathode 阴极 is the negative ($-$) electrode.

    The electrodes are often inert 惰性 (they do not react), such as platinum 铂 or carbon/graphite 石墨.

    What is formed at each electrode

    There is a simple rule:

    Metals/hydrogen form at the cathode; non-metals form at the anode
    Metals and hydrogen form at the cathode; non-metals at the anode
    • Metals 金属 or hydrogen 氢气 are formed at the cathode.
    • Non-metals 非金属 (other than hydrogen) are formed at the anode.

    For aqueous solutions, the product can depend on whether the solution is dilute 稀 or concentrated 浓. Here are the three Core examples:

    Electrolyte At the cathode ($-$) At the anode ($+$)
    molten lead(II) bromide, $\text{PbBr}_2$ lead 铅 (silvery liquid) bromine 溴 (red-brown vapour)
    concentrated aqueous sodium chloride hydrogen (bubbles of gas) chlorine 氯气 (pale green gas)
    dilute sulfuric acid 硫酸 hydrogen (bubbles of gas) oxygen 氧气 (bubbles of gas)

    Worked example. Predict the products of electrolysing molten zinc chloride, $\text{ZnCl}_2$. Because it is molten, only zinc ions and chloride ions are present - there is no water to complicate things. Zinc is a metal, so it forms at the cathode: $\text{Zn}^{2+}$ ions gain electrons and silvery zinc appears. Chlorine is a non-metal, so it forms at the anode: pale green chlorine gas bubbles off. Always check the state first. A molten compound simply gives you its own two elements, while an aqueous one brings water into the competition - which is why concentrated sodium chloride solution gives hydrogen at the cathode rather than sodium.

    How the charge moves

    During electrolysis:

    • Electrons 电子 move through the wires (the external circuit 外电路) from the power supply.
    • At the cathode, positive ions gain electrons. Gaining electrons is reduction 还原.
    • At the anode, negative ions lose electrons. Losing electrons is oxidation 氧化.
    • Inside the electrolyte, the ions move: positive ions go to the cathode and negative ions go to the anode.

    You can write a half-equation 半反应式 for each electrode. For molten lead(II) bromide:

    $$\text{Pb}^{2+} + 2e^{-} \rightarrow \text{Pb} \quad (\text{cathode, reduction})$$
    $$2\text{Br}^{-} \rightarrow \text{Br}_2 + 2e^{-} \quad (\text{anode, oxidation})$$
    An electrolysis cell for molten lead bromide: a d.c. supply connected to a cathode and anode dipping into the melt, with lead ions moving to the cathode and bromide ions to the anode
    In molten lead(II) bromide, $\text{Pb}^{2+}$ moves to the cathode and $\text{Br}^-$ to the anode, where each is discharged

    Electrolysis of copper(II) sulfate

    The product at the anode depends on the electrode:

    • With inert carbon electrodes: copper 铜 forms at the cathode and oxygen forms at the anode. The blue colour of the solution slowly fades as copper is removed.
    • With copper electrodes: copper forms at the cathode, while the anode itself dissolves into the solution. The blue colour stays the same. This is used to purify copper.

    Electroplating

    Electroplating 电镀 means covering a metal object with a thin layer of another metal. This improves its appearance and its resistance to corrosion 腐蚀 (rusting and wearing away).

    To electroplate an object:

    • the object to be coated is made the cathode;
    • the plating metal is made the anode;
    • the electrolyte is a solution containing ions of the plating metal.
    An electroplating cell with the object as the cathode and a bar of pure copper as the anode in copper(II) sulfate solution
    To electroplate, make the object the cathode and the plating metal the anode, in a solution of the plating-metal ions
    Explore · ⁨探索⁩

    Inside an electrolysis cell · ⁨電気分解セル内部⁩

    Pick an electrolyte and watch ions move: positive ions to the cathode, negative ions to the anode. · ⁨電解質を選び、イオンが動く様子を見よ:陽イオンは陰極へ、陰イオンは陽極へ向かう。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    electrolysis/ɪlekˈtrɒləsɪs/ 電気分解
    decomposition/ˌdiːkɒmpəˈzɪʃn/ 分解反応
    ionic compound/aɪˈɒnɪk ˈkɒmpaʊnd/ イオン化合物
    molten/ˈməʊltn/ 溶融状態(_molten_)
    aqueous/ˈeɪkwɪəs/ 水溶液(_aqueous_)
    electric current/ɪˈlektrɪk ˈkʌrənt/ 電流
    ions/ˈaɪɒnz/ イオン
    electrolytic cell/ɪˌlektrəˈlɪtɪk sel/ 電解セル
    electrodes/ɪˈlektrəʊdz/ 電極
    electrolyte/ɪˈlektrəlaɪt/ 電気分解液
    anode/ˈænəʊd/ アノード
    cathode/ˈkæθəʊd/ カソード
    inert/ɪˈnɜːt/ 不活性
    platinum/ˈplætɪnəm/ 白金
    graphite/ˈɡræfaɪt/ グラファイト
    metals/ˈmetlz/ 金属
    hydrogen/ˈhaɪdrədʒn/ 水素
    non-metals/nɒn ˈmetlz/ 非金属
    dilute/daɪˈluːt/ 希溶液
    concentrated/ˈkɒnsəntreɪtɪd/ 濃密な
    lead/liːd/ リード
    bromine/ˈbrəʊmaɪn/ 臭素
    chlorine/ˈklɔːriːn/ 塩素
    sulfuric acid/sʌlˈfjʊərɪk ˈæsɪd/ 硫酸
    oxygen/ˈɒksɪdʒn/ 酸素
    electrons/ɪˈlektrɒnz/ 電子
    external circuit/ekˈstɜːnl ˈsɜːkɪt/ 外部回路
    reduction/rɪˈdʌkʃn/ 還元
    oxidation/ˌɒksɪˈdeɪʃn/ 酸化反応
    half-equation/hɑːf ɪˈkweɪʒn/ 半反応式
    copper/ˈkɒpə/ 銅
    electroplating/ɪˌlektrəʊˈpleɪtɪŋ/ 電着
    corrosion/kəˈrəʊʒn/ 腐食
    4.2

    Hydrogen–oxygen fuel cells

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State that a hydrogen–oxygen fuel cell uses hydrogen and oxygen to produce electricity with water as the only chemical product 2 Describe the advantages and disadvantages of using hydrogen–oxygen fuel cells in comparison with gasoline/petrol engines in vehicles
    日本語
    コア サプリメント
    1 水素-酸素燃料電池は、水素と酸素を用いて電力を生み出し、唯一の化学的生成物は水であることを述べる 2 車両用ガソリンエンジンと比較した際の水素-酸素燃料電池の使用上の利点と欠点を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A fuel cell 燃料电池 uses hydrogen and oxygen to make electricity directly. The only chemical product is water.

    $$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$

    It is useful to compare a fuel cell with a normal petrol (gasoline) engine in a vehicle.

    Hydrogen–oxygen fuel cell Petrol engine
    Main product water only carbon dioxide and pollutants 污染物
    Effect on air clean adds to air pollution
    Fuel storage hydrogen is hard and dangerous to store (flammable, needs high pressure) petrol is easy to store
    Source hydrogen may be made using fossil fuels made from crude oil

    Advantages of the fuel cell: the only product is water, so it does not pollute the air, and it changes chemical energy into electricity efficiently. Disadvantages: hydrogen is hard to store and transport safely, and producing the hydrogen can still use energy from fossil fuels.

    日本語

    A fuel cell 燃料电池 uses hydrogen and oxygen to make electricity directly. The only chemical product is water.

    $$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$
    A hydrogen-oxygen fuel cell with hydrogen and oxygen fed to two electrodes, an external circuit running a motor, and water leaving as the only product
    A hydrogen–oxygen fuel cell turns chemical energy straight into electricity, with water as the only product

    It is useful to compare a fuel cell with a normal petrol (gasoline) engine in a vehicle.

    Hydrogen–oxygen fuel cell Petrol engine
    Main product water only carbon dioxide and pollutants 污染物
    Effect on air clean adds to air pollution
    Fuel storage hydrogen is hard and dangerous to store (flammable, needs high pressure) petrol is easy to store
    Source hydrogen may be made using fossil fuels made from crude oil

    Advantages of the fuel cell: the only product is water, so it does not pollute the air, and it changes chemical energy into electricity efficiently. Disadvantages: hydrogen is hard to store and transport safely, and producing the hydrogen can still use energy from fossil fuels.

    Explore · ⁨探索⁩

    Inside a hydrogen–oxygen fuel cell · ⁨水素-酸素燃料電池の内部⁩

    Step through a fuel cell. Hydrogen and oxygen combine to make water, and the energy of that reaction is released as electricity — with no carbon dioxide. · ⁨燃料電池の仕組み。水素と酸素が結合して水となり、この反応エネルギーが電気として放出されます——二酸化炭素は発生しません。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    fuel cell/ˈfjuːəl sel/ 燃料電池
    pollutants/pəˈluːtənts/ 汚染物質
    4.2

    Exam tips

    • At the cathode (negative) you get a metal or hydrogen; at the anode (positive) you get a non-metal. Positive ions move to the cathode, negative ions to the anode.
    • A molten compound just gives its two elements. An aqueous solution can give something different (concentrated sodium chloride gives hydrogen and chlorine, not sodium).
    • Reduction is gain of electrons (at the cathode); oxidation is loss of electrons (at the anode). Remember OIL RIG.
    • In a half-equation the atoms and the charges must balance, with the electrons on the correct side: $\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$ (cathode).
    • To electroplate an object, make it the cathode, make the plating metal the anode, and use a solution containing ions of the plating metal.
  • 5

    Chemical energetics

    Watch lesson · ⁨レッスンを視聴⁩
    5.1

    Exothermic and endothermic reactions · ⁨発熱反応と吸熱反応⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State that an exothermic reaction transfers thermal energy to the surroundings leading to an increase in the temperature of the surroundings 4 State that the transfer of thermal energy during a reaction is called the enthalpy change, $\Delta H$, of the reaction. $\Delta H$ is negative for exothermic reactions and positive for endothermic reactions
    2 State that an endothermic reaction takes in thermal energy from the surroundings leading to a decrease in the temperature of the surroundings
    5 Define activation energy, $E_a$, as the minimum energy that colliding particles must have to react
    3 Interpret reaction pathway diagrams showing exothermic and endothermic reactions 6 Draw and label reaction pathway diagrams for exothermic and endothermic reactions using information provided, to include: (a) reactants (b) products (c) enthalpy change of the reaction, $\Delta H$ (d) activation energy, $E_a$
    7 State that bond breaking is an endothermic process and bond making is an exothermic process and explain the enthalpy change of a reaction in terms of bond breaking and bond making
    8 Calculate the enthalpy change of a reaction using bond energies
    日本語
    コア サプリメント
    1 発熱反応は熱エネルギーを周囲に放出し、周囲の温度を上昇させることを述べる 4 反応中の熱エネルギーの移動を反応エンタルピー変化 $\Delta H$ と呼ぶことを述べる。 $\Delta H$ は発熱反応では負であり、吸熱反応では正である
    2 吸熱反応は周囲から熱エネルギーを取り込み、周囲の温度を低下させることを述べる
    5 衝突粒子が反応するために必要な最小エネルギーである活性化エネルギー $E_a$ を定義する
    3 発熱反応および吸熱反応を示す反応経路図を解釈する 6 与えられた情報を使用して、発熱反応および吸熱反応の反応経路図を描き、以下を含めてラベル付けせよ:(a) 反応物 (b) 生成物 (c) 反応エンタルピー変化 $\Delta H$ (d) 活性化エネルギー $E_a$
    7 結合の切断は吸熱過程であり、結合の形成は発熱過程であることを述べ、反応のエンタルピー変化を結合の切断および形成の観点から説明する
    8 結合エネルギーを用いて反応のエンタルピー変化を計算する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Reaction profile: exothermic energy change

    In every chemical reaction, energy is transferred. The reaction is either exothermic or endothermic, depending on which way the energy moves.

    • An exothermic reaction 放热反应 gives out thermal energy 热能 to the surroundings 环境. So the temperature of the surroundings goes up.
    • An endothermic reaction 吸热反应 takes in thermal energy from the surroundings. So the temperature of the surroundings goes down.

    Examples of exothermic reactions are combustion 燃烧 (burning a fuel) and neutralisation 中和 (an acid reacting with an alkali). An example of an endothermic reaction is the thermal decomposition 分解 of a compound (breaking it down using heat).

    日本語
    反応プロファイル:発熱によるエネルギー変化
    キャンプファイアで薪が燃えている様子
    薪の燃焼は発熱反応であり、周囲へ熱を放出する。

    すべての化学反応において、エネルギーの移動が生じる。エネルギーの向きによって、反応は発熱反応または吸熱反応となる。

    • 発熱反応は熱エネルギーを周囲へ放出するため、周囲の温度は上昇する。
    • 吸熱反応は周囲から熱エネルギーを取り込むため、周囲の温度は低下する。

    発熱反応の例として、燃焼(燃料の燃焼)や中和反応(酸とアルカリの反応)がある。吸熱反応の例として、化合物の熱的分解(熱を用いて分解すること)がある。

    2本の温度計:発熱反応では目盛りが上がり、吸熱反応では下がっている様子
    発熱反応は周囲を温め、吸熱反応は周囲を冷やす
    Explore · ⁨探索⁩

    Exothermic & endothermic · ⁨発熱反応 & 吸熱反応⁩

    ΔH = products − reactants · ⁨ΔH = 生成物 − 反応物⁩

    Exothermic drops to lower-energy products (releases heat); endothermic climbs to higher. · ⁨発熱反応は低エネルギーの生成物へと下がり(熱を放出)、吸熱反応は高エネルギー側へ上昇します。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    exothermic reaction/eɡzəˈðɜːmɪk rɪˈækʃn/ 発熱反応
    thermal energy/ˈθɜːml ˈenədʒi/ 熱エネルギーに変換する
    surroundings/səˈraʊndɪŋz/ 周囲
    endothermic reaction/ˌendəʊˈθɜːmɪk rɪˈækʃn/ 吸熱反応
    combustion/kəmˈbʌstʃn/ 燃焼反応
    neutralisation/ˌnjuːtrəlaɪˈzeɪʃn/ 中和反応
    decomposition/ˌdiːkɒmpəˈzɪʃn/ 分解反応
    enthalpy change/enˈθælpi tʃeɪndʒ/ エンタルピー変化
    activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ 活性化エネルギー
    colliding/kəˈlaɪdɪŋ/ 衝突
    particles/ˈpɑːtɪklz/ 粒子
    reaction pathway diagram/rɪˈækʃn ˈpæθweɪ ˈdaɪəɡræm/ 反応経路図
    reactants/rɪˈæktənts/ 反応物である
    products/ˈprɒdʌkts/ 生成物
    bond breaking/bɒnd ˈbreɪkɪŋ/ 結合切断
    bond making/bɒnd ˈmeɪkɪŋ/ 結合形成
    5.1

    Enthalpy change, ΔH · ⁨エンタルピー変化、ΔH⁩

    English

    The amount of thermal energy transferred in a reaction is called the enthalpy change 焓变. It is written as $\Delta H$ and has these signs:

    • $\Delta H$ is negative for an exothermic reaction, because energy leaves the chemicals.
    • $\Delta H$ is positive for an endothermic reaction, because energy is taken in.
    日本語

    反応における熱エネルギーの移動量をエンタルピー変化と呼ぶ。記号は $\Delta H$ で表し、次のような符号を持つ:

    • 発熱反応では化学物質からエネルギーが放出されるため、$\Delta H$ は負となる。
    • 吸熱反応ではエネルギーが取り込まれるため、$\Delta H$ は正となる。
    バーンバーン炎がエネルギーを供給:エンタルピー変化 ΔH は定圧での熱移動量
    バーンバーン炎がエネルギーを供給:エンタルピー変化 ΔH は定圧での熱移動量
    5.1

    Activation energy · ⁨活性化エネルギー⁩

    English

    Particles do not react every time they meet. The activation energy 活化能 ($E_a$) is the smallest amount of energy that colliding 碰撞 particles 粒子 must have before they can react. It is like a hill the particles must get over before the reaction can happen.

    日本語

    粒子が衝突しても必ず反応するわけではない。活性化エネルギー($E_a$)とは、衝突する粒子が反応するために必要な最小限のエネルギーのことである。これは、反応が起こる前に粒子が越えなければならない「山」のようなものと考えられる。

    Explore · ⁨探索⁩

    Activation energy · ⁨活性化エネルギー⁩

    Ea = the energy barrier · ⁨Ea = 能量障壁(エネルギーバリア)⁩

    Every reaction must climb the activation-energy barrier before products can form. · ⁨すべての反応では、生成物が形成される前に活性化エネルギーの障壁を越えなければなりません。⁩

    5.1

    Reaction pathway diagrams · ⁨反応経路図⁩

    English

    A reaction pathway diagram 反应进程图 shows how the energy changes as a reaction happens. Energy is on the vertical axis, and the progress of the reaction is on the horizontal axis.

    • The line starts at the energy level of the reactants 反应物.
    • It rises over a 'hill' — the height of this hill is the activation energy $E_a$.
    • It then falls or rises to the energy level of the products 生成物.
    • The gap between the reactant level and the product level is the enthalpy change $\Delta H$.

    For an exothermic reaction, the products are lower than the reactants, so energy is given out and $\Delta H$ is negative.

    For an endothermic reaction, the products are higher than the reactants, so energy is taken in and $\Delta H$ is positive.

    日本語

    反応経路図は、反応に伴うエネルギーの変化を示す。縦軸にエネルギー、横軸に反応の進行を表す。

    • 線は反応物のエネルギーレベルから始まる。
    • 「山」の上へ立ち上がり、この山の高さが活性化エネルギー $E_a$ である。
    • その後、生成物のエネルギーレベルまで下がるか、あるいは上がる。
    • 反応物レベルと生成物レベルの差がエンタルピー変化 $\Delta H$ である。

    発熱反応では、生成物のエネルギーレベルは反応物より低くなるため、エネルギーが放出され、$\Delta H$ は負となる。

    吸熱反応では、生成物のエネルギーレベルは反応物より高くなるため、エネルギーが取り込まれ、$\Delta H$ は正となる。

    2つの反応経路図:生成物が反応物より低い発熱反応の図と、生成物が反応物より高い吸熱反応の図
    発熱反応は始点より低点で終わり ($\Delta H<0$);吸熱反応は高点で終わる ($\Delta H>0$)
    Explore · ⁨探索⁩

    Reaction pathway diagrams · ⁨反応経路図⁩

    Drag ΔH and the activation energy. Exothermic releases energy; endothermic takes it in; the hump is the energy barrier. · ⁨ΔHと活性化エネルギーをドラッグせよ。発熱反応はエネルギーを放出し、吸熱反応は取り込む。山のような形がエネルギー障壁である。⁩

    5.1

    Bonds and energy · ⁨結合とエネルギー⁩

    English

    A reaction involves breaking the bonds in the reactants and making new bonds in the products.

    • Bond breaking 断键 takes in energy, so it is an endothermic step.
    • Bond making 成键 gives out energy, so it is an exothermic step.

    The bond energy 键能 is the energy needed to break one mole of a particular bond. You can use bond energies to find the enthalpy change:

    $$\Delta H = (\text{energy to break all bonds}) - (\text{energy released making all bonds})$$

    If more energy is given out making bonds than is taken in breaking bonds, the reaction is exothermic ($\Delta H$ negative). If less energy is given out, it is endothermic ($\Delta H$ positive).

    Worked example

    For the reaction $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$, use these bond energies (in kJ/mol): H–H $= 436$, Cl–Cl $= 242$, H–Cl $= 431$.

    Bonds broken: one H–H and one Cl–Cl $= 436 + 242 = 678$.

    Bonds made: two H–Cl $= 2 \times 431 = 862$.

    $$\Delta H = 678 - 862 = -184 \text{ kJ/mol}$$

    The answer is negative, so this reaction is exothermic.

    日本語
    青い炎を出しているバーンバーンバーナー
    結合の切断と形成によりエネルギーが移動する—ここでは炎がそれを供給している。

    反応は、反応物の結合を切断して、生成物の新しい結合を形成することを含む。

    • 結合の切断はエネルギーを取り込むため、吸熱的な段階である。
    • 結合の形成はエネルギーを放出するため、発熱的な段階である。

    結合エネルギーとは、ある特定の結合1モールを切断するために必要なエネルギーのことである。結合エネルギーを使ってエンタルピー変化を求めることができる:

    $$\Delta H = (\text{energy to break all bonds}) - (\text{energy released making all bonds})$$

    結合の形成時に放出されるエネルギーが、結合の切断時に吸収されるエネルギーよりも多い場合、反応は発熱反応($\Delta H$ 負)となる。逆の場合、吸熱反応($\Delta H$ 正)となる。

    worked example

    反応 $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$ に対して、以下の結合エネルギー(単位:kJ/mol)を用いる:H–H $= 436$、Cl–Cl $= 242$、H–Cl $= 431$。

    切断された結合:H–H が1つ、Cl–Cl が1つ $= 436 + 242 = 678$。

    形成された結合:H–Cl が2つ $= 2 \times 431 = 862$。

    $$\Delta H = 678 - 862 = -184 \text{ kJ/mol}$$

    答えが負であるため、この反応は発熱反応である。

    水素と塩素から塩化水素が生成されるためのエネルギー準位図:結合切断による上昇、その後結合形成によるより大きな下降、最終的に開始点より低い位置で終わる
    結合の切断はエネルギーを取り込み($+678$)、結合の形成はより多く放出する($-862$)ため、$\Delta H = -184$ kJ/mol
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    bond energy/bɒnd ˈenədʒi/ 結合エネルギー
    5.1

    Exam tips · ⁨試験対策⁩

    English
    • Exothermic gives out heat, so the surroundings warm up and $\Delta H$ is negative; endothermic takes in heat, so the surroundings cool and $\Delta H$ is positive.
    • Bond breaking takes energy in (endothermic); bond making gives energy out (exothermic). Use $\Delta H = (\text{energy to break bonds}) - (\text{energy released making bonds})$.
    • If more energy is released making bonds than is used breaking them, the reaction is exothermic. Always check the sign of your final answer.
    • On a reaction pathway diagram, an exothermic reaction ends lower than it starts; an endothermic one ends higher. The height of the hill is the activation energy.
    日本語
    • 発熱反応は熱を放出するため周囲が暖まり、$\Delta H$ は負;吸熱反応は熱を取り込むため周囲が冷え、$\Delta H$ は正となる。
    • 結合の切断はエネルギーを取り込み(吸熱)、結合の形成はエネルギーを放出(発熱)する。$\Delta H = (\text{energy to break bonds}) - (\text{energy released making bonds})$ を用いる。
    • 結合の形成時に放出されるエネルギーが、切断時に消費されるエネルギーよりも多い場合、反応は発熱反応となる。常に最終的な答えの符号を確認しよう。
    • 反応経路図において、発熱反応は開始点より低い位置で終わり、吸熱反応は高い位置で終わる。山のの高さが活性化エネルギーである。
  • 6

    Chemical reactions

    Watch lesson · ⁨レッスンを視聴⁩
    6.1

    Physical and chemical changes · ⁨物理変化と化学変化⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Identify physical and chemical changes, and describe the differences between them
    日本語
    コア サプリメント
    1 物理変化および化学変化を識別し、それらの違いを説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A physical change 物理变化 does not make a new substance. The substance only changes its state or shape, and the change can usually be reversed. Melting ice and dissolving sugar are physical changes.

    A chemical change 化学变化 (a chemical reaction) makes one or more new substances and is usually hard to reverse. Signs of a chemical change include a colour change, a gas being given off, an energy change, or a precipitate 沉淀 (a solid) forming.

    Physical change Chemical change
    no new substance made new substance(s) made
    easy to reverse usually hard to reverse
    e.g. melting, boiling, dissolving e.g. burning, rusting
    日本語
    鉄の表面の錆
    鉄のサビは化学変化であり、新たな物質を生成する。

    物理変化では新たな物質は生成されない。物質の状態や形状のみが変化し、通常は逆転可能である。氷の溶解や砂糖の溶解は物理変化である。

    化学変化(化学反応)は、1つ以上の新しい物質を生成し、通常は逆転させることが困難です。化学変化の兆候には、色の変化、ガスの発生、エネルギーの変化、または沈殿(固体)の生成が含まれます。

    物理変化 化学変化
    新しい物質は生成されない 新しい物質が生成される
    容易に逆転できる 通常は逆転が困難
    例:融解、沸騰、溶解 例:燃焼、錆び
    Explore · ⁨探索⁩

    Physical change vs chemical change · ⁨物理変化 vs 化学変化⁩

    Step through the difference. A physical change makes no new substance and is usually easy to reverse; a chemical change makes a new substance and is hard to undo. · ⁨違いを確認しましょう。物理変化では新しい物質は生成せず、通常容易に元に戻せます;化学変化では新しい物質が生成され、元に戻すのは困難です。⁩

    6.2

    Rate of reaction · ⁨反応速度⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    5 Describe collision theory in terms of: (a) number of particles per unit volume (b) frequency of collisions between particles (c) kinetic energy of particles (d) activation energy, $E_a$
    1 Describe the effect on the rate of reaction of: (a) changing the concentration of solutions (b) changing the pressure of gases (c) changing the surface area of solids (d) changing the temperature (e) adding or removing a catalyst, including enzymes 6 Describe and explain the effect on the rate of reaction of: (a) changing the concentration of solutions (b) changing the pressure of gases (c) changing the surface area of solids (d) changing the temperature (e) adding or removing a catalyst, including enzymes using collision theory
    2 State that a catalyst increases the rate of a reaction and is unchanged at the end of a reaction 7 State that a catalyst decreases the activation energy, $E_a$, of a reaction
    3 Describe practical methods for investigating the rate of a reaction including change in mass of a reactant or a product and the formation of a gas 8 Evaluate practical methods for investigating the rate of a reaction including change in mass of a reactant or a product and the formation of a gas
    4 Interpret data, including graphs, from rate of reaction experiments
    日本語
    コア サプリメント
    5 衝突理論を以下の観点から説明せよ:(a) 単位体積あたりの粒子数 (b) 粒子間の衝突頻度 (c) 粒子の運動エネルギー (d) 活性化エネルギー $E_a$
    1 反応速度に与える影響を記述せよ:(a) 溶液の濃度の変更 (b) 気体の圧力の変更 (c) 固体の表面積の変更 (d) 温度の変更 (e)触媒(酵素を含む)の添加または除去 6 衝突理論を用いて、反応速度に与える影響を説明せよ:(a) 溶液の濃度の変更 (b) 気体の圧力の変更 (c) 固体の表面積の変更 (d) 温度の変更 (e)触媒(酵素を含む)の添加または除去
    2 触媒は反応の速度を増加させ、反応の終了後には変化しないことを述べよ 7 触媒は反応の活性化エネルギー $E_a$ を減少させることを述べよ
    3 反応速度を調べるための実験手法を記述せよ。反応物または生成物の質量の変化および気体の生成を含む 8 反応速度を調べるための実験手法を評価せよ。反応物または生成物の質量の変化および気体の生成を含む
    4 反応速度の実験から得られたデータ(グラフを含む)を解釈せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Collision theory: energy and orientation

    The rate of reaction 反应速率 tells you how fast the reactants 反应物 change into products 生成物.

    Collision theory

    Collision theory 碰撞理论 explains what controls the rate. For a reaction to happen, the particles 粒子 must collide 碰撞, and they must collide with enough energy. The least energy they need is the activation energy 活化能 ($E_a$). A faster rate happens when the particles have successful collisions more often.

    What changes the rate

    Change Effect Reason (collision theory)
    increase concentration 浓度 of a solution faster more particles in the same volume, so collisions happen more often
    increase pressure 压强 of gases faster particles are pushed closer, so collisions happen more often
    increase surface area 表面积 of a solid faster more particles are exposed, so collisions happen more often
    increase temperature faster particles gain kinetic energy 动能 and move faster, so collisions are more frequent and more of them have enough energy
    add a catalyst 催化剂 faster the catalyst lowers the activation energy, so more collisions are successful

    Catalysts

    A catalyst speeds up a reaction but is not used up — it is unchanged at the end. It works by lowering the activation energy. Enzymes 酶 are biological catalysts (catalysts that work in living things).

    Measuring the rate

    You can follow a reaction over time in these ways:

    • Change in mass: stand the flask on a balance. If a gas escapes, the mass falls. Record the mass at regular times.
    • Volume of gas: collect the gas in a gas syringe 注射器 and read its volume at regular times.
    • Formation of a precipitate: in a reaction that turns cloudy, time how long it takes for a mark under the flask to disappear.

    On a graph of product against time, the line is steepest at the start (fastest rate), becomes less steep as reactants are used up, and goes flat when the reaction has finished.

    日本語
    衝突理論:エネルギーと配向性

    反応速度は、反応物から生成物への変化がどの速さで起こるかを示します。

    衝突理論

    衝突理論は、反応速度を支配する要因を説明します。反応が起こるためには、粒子同士が衝突し、かつ十分なエネルギーを持って衝突する必要があります。必要な最小限のエネルギーを活性エネルギー($E_a$)といいます。粒子の成功する衝突回数が増えると、反応速度は速くなります。

    反応速度を変える要因

    変化 効果 理由(衝突理論)
    溶液の濃度を上げる 速くなる 同じ体積内の粒子数が増えるため、衝突の機会が増える
    気体の圧力を上げる 速くなる 粒子が押し寄せ合うため、衝突の機会が増える
    固体の表面積を broaden 速くなる 露出した粒子数が増えるため、衝突の機会が増える
    温度を上げる 速くなる 粒子は運動エネルギーを得て速く動くため、衝突の頻度が高まり、かつ十分のエネルギーを持つ衝突が増える
    触媒を加える 速くなる 触媒は活性エネルギーを下げるため、成功する衝突の割合が増える
    2つの粒子箱:高濃度の箱には粒子が多く含まれているため、衝突の機会が多い
    同じ体積内の粒子数が増えるため衝突の機会が多くなり、反応速度は速くなる
    酸に触れた大塊の固体と、粉末状にして酸が全体に触れる同一の固体
    固体を粉末にすると表面積が大きく増えるため、反応速度は速くなる

    触媒

    触媒は反応を加速しますが、消費されません。反応の最終的に変化することはありません。活性エネルギーを下げることで機能します。酵素は生物学的な触媒(生体内で機能する触媒)です。

    触媒なしでは高い山、触媒ありでは低い山の反応経路、どちらも同じ生成物で終わる
    触媒は活性エネルギーを下げるため成功する衝突が増え、$\Delta H$は変わらない

    反応速度の測定

    反応を時間経過とともに追跡する方法として、以下のようなものがあります:

    • 質量の変化:フラスコを天秤に乗せます。ガスが逸散すれば質量は減少します。一定時間ごとに質量を記録します。
    • ガスの体積:ガスをガスシリンジに集め、一定時間ごとにその体積を読みます。
    • 沈殿の生成:濁りを伴う反応において、フラスコの下の印が消えるまでの時間を測ります。
    反応混合物が入った三角フラスコが、ガスで満たされていくガスシリンジへ導管で接続されている
    ガスシリンジは発生したガスを収集し、一定時間ごとに体積を読むことで反応速度を追跡する

    生成物を横軸に時間をとったグラフにおいて、曲線は始点で最も急(反応速度が最大)になり、反応物が消費されるにつれて緩やかになり、反応が終わると水平になります。

    気体量と時間のグラフ:最初は急勾配でその後水平になり、速い曲線と遅い曲線が同じ最終値に達する
    反応速度は始点で最も速く(最も急)、反応が終わると曲線は水平になる
    Explore · ⁨探索⁩

    Rate from a gas-volume graph · ⁨気体体積グラフからの速度⁩

    The gradient of the volume–time curve is the rate; it is steepest at the start and flattens as reactants run out. · ⁨体積-時間曲線の傾きは速度であり、始点で最も急で、試薬が尽きるにつれて緩やかになる。⁩

    Explore · ⁨探索⁩

    Rate of reaction · ⁨反応速度⁩

    conc = c₀·bᵗ

    Concentration falls over time — fast at first, then slower. · ⁨濃度は時間とともに 低下 する — 最初は速く、後に遅くなる。⁩

    Explore · ⁨探索⁩

    Collision theory in action · ⁨衝突理論の実践⁩

    Watch the particles collide. Heat speeds them up so they collide more often and harder; more particles or a catalyst give more successful collisions — the rate climbs. · ⁨粒子が衝突する様子を見る。熱は粒子を加速させ、衝突の頻度と強さを増します;より多くの粒子や触媒はより成功した衝突をもたらす — 速度は上昇する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    physical change/ˈfɪzɪkl tʃeɪndʒ/ 物理変化
    chemical change/ˈkemɪkl tʃeɪndʒ/ 化学変化
    precipitate/prɪˈsɪpɪteɪt/ 沈殿物
    rate of reaction/reɪt ɒv rɪˈækʃn/ 反応速度
    reactants/rɪˈæktənts/ 反応物である
    products/ˈprɒdʌkts/ 生成物
    collision theory/kəˈlɪʒn ˈθɪəri/ 衝突理論
    particles/ˈpɑːtɪklz/ 粒子
    collide/kəˈlaɪd/ 衝突
    activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ 活性化エネルギー
    concentration/ˌkɒnsənˈtreɪʃn/ 濃度
    pressure/ˈpreʃə/ 圧力
    surface area/ˈsɜːfɪs ˈeərɪə/ 表面積
    kinetic energy/kɪˈnetɪk ˈenədʒi/ 運動エネルギー
    catalyst/ˈkætəlɪst/ 触媒
    enzymes/ˈenzaɪmz/ 酵素
    syringe/sɪˈrɪndʒ/ 注射器
    reversible reaction/rɪˈvɜːsɪbl rɪˈækʃn/ 可逆反応
    hydrated/haɪˈdreɪtɪd/ 水和物(_attrated_)
    anhydrous/ænˈhaɪdrəs/ 無水物(_anhydrous_)
    closed system/kləʊzd ˈsɪstəm/ 閉じた系
    Watch lesson · ⁨レッスンを視聴⁩
    6.3

    Reversible reactions and equilibrium · ⁨可逆反応と平衡⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State that some chemical reactions are reversible as shown by the symbol $\rightleftharpoons$ 3 State that a reversible reaction in a closed system is at equilibrium when: (a) the rate of the forward reaction is equal to the rate of the reverse reaction (b) the concentrations of reactants and products are no longer changing
    2 Describe how changing the conditions can change the direction of a reversible reaction for: (a) the effect of heat on hydrated compounds (b) the addition of water to anhydrous compounds limited to copper(II) sulfate and cobalt(II) chloride 4 Predict and explain, for a reversible reaction, how the position of equilibrium is affected by: (a) changing temperature (b) changing pressure (c) changing concentration (d) using a catalyst using information provided
    5 State the symbol equation for the production of ammonia in the Haber process, $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$
    6 State the sources of the hydrogen (methane) and nitrogen (air) in the Haber process
    7 State the typical conditions in the Haber process as $450\text{ }^{\circ}\text{C}$, $20\,000\text{ kPa}/200\text{ atm}$ and an iron catalyst
    8 State the symbol equation for the conversion of sulfur dioxide to sulfur trioxide in the Contact process, $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$
    9 State the sources of the sulfur dioxide (burning sulfur or roasting sulfide ores) and oxygen (air) in the Contact process
    10 State the typical conditions for the conversion of sulfur dioxide to sulfur trioxide in the Contact process as $450\text{ }^{\circ}\text{C}$, $200\text{ kPa}/2\text{ atm}$ and a vanadium(V) oxide catalyst
    11 Explain, in terms of rate of reaction and position of equilibrium, why the typical conditions stated are used in the Haber process and in the Contact process, including safety considerations and economics
    日本語
    コア サプリメント
    1 化学反応の一部是可逆であること、その表現に使用される記号 $\rightleftharpoons$ を述べよ 3 密閉系における可逆反応が平衡状態にあるとき、以下の条件が満たされることを述べよ:(a) 正反応の速度が逆反応の速度と等しい (b) 反応物および生成物の濃度が一定となる
    2 可逆反応の進行方向が条件の変化によってどのように変化するかを記述せよ:(a) 熱による水素化物の影響 (b) 無水化合物への水の添加(銅(II)硫酸塩およびコバルト(II)塩化物のみ) 4 提供された情報を用いて、可逆反応において平衡位置が以下のように影響を受けることを予測し説明せよ:(a) 温度の変更 (b) 圧力の変更 (c) 濃度の変更 (d) 触媒の使用
    5 ハーバー法によるアンモニア合成の記号方程式 $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ を述べよ
    6 ハーバー法における水素(メタン)および窒素(空気)の原料を述べよ
    7 ハーバー法の一般的な反応条件として $450\text{ }^{\circ}\text{C}$、$20\,000\text{ kPa}/200\text{ atm}$ および鉄触媒を述べよ
    8 コンタクト法による二酸化硫黄から三酸化硫黄への変換の記号方程式 $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$ を述べよ
    9 コンタクト法における二酸化硫黄(硫黄の燃焼または硫化鉱石の焙煎)および酸素(空気)の原料を述べよ
    10 コンタクト法における二酸化硫黄から三酸化硫黄への変換のための一般的な反応条件として $450\text{ }^{\circ}\text{C}$、$200\text{ kPa}/2\text{ atm}$ および五酸化二バナジウム触媒を述べよ
    11 ハーバー法およびコンタクト法で用いられる一般的な反応条件が採用される理由を、反応速度と平衡位置の観点から説明せよ。安全性および経済性の考慮点を含む

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Le Chatelier's principle
    Dynamic equilibrium: the rates converge

    Some reactions are reversible 可逆反应: the products can react to form the reactants again. We show this with the symbol $\rightleftharpoons$.

    Changing the direction

    A clear example uses hydrated 水合 and anhydrous 无水 compounds:

    • Blue hydrated copper(II) sulfate, when heated, loses its water to become white anhydrous copper(II) sulfate. Adding water turns it blue again.
    • Pink hydrated cobalt(II) chloride loses water when heated to become blue anhydrous cobalt(II) chloride. Adding water turns it pink again.

    Adding water to the anhydrous solid (and seeing the colour return) is used as a test for water.

    Equilibrium

    In a closed system 密闭系统 (where nothing enters or leaves), a reversible reaction reaches equilibrium 平衡 when:

    • the rate of the forward reaction 正反应 equals the rate of the reverse reaction 逆反应, and
    • the concentrations of the reactants and products are no longer changing.

    Changing the position of equilibrium

    The position of equilibrium 平衡位置 tells you whether there are more reactants or more products. You can move it:

    • Temperature: heating moves the equilibrium in the direction that takes in heat (the endothermic 吸热反应 direction); cooling moves it in the direction that gives out heat (the exothermic 放热反应 direction).
    • Pressure (gases): more pressure moves the equilibrium to the side with fewer gas molecules.
    • Concentration: adding more of a substance moves the equilibrium to the other side, to use it up.
    • A catalyst does not move the position of equilibrium. It only helps the reaction reach equilibrium faster.

    The Haber process

    The Haber process 哈伯法 makes ammonia:

    $$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$
    • The hydrogen 氢气 comes from methane 甲烷 (natural gas); the nitrogen 氮气 comes from the air.
    • Typical conditions: a temperature of $450\,{}^{\circ}\text{C}$, a pressure of about $200$ atm ($20\,000$ kPa), and an iron 铁 catalyst.

    The Contact process

    The Contact process 接触法 turns sulfur dioxide into sulfur trioxide:

    $$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$$
    • The sulfur dioxide comes from burning sulfur 硫 (or roasting sulfide ores); the oxygen 氧气 comes from the air.
    • Typical conditions: a temperature of $450\,{}^{\circ}\text{C}$, a pressure of about $2$ atm ($200$ kPa), and a vanadium(V) oxide 五氧化二钒 catalyst.

    Why these conditions are chosen

    The conditions are a compromise 折中:

    • A higher pressure would give more product, but very high pressure is dangerous and expensive, so a medium pressure is used.
    • A lower temperature would give more product (both forward reactions are exothermic), but the reaction would be too slow, so a fairly high temperature is used to keep a good rate.
    • The catalyst speeds up the reaction without changing the position of equilibrium, which lowers cost.
    日本語
    ルシャトリエの原理
    動態平衡:速度が収束する

    ある反応は可逆的です。生成物が再び反応して反応物に戻ることができます。これを記号 $\rightleftharpoons$ で表します。

    青色の水合硫酸銅(II)は加熱により脱水して無水白色となり、水を加えると再び青色に戻る
    青色の硫酸銅に熱を加えて水を蒸発させ、水を加えると再び青色に戻る

    方向の変更

    明確な例として、水合化合物と無水化合物が挙げられます:

    • 青色の水合硫酸銅(II)を加熱すると水分を失って無水白色の硫酸銅(II)になります。水を加えると再び青色に戻ります。
    • 粉色の水合塩化コバルト(II)を加熱すると水分を失って無水青色の塩化コバルト(II)になります。水を加えると再び粉色に戻ります。

    無水固体に水を加えて色が変わることを確認することは、水の検出法として用いられます。

    均衡点

    外部との物質交換がない閉鎖系において、可逆反応は次の条件が満たされたときに平衡状態になります:

    • 正反応の速度が逆反応の速度と等しくなり、かつ
    • 反応物および生成物の濃度が変化しなくなります。

    平衡位置の変更

    平衡位置は、反応物と生成物のどちらが多いかを示します。これを以下のように変更できます:

    • 温度:加熱すると熱を吸収する方向(吸熱側)へ平衡が移動し、冷却すると熱を放出する方向(発熱側)へ平衡が移動します。
    • 圧力(気体):圧力を上げると、気体分子数が少ない側へ平衡が移動します。
    • 濃度:ある物質を増やすと、それを消費するために平衡は反対側へ移動します。
    • 触媒は平衡位置を変えません。反応が平衡に達するのを早めるだけですが。

    ハーバー法

    ハーバー法はアンモニアを製造します:

    $$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$
    • 水素はメタン(天然ガス)由来であり、窒素は空気由来です。
    • 一般的な条件:温度 $450\,{}^{\circ}\text{C}$、圧力約 $200$ atm($20\,000$ kPa)、および鉄触媒を使用します。

    コンタクト法

    コンタクト法は、二酸化硫黄を三酸化硫黄に変換します:

    $$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$$
    • 二酸化硫黄の原料は硫黄(または硫化鉱石の焙焼)で、酸素は空気から得られます。
    • 一般的な条件:温度 $450\,{}^{\circ}\text{C}$、圧力約 $2$ atm($200$ kPa)、および**五酸化バナジウム(V)**触媒。

    なぜこれらの条件が採用されるか

    これらの条件は妥協点です:

    • 圧力を上げると生成物が増えますが、非常に高い圧力は危険で高価なため、中程度の圧力が用いられます。
    • 温度を下げると生成物が増えます(正反応は発熱反応であるため)、反応速度が速すぎるため、良好な反応速度を維持するために比較的高い温度が用いられます。
    • 触媒は平衡の位置を変えずに反応を速くし、コストを削減します。
    Explore · ⁨探索⁩

    Shifting an equilibrium · ⁨平衡の移動⁩

    Change temperature, pressure or concentration and watch a reversible reaction shift to a new balance. · ⁨温度、圧力、または濃度を変化させ、可逆反応が新しい平衡へシフトする様子を見よ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    equilibrium/ˌiːkwɪˈlɪbrɪəm/ 平衡
    forward reaction/ˈfɔːwəd rɪˈækʃn/ 順反応
    reverse reaction/rɪˈvɜːs rɪˈækʃn/ 逆反応
    position of equilibrium/pəˈzɪʃn ɒv ˌiːkwɪˈlɪbrɪəm/ 平衡の位置
    endothermic/ˌendəʊˈθɜːmɪk/ 吸熱反応
    exothermic/eɡzəˈðɜːmɪk/ 発熱反応
    Haber process/ˈheɪbə ˈprəʊses/ ハーバー法
    hydrogen/ˈhaɪdrədʒn/ 水素
    methane/ˈmiːθeɪn/ メタン
    nitrogen/ˈnaɪtrədʒn/ 窒素
    iron/ˈaɪən/ 鉄
    Contact process/ˈkɒntækt ˈprəʊses/ コンタクト法
    sulfur/ˈsʌlfɜː/ 硫黄
    oxygen/ˈɒksɪdʒn/ 酸素
    vanadium(V) oxide/væˈneɪdɪəm ˈɒksaɪd/ 五酸化バナジウム
    compromise/ˈkɒmprəmaɪz/ 妥協
    Watch lesson · ⁨レッスンを視聴⁩
    6.4

    Redox · ⁨酸化還元⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Use a Roman numeral to indicate the oxidation number of an element in a compound
    2 Define redox reactions as involving simultaneous oxidation and reduction
    3 Define oxidation as gain of oxygen and reduction as loss of oxygen 6 Define oxidation in terms of: (a) loss of electrons (b) an increase in oxidation number
    7 Define reduction in terms of: (a) gain of electrons (b) a decrease in oxidation number
    4 Identify redox reactions as reactions involving gain and loss of oxygen 8 Identify redox reactions as reactions involving gain and loss of electrons
    5 Identify oxidation and reduction in redox reactions 9 Identify redox reactions by changes in oxidation number using: (a) the oxidation number of elements in their uncombined state is zero (b) the oxidation number of a monatomic ion is the same as the charge on the ion (c) the sum of the oxidation numbers in a compound is zero (d) the sum of the oxidation numbers in an ion is equal to the charge on the ion
    10 Identify redox reactions by the colour changes involved when using acidified aqueous potassium manganate(VII) or aqueous potassium iodide
    11 Define an oxidising agent as a substance that oxidises another substance and is itself reduced
    12 Define a reducing agent as a substance that reduces another substance and is itself oxidised
    13 Identify oxidising agents and reducing agents in redox reactions
    日本語
    コア サプリメント
    1 化合物中の元素の酸化数を示すためにローマ数字を使用せよ
    2 酸化還元反応が酸化的反応と還元的反応が同時に起こる反応であることを定義せよ
    3 酸化を酸素の付加、還元を酸素の脱離として定義せよ 6 酸化を以下の観点から定義せよ:(a) 電子の放出 (b) 酸化数の増加
    7 還元を以下の観点から定義せよ:(a) 電子の取得 (b) 酸化数の減少
    4 酸化還元反応が酸素の付加および脱離を伴う反応であることを特定せよ 8 酸化還元反応が電子の付加および脱離を伴う反応であることを特定せよ
    5 酸化還元反応における酸化および還元の特定 9 酸化数の変化を用いて酸化還元反応を特定せよ:(a) 単体状態の元素の酸化数はゼロである (b) 単原子イオンの酸化数はイオンの電荷と同じである (c) 化合物内の酸化数の総和はゼロである (d) イオン内の酸化数の総和はイオンの電荷に等しい
    10 酸性化aqueousカリウムマンガン酸(VII)またはaqueousカリウムヨウ化物を用いた際の色変化に基づき、酸化還元反応を特定せよ
    11 酸化剤を、他の物質を酸化自身は還元される物質として定義せよ
    12 還元剤を、他の物質を還元自身は酸化される物質として定義せよ
    13 酸化還元反応における酸化剤および還元剤の特定

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Oxidation 氧化 and reduction 还原 always happen at the same time, in what is called a redox reaction 氧化还原反应. ('Redox' is short for reduction–oxidation.)

    There are three ways to describe oxidation and reduction:

    • Oxygen: oxidation is the gain of oxygen; reduction is the loss of oxygen.
    • Electrons 电子: oxidation is the loss of electrons; reduction is the gain of electrons.
    • Oxidation number 氧化数: in oxidation the oxidation number goes up; in reduction it goes down.

    A useful memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain — of electrons.

    Oxidation number rules

    The oxidation number is shown by a Roman numeral, as in iron(II) and iron(III). The rules are:

    • An element that is not combined has an oxidation number of $0$.
    • A single-atom ion has an oxidation number equal to its charge (so $\text{Na}^{+}$ is $+1$).
    • The oxidation numbers in a compound add up to $0$.
    • The oxidation numbers in an ion add up to the charge on the ion.

    Worked example. Find the oxidation number of manganese in the manganate(VII) ion, $\text{MnO}_4^{-}$. Each oxygen is $-2$, and there are four of them, giving $4 \times (-2) = -8$. The oxidation numbers in an ion must add up to the charge on the ion, which here is $-1$. So if manganese is $x$, then $x + (-8) = -1$, giving $x = +7$ - exactly what the (VII) in the name tells you. Set the total to the ion's charge, not to zero: zero is only for a neutral compound.

    Oxidising and reducing agents

    • An oxidising agent 氧化剂 oxidises another substance, and is itself reduced.
    • A reducing agent 还原剂 reduces another substance, and is itself oxidised.

    Some redox reactions show clear colour changes:

    • Acidified potassium manganate(VII) 高锰酸钾 is purple. When it acts as an oxidising agent it is reduced, and the purple colour fades to colourless.
    • Potassium iodide 碘化钾 is colourless. When it is oxidised, red-brown iodine 碘 is formed.
    日本語

    酸化と還元は常に同時に起こり、これを酸化還元反応と呼びます。(「Redox」は reduction–oxidation の略語です。)

    酸化と還元を説明する方法には3つあります:

    • 酸素: 酸化は酸素の獲得、還元は酸素の喪失です。
    • 電子: 酸化は電子の喪失、還元は電子の獲得です。
    • 酸化数: 酸化では酸化数が上昇し、還元では低下します。

    便利な暗記法としてOIL RIGがあります:Oxidation Is Loss, Reduction Is Gain — 電子についてです。

    一つの物質が他の物質へ電子を渡し、前者が酸化され後者が還元される
    電子は酸化された物質(喪失する側)から還元された物質(獲得する側)へ移ります — これらは常にセットで起こります

    酸化数のルール

    酸化数はローマ数字で示され、例えば鉄(II)や鉄(III)のように表されます。ルールは以下の通りです:

    • 単体(元素)の酸化数は $0$ です。
    • 単原子イオンの酸化数はその電荷と等しい(つまり $\text{Na}^{+}$ は $+1$ となる)。
    • 化合物内の酸化数の総和は $0$ です。
    • イオン内の酸化数の総和はそのイオンの電荷と等しい。

    計算例。 マンガン酸(VII)イオン $\text{MnO}_4^{-}$ 中のマンガンの酸化数を求めよ。各酸素は $-2$ で、4つあるため $4 \times (-2) = -8$ となる。イオン内の酸化数の総和はイオンの電荷と等しく、ここでは $-1$ である。したがって、マンガンが $x$ なら、$x + (-8) = -1$ となり、$x = +7$ となる — これは名前の (VII) が示す値と一致する。合計はゼロではなくイオンの電荷に設定すること:ゼロは中性化合物の場合のみである。

    酸化剤と還元剤

    • 酸化剤は他の物質を酸化させ、自身は還元される。
    • 還元剤は他の物質を還元させ、自身は酸化される。

    いくつかの酸化還元反応では明確な色の変化が見られます:

    • 酸性化したマンガン酸(VII)カリウムは紫色である。酸化剤として作用すると還元され、紫色是无色になる。
    • ヨウ化カリウムは無色である。酸化されると赤褐色のヨウ素が生成される。
    右側が濃い紫色から左側へ向かって希釈されるにつれ無色になるマンガン酸(VII)カリウムの試験管ラック
    マンガン酸(VII)カリウムは濃い紫色ですが、還元(または希釈)されると紫色が無色になります
    Explore · ⁨探索⁩

    Oxidation and reduction happen together · ⁨酸化と還元は同時に起こる⁩

    Step through a redox reaction. One substance loses electrons while another gains them — you can never have one without the other. · ⁨酸化還元反応をステップごとに確認する。一方の物質が電子を失う間にもう一方が得る — どちらか片方だけが存在することは never ない。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    oxidation/ˌɒksɪˈdeɪʃn/ 酸化反応
    reduction/rɪˈdʌkʃn/ 還元
    redox reaction/rɪˈdɒks rɪˈækʃn/ 酸化還元反応
    potassium manganate(VII)/pəˈtæsɪəm ˈmæŋɡəneɪt/ 高マンガン酸カリウム
    potassium iodide/pəˈtæsɪəm ˈaɪədaɪd/ ヨウ化カリウム
    iodine/ˈaɪədiːn/ ヨウ素
    6.4

    Exam tips · ⁨試験対策⁩

    English
    • Explain a faster rate with collision theory: successful collisions happen more often. Concentration, pressure and surface area all raise how often particles collide; a higher temperature does that and gives more particles enough energy.
    • A catalyst lowers the activation energy and speeds the reaction up, but is not used up and does not change $\Delta H$ or the position of equilibrium.
    • At equilibrium the forward and reverse rates are equal and the concentrations stop changing — it does not mean the amounts of reactant and product are equal.
    • Use Le Chatelier's principle: raising temperature favours the endothermic direction; raising pressure favours the side with fewer gas molecules; adding a substance shifts the equilibrium away from it.
    • Redox — OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons. They always happen together, and an oxidising agent is itself reduced.
    日本語
    • 衝突理論で反応速度の増加を説明する:有効衝突の頻度が増える。濃度、圧力、表面積は粒子間の衝突頻度を上げ、高温はそれに加え、十分なエネルギーを持つ粒子の割合も増やす。
    • 触媒は活性化エネルギーを下げて反応を速くするが、消費されず、$\Delta H$ や平衡の位置を変えない。
    • 平衡状態では正反応と逆反応の速度が等しくなり、濃度が変化しなくなる — 反応物と生成物の量が等しいことを意味しない。
    • ル・シャトリエの原理を使用する:温度を上げると吸熱方向が優位になる、圧力を上げると気体分子数が少ない側が優位になる、物質を追加すると平衡はその物質から遠ざかる方向に移動する。
    • 酸化還元 — OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons. 常にセットで起こり、酸化剤自身は還元される。
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    electrons/ɪˈlektrɒnz/ 電子
    oxidation number/ˌɒksɪˈdeɪʃn ˈnʌmbə/ 酸化数
    oxidising agent/ˈɒksɪdaɪzɪŋ ˈeɪdʒənt/ 酸化剤
    reducing agent/rɪˈdjuːsɪŋ ˈeɪdʒənt/ 還元剤
  • 7

    Acids, bases and salts

    Watch lesson · ⁨レッスンを視聴⁩
    7.1

    Acids

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the characteristic properties of acids in terms of their reactions with: (a) metals (b) bases (c) carbonates
    2 Describe acids in terms of their effect on: (a) litmus (b) thymolphthalein (c) methyl orange
    3 State that bases are oxides or hydroxides of metals and that alkalis are soluble bases
    4 Describe the characteristic properties of bases in terms of their reactions with: (a) acids (b) ammonium salts
    5 Describe alkalis in terms of their effect on: (a) litmus (b) thymolphthalein (c) methyl orange
    6 State that aqueous solutions of acids contain $\text{H}^+$ ions and aqueous solutions of alkalis contain $\text{OH}^-$ ions 9 Define acids as proton donors and bases as proton acceptors
    10 Define a strong acid as an acid that is completely dissociated in aqueous solution and a weak acid as an acid that is partially dissociated in aqueous solution
    11 State that hydrochloric acid is a strong acid, as shown by the symbol equation, $\text{HCl}(\text{aq}) \rightarrow \text{H}^+(\text{aq}) + \text{Cl}^-(\text{aq})$
    12 State that ethanoic acid is a weak acid, as shown by the symbol equation, $\text{CH}_3\text{COOH}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{CH}_3\text{COO}^-(\text{aq})$
    7 Describe how to compare hydrogen ion concentration, neutrality, relative acidity and relative alkalinity in terms of colour and pH using universal indicator paper
    8 Describe the neutralisation reaction between an acid and an alkali to produce water, $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(l)$
    日本語
    コア サプリメント
    1 酸の特性を、以下の物質との反応の観点から記述せよ:(a) 金属 (b) 塩基 (c) カルボネート
    2 酸を以下の試薬に対する作用の観点から記述せよ:(a) リトマス (b) トルフェンファレイン (c) メチルオレンジ
    3 塩基は金属の酸化物または水酸化物であり、アルカリは可溶性塩基であることを述べよ
    4 塩基の特性を、以下の物質との反応の観点から記述せよ:(a) 酸 (b) アンモニウム塩
    5 アルカリを以下の試薬に対する作用の観点から記述せよ:(a) リトマス (b) トルフェンファレイン (c) メチルオレンジ
    6 酸のaqueous溶液は $\text{H}^+$ イオンを含み、アルカリのaqueous溶液は $\text{OH}^-$ イオンを含んでいることを述べよ 9 酸をプロトン供与体、塩基をプロトン受容体として定義せよ
    10 強酸をaqueous溶液中で完全に解離する酸、弱酸をaqueous溶液中で部分的に解離する酸として定義せよ
    11 塩酸は強酸であることを述べよ。記号方程式 $\text{HCl}(\text{aq}) \rightarrow \text{H}^+(\text{aq}) + \text{Cl}^-(\text{aq})$ で示される
    12 シュウ酸が弱酸であることを記号方程式 $\text{CH}_3\text{COOH}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{CH}_3\text{COO}^-(\text{aq})$ で示されることから述べる
    7 万能指示薬紙を用いて、水素イオン濃度、中性、相対的な酸性・アルカリ性を色とpHの観点から比較する方法を説明せよ。
    8 酸と塩基との中和反応について、水と $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(l)$ が生成されることを説明せよ。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    An acid 酸 is a substance that forms hydrogen ions ($\text{H}^{+}$) when dissolved in water. Acids have three typical reactions:

    • with metals 金属: acid + metal → a salt 盐 + hydrogen 氢气
    • with bases: acid + base 碱 → a salt + water (this is neutralisation 中和)
    • with carbonates 碳酸盐: acid + carbonate → a salt + water + carbon dioxide 二氧化碳

    Describing what you see

    Naming the products is only half of what the exam wants. "State the observations" asks what you would see with your eyes, and it carries its own marks - so learn the words:

    You see Say
    bubbles of gas coming off effervescence 泡腾 (also accepted: fizzing, bubbling)
    the solid getting smaller and vanishing the solid dissolves / disappears
    the colour of the solution changing name the colour, e.g. turns blue (copper salts)
    bubbling that stops no more effervescence - the acid is used up

    So for magnesium added to hydrochloric acid, a full answer is: effervescence, and the solid dissolves (and the tube gets warm). For copper(II) oxide added to warm sulfuric acid: the black solid dissolves and the solution turns blue.

    Two rules save most of the lost marks. Never write "hydrogen is given off" as an observation - you cannot see hydrogen, you can only see the bubbles, so write effervescence. And do not write a name you cannot see: "a salt forms" is a conclusion, not an observation.

    Indicators

    An indicator 指示剂 is a dye that changes colour to show whether a solution is acidic or alkaline.

    Indicator In acid In alkali
    litmus 石蕊 red blue
    thymolphthalein 百里酚酞 colourless blue
    methyl orange 甲基橙 red yellow

    Aqueous solutions of acids contain hydrogen ions 离子 ($\text{H}^{+}$). Aqueous solutions of alkalis contain hydroxide ions ($\text{OH}^{-}$).

    日本語

    An acid 酸 is a substance that forms hydrogen ions ($\text{H}^{+}$) when dissolved in water. Acids have three typical reactions:

    An acid reacts with a metal, a base and a carbonate to give different products
    Acids react with metals, bases and carbonates to give different products
    • with metals 金属: acid + metal → a salt 盐 + hydrogen 氢气
    • with bases: acid + base 碱 → a salt + water (this is neutralisation 中和)
    • with carbonates 碳酸盐: acid + carbonate → a salt + water + carbon dioxide 二氧化碳

    Describing what you see

    Naming the products is only half of what the exam wants. "State the observations" asks what you would see with your eyes, and it carries its own marks - so learn the words:

    You see Say
    bubbles of gas coming off effervescence 泡腾 (also accepted: fizzing, bubbling)
    the solid getting smaller and vanishing the solid dissolves / disappears
    the colour of the solution changing name the colour, e.g. turns blue (copper salts)
    bubbling that stops no more effervescence - the acid is used up

    So for magnesium added to hydrochloric acid, a full answer is: effervescence, and the solid dissolves (and the tube gets warm). For copper(II) oxide added to warm sulfuric acid: the black solid dissolves and the solution turns blue.

    Two rules save most of the lost marks. Never write "hydrogen is given off" as an observation - you cannot see hydrogen, you can only see the bubbles, so write effervescence. And do not write a name you cannot see: "a salt forms" is a conclusion, not an observation.

    Indicators

    An indicator 指示剂 is a dye that changes colour to show whether a solution is acidic or alkaline.

    Indicator In acid In alkali
    litmus 石蕊 red blue
    thymolphthalein 百里酚酞 colourless blue
    methyl orange 甲基橙 red yellow

    Aqueous solutions of acids contain hydrogen ions 离子 ($\text{H}^{+}$). Aqueous solutions of alkalis contain hydroxide ions ($\text{OH}^{-}$).

    Blue litmus paper strips and a booklet of red litmus paper on a white background
    Litmus paper: blue litmus turns red in an acid, and red litmus turns blue in an alkali
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    acid/ˈæsɪd/ 酸
    metals/ˈmetlz/ 金属
    salt/sɒlt/ 塩
    hydrogen/ˈhaɪdrədʒn/ 水素
    base/beɪs/ 底
    neutralisation/ˌnjuːtrəlaɪˈzeɪʃn/ 中和反応
    carbonates/ˈkɑːbəneɪts/ 炭酸塩
    carbon dioxide/ˈkɑːbən daɪˈɒksaɪd/ 二酸化炭素
    indicator/ˈɪndɪkeɪtə/ 指示薬
    litmus/ˈlɪtməs/ リトマス試薬
    thymolphthalein/ˈθaɪməlfθæliːn/ チムフォルファレイン
    methyl orange/ˈmiːθaɪl ˈɒrɪndʒ/ メチルオレンジ
    ions/ˈaɪɒnz/ イオン
    effervescence/ˌefəˈvesəns/ 泡立ち
    7.1

    Bases and alkalis

    English

    Bases are oxides 氧化物 or hydroxides 氢氧化物 of metals. An alkali 可溶性碱 is a base that dissolves in water (a soluble base).

    Bases have two typical reactions:

    • with acids: base + acid → a salt + water (neutralisation again)
    • with ammonium salts 铵盐: this releases ammonia gas.
    日本語

    Bases are oxides 氧化物 or hydroxides 氢氧化物 of metals. An alkali 可溶性碱 is a base that dissolves in water (a soluble base).

    Bases have two typical reactions:

    • with acids: base + acid → a salt + water (neutralisation again)
    • with ammonium salts 铵盐: this releases ammonia gas.
    pH strips: alkalis turn universal indicator purple or blue (pH > 7)
    pH strips: alkalis turn universal indicator purple or blue (pH > 7)
    Explore · ⁨探索⁩

    Watch the pH during neutralisation

    Adding alkali to acid raises the pH. Near the end point the pH leaps up sharply — that steep jump is where the acid has just been neutralised.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    oxides/ˈɒksaɪdz/ 酸化物
    hydroxides/haɪˈdrɒksaɪdz/ 水酸化物
    alkali/ˈælkəlaɪ/ 塩基
    ammonium salts/æˈməʊnɪəm sɒlts/ アンモニウム塩
    7.1

    Strong and weak acids

    English

    An acid is a proton 质子 donor — it gives away $\text{H}^{+}$ ions (a hydrogen ion is just a proton). A base is a proton acceptor.

    How strong an acid is depends on how much of it splits into ions in water:

    • A strong acid 强酸 is completely dissociated 电离 (fully split into ions) in water. Hydrochloric acid 盐酸 is strong:
    $$\text{HCl}(aq) \rightarrow \text{H}^{+}(aq) + \text{Cl}^{-}(aq)$$
    • A weak acid 弱酸 is only partly dissociated. Ethanoic acid 乙酸 is weak, so the equation uses the reversible arrow:
    $$\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{H}^{+}(aq) + \text{CH}_3\text{COO}^{-}(aq)$$

    Note: 'strong' and 'weak' are about dissociation, not about being dilute or concentrated.

    日本語

    An acid is a proton 质子 donor — it gives away $\text{H}^{+}$ ions (a hydrogen ion is just a proton). A base is a proton acceptor.

    How strong an acid is depends on how much of it splits into ions in water:

    • A strong acid 强酸 is completely dissociated 电离 (fully split into ions) in water. Hydrochloric acid 盐酸 is strong:
    $$\text{HCl}(aq) \rightarrow \text{H}^{+}(aq) + \text{Cl}^{-}(aq)$$
    • A weak acid 弱酸 is only partly dissociated. Ethanoic acid 乙酸 is weak, so the equation uses the reversible arrow:
    $$\text{CH}_3\text{COOH}(aq) \rightleftharpoons \text{H}^{+}(aq) + \text{CH}_3\text{COO}^{-}(aq)$$

    Note: 'strong' and 'weak' are about dissociation, not about being dilute or concentrated.

    Two beakers: a strong acid full of separate positive and negative ions, and a weak acid that is mostly intact molecules with only a few ions
    A strong acid is fully split into ions; a weak acid stays mostly as molecules (only partly dissociated)
    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    proton/ˈprəʊtɒn/ 陽子
    strong acid/strɒŋ ˈæsɪd/ 強酸
    dissociated/dɪˈsəʊsɪeɪtɪd/ 解離
    hydrochloric acid/ˌhaɪdrəˈklɔːrɪk ˈæsɪd/ 塩酸
    weak acid/wiːk ˈæsɪd/ 弱酸
    ethanoic acid/ˌeθəˈnəʊɪk ˈæsɪd/ 酢酸
    7.1

    The pH scale

    English

    The pH scale runs from 0 to 14 and tells you how acidic or alkaline a solution is. A higher hydrogen ion concentration means a lower pH.

    You can find pH using universal indicator 通用指示剂, which turns different colours:

    pH Type Colour of universal indicator
    below 7 acidic 酸性 red / orange / yellow
    exactly 7 neutral 中性 green
    above 7 alkaline 碱性 blue / purple

    When an acid and an alkali react, the $\text{H}^{+}$ and $\text{OH}^{-}$ ions join to make water:

    $$\text{H}^{+}(aq) + \text{OH}^{-}(aq) \rightarrow \text{H}_2\text{O}(l)$$
    日本語

    The pH scale runs from 0 to 14 and tells you how acidic or alkaline a solution is. A higher hydrogen ion concentration means a lower pH.

    You can find pH using universal indicator 通用指示剂, which turns different colours:

    pH Type Colour of universal indicator
    below 7 acidic 酸性 red / orange / yellow
    exactly 7 neutral 中性 green
    above 7 alkaline 碱性 blue / purple
    The pH scale from 0 to 14 coloured like universal indicator: red at the acidic end, green at neutral, purple at the alkaline end
    Universal indicator turns red in a strong acid, green at neutral (pH 7), and purple in a strong alkali

    When an acid and an alkali react, the $\text{H}^{+}$ and $\text{OH}^{-}$ ions join to make water:

    $$\text{H}^{+}(aq) + \text{OH}^{-}(aq) \rightarrow \text{H}_2\text{O}(l)$$
    Explore · ⁨探索⁩

    The pH scale

    Slide the pH or tap a substance — lemon, water, soap, bleach. The indicator goes red→green→purple, and each step down in pH is ten times more H⁺ ions.

    Explore · ⁨探索⁩

    A titration curve

    Add alkali to acid and watch the pH rise. The steep jump is where the acid is just neutralised.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    universal indicator/ˌjuːnɪˈvɜːsl ˈɪndɪkeɪtə/ 万能指示薬
    acidic/əˈsɪdɪk/ 酸性であるため
    neutral/ˈnjuːtrəl/ 中性
    alkaline/ˈælkəlaɪn/ アルカリ性である
    7.2

    Oxides

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Classify oxides as acidic, including $\text{SO}_2$ and $\text{CO}_2$, or basic, including $\text{CuO}$ and $\text{CaO}$, related to metallic and non-metallic character 2 Describe amphoteric oxides as oxides that react with acids and with bases to produce a salt and water
    3 Classify $\text{Al}_2\text{O}_3$ and $\text{ZnO}$ as amphoteric oxides
    日本語
    コア サプリメント
    1 非金属性に関連する酸性酸化物 $\text{SO}_2$ および $\text{CO}_2$ 、あるいは金属性に関連する塩基性酸化物 $\text{CuO}$ および $\text{CaO}$ に分類する 2 酸および塩基の両方と反応して塩と水を生成する酸化物をアンフィテリック酸化物として説明する
    3 $\text{Al}_2\text{O}_3$ と $\text{ZnO}$ をアムフェテリック酸化物として分類せよ。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Oxides can be sorted by how they behave:

    • Acidic oxides are oxides of non-metals, such as $\text{SO}_2$ and $\text{CO}_2$.
    • Basic oxides are oxides of metals, such as $\text{CuO}$ and $\text{CaO}$.
    • Amphoteric 两性 oxides react with both acids and bases to make a salt and water. $\text{Al}_2\text{O}_3$ and $\text{ZnO}$ are amphoteric.

    So metal oxides tend to be basic and non-metal oxides tend to be acidic.

    日本語

    Oxides can be sorted by how they behave:

    • Acidic oxides are oxides of non-metals, such as $\text{SO}_2$ and $\text{CO}_2$.
    • Basic oxides are oxides of metals, such as $\text{CuO}$ and $\text{CaO}$.
    • Amphoteric 两性 oxides react with both acids and bases to make a salt and water. $\text{Al}_2\text{O}_3$ and $\text{ZnO}$ are amphoteric.

    So metal oxides tend to be basic and non-metal oxides tend to be acidic.

    Limestone is mostly calcium carbonate — a metal carbonate that acts as a base and neutralises acids
    Limestone is mostly calcium carbonate — a metal carbonate that acts as a base and neutralises acids
    Explore · ⁨探索⁩

    Oxide type lab

    Classify oxides by how they behave with acids and bases.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    amphoteric/ˌæmfəʊˈterɪk/ 両性である
    7.3

    Preparing salts

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the preparation, separation and purification of soluble salts by reaction of an acid with: (a) an alkali by titration (b) excess metal (c) excess insoluble base (d) excess insoluble carbonate 4 Describe the preparation of insoluble salts by precipitation
    2 Describe the general solubility rules for salts: (a) sodium, potassium and ammonium salts are soluble (b) nitrates are soluble (c) chlorides are soluble, except lead and silver (d) sulfates are soluble, except barium, calcium and lead (e) carbonates are insoluble, except sodium, potassium and ammonium (f) hydroxides are insoluble, except sodium, potassium, ammonium and calcium (partially)
    3 Define a hydrated substance as a substance that is chemically combined with water and an anhydrous substance as a substance containing no water 5 Define the term water of crystallisation as the water molecules present in hydrated crystals, including $\text{CuSO}_4\bullet5\text{H}_2\text{O}$ and $\text{CoCl}_2\bullet6\text{H}_2\text{O}$
    日本語
    コア サプリメント
    1 酸との反応による可溶性塩の調製、分離および精製を説明せよ:(a) 塩基による滴定 (b) 過剰な金属 (c) 過剰な難溶性塩基 (d) 過剰な難溶性炭酸塩 4 沈殿反応による難溶性塩の調製を説明する
    2 塩の一般的な溶解性の法則を説明せよ:(a) ナトリウム、カリウムおよびアンモニウムの塩は可溶である。(b) ニトレートは可溶である。(c) クロリドは可溶であるが、鉛と銀を除く。(d) スルフェートは可溶であるが、バリウム、カルシウムおよび鉛を除く。(e) カルボネートは不溶であるが、ナトリウム、カリウムおよびアンモニウムを除く。(f) ハイドロキシドは不溶であるが、ナトリウム、カリウム、アンモニウムおよびカルシウム(部分的に)を除く。
    3 水和物を化学的に水と結合した物質、無水物を水含まない物質として定義する 5 結晶水という用語を定義する(水和結晶に含まれる水分子であり、$\text{CuSO}_4\bullet5\text{H}_2\text{O}$ および $\text{CoCl}_2\bullet6\text{H}_2\text{O}$ を含む)

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Whether a salt can be made by a certain method depends on whether it is soluble 可溶 (dissolves) or insoluble 不溶 (does not dissolve).

    Solubility rules

    Salt type Rule
    sodium, potassium, ammonium salts all soluble
    nitrates 硝酸盐 all soluble
    chlorides 氯化物 soluble, except lead and silver
    sulfates 硫酸盐 soluble, except barium, calcium and lead
    carbonates insoluble, except sodium, potassium and ammonium
    hydroxides insoluble, except sodium, potassium, ammonium and (partly) calcium

    Making a soluble salt

    You react an acid with one of these:

    • an alkali, using titration 滴定 (since both are solutions, you must measure the exact volumes);
    • an excess 过量 of a metal, an insoluble base, or an insoluble carbonate.

    When you use an excess of a solid, you then filter 过滤 to remove the leftover solid. Finally you evaporate 蒸发 some water and let the solution crystallise 结晶 to get the salt.

    Worked example. Which method would you use to prepare (a) copper(II) sulfate and (b) barium sulfate? Check the solubility rules first. (a) Copper(II) sulfate is soluble (only barium, calcium and lead sulfates are not), so make it from an insoluble solid and an acid: add excess copper(II) oxide to warm dilute sulfuric acid, filter off the leftover solid, then evaporate some water and crystallise. (b) Barium sulfate is insoluble, so make it by precipitation: mix two solutions that each carry one of the needed ions, then filter, wash and dry the solid. The solubility of the salt you want is what decides the method - consult the rules before you choose.

    Making an insoluble salt

    An insoluble salt is made by precipitation 沉淀: mix two solutions that each contain one of the needed ions, and the insoluble salt forms as a solid. You then filter, wash and dry it.

    Water in salts

    • A hydrated 水合 substance is chemically joined with water.
    • An anhydrous 无水 substance contains no water.

    The water molecules inside hydrated crystals are called the water of crystallisation 结晶水. For example, $\text{CuSO}_4 \bullet 5\text{H}_2\text{O}$ has five water molecules for each formula unit.

    日本語

    Whether a salt can be made by a certain method depends on whether it is soluble 可溶 (dissolves) or insoluble 不溶 (does not dissolve).

    Solubility rules

    Salt type Rule
    sodium, potassium, ammonium salts all soluble
    nitrates 硝酸盐 all soluble
    chlorides 氯化物 soluble, except lead and silver
    sulfates 硫酸盐 soluble, except barium, calcium and lead
    carbonates insoluble, except sodium, potassium and ammonium
    hydroxides insoluble, except sodium, potassium, ammonium and (partly) calcium

    Making a soluble salt

    You react an acid with one of these:

    • an alkali, using titration 滴定 (since both are solutions, you must measure the exact volumes);
    • an excess 过量 of a metal, an insoluble base, or an insoluble carbonate.

    When you use an excess of a solid, you then filter 过滤 to remove the leftover solid. Finally you evaporate 蒸发 some water and let the solution crystallise 结晶 to get the salt.

    A four-step flowchart: add excess solid to the acid, filter, evaporate some water, then leave to crystallise
    Making a soluble salt from an insoluble solid: react with excess, filter off the excess, evaporate, then crystallise
    Bright blue geometric crystals of copper(II) sulfate against a white background
    Letting the solution crystallise slowly gives well-shaped crystals — here, the blue crystals of copper(II) sulfate

    Worked example. Which method would you use to prepare (a) copper(II) sulfate and (b) barium sulfate? Check the solubility rules first. (a) Copper(II) sulfate is soluble (only barium, calcium and lead sulfates are not), so make it from an insoluble solid and an acid: add excess copper(II) oxide to warm dilute sulfuric acid, filter off the leftover solid, then evaporate some water and crystallise. (b) Barium sulfate is insoluble, so make it by precipitation: mix two solutions that each carry one of the needed ions, then filter, wash and dry the solid. The solubility of the salt you want is what decides the method - consult the rules before you choose.

    Making an insoluble salt

    An insoluble salt is made by precipitation 沉淀: mix two solutions that each contain one of the needed ions, and the insoluble salt forms as a solid. You then filter, wash and dry it.

    Two clear solutions poured together to give a beaker with a solid forming at the bottom
    Mixing two solutions that each carry one of the needed ions makes the insoluble salt appear as a solid precipitate, which you then filter off

    Water in salts

    • A hydrated 水合 substance is chemically joined with water.
    • An anhydrous 无水 substance contains no water.

    The water molecules inside hydrated crystals are called the water of crystallisation 结晶水. For example, $\text{CuSO}_4 \bullet 5\text{H}_2\text{O}$ has five water molecules for each formula unit.

    Explore · ⁨探索⁩

    Preparing a soluble salt

    Step through the method. React an acid with excess insoluble base, filter off what's left over, then evaporate and crystallise to get pure salt crystals.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    soluble/ˈsɒljuːbl/ 可溶
    insoluble/ɪnˈsɒljuːbl/ 不溶である
    nitrates/ˈnaɪtreɪts/ 硝酸塩
    chlorides/ˈklɔːraɪdz/ 塩化物
    sulfates/ˈsʌlfeɪts/ 硫酸塩
    titration/taɪˈtreɪʃn/ 滴定
    excess/ekˈses/ 過剰
    filter/ˈfɪltə/ フィルタリング
    evaporate/ɪˈvæpəreɪt/ 蒸発する
    crystallise/ˈkrɪstəlaɪz/ 結晶化する
    precipitation/prɪˌsɪpɪˈteɪʃn/ 沈殿反応
    hydrated/haɪˈdreɪtɪd/ 水和物(_attrated_)
    anhydrous/ænˈhaɪdrəs/ 無水物(_anhydrous_)
    water of crystallisation/ˈwɔːtə ɒv ˌkrɪstəlaɪˈzeɪʃn/ 結晶水
    7.3

    Exam tips

    • Learn the three acid reactions: acid + metal → salt + hydrogen; acid + base → salt + water (neutralisation); acid + carbonate → salt + water + carbon dioxide.
    • Strong and weak describe how much the acid splits into ions (dissociation), not how concentrated it is. A strong acid is fully dissociated; a weak acid only partly.
    • A lower pH means a higher hydrogen-ion concentration. pH 7 is neutral, below 7 is acidic, above 7 is alkaline.
    • Choose the salt method by solubility: a soluble salt is made from an acid + excess solid (then filter, evaporate, crystallise), or by titration with an alkali; an insoluble salt is made by precipitation (mix two solutions, then filter).
    • Learn the solubility rules — all sodium, potassium, ammonium and nitrate salts are soluble.
  • 8

    The Periodic Table

    Watch lesson · ⁨レッスンを視聴⁩
    8.1

    Arrangement of the elements

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the Periodic Table as an arrangement of elements in periods and groups and in order of increasing proton number/atomic number
    2 Describe the change from metallic to non‑metallic character across a period
    3 Describe the relationship between group number and the charge of the ions formed from elements in that group
    4 Explain similarities in the chemical properties of elements in the same group of the Periodic Table in terms of their electronic configuration
    5 Explain how the position of an element in the Periodic Table can be used to predict its properties 6 Identify trends in groups, given information about the elements
    日本語
    コア サプリメント
    1 周期表を、原子番号/陽子数が増加する順に、周期および族で配置された元素の配列として説明せよ。
    2 周期に沿った金属性から非金属性への変化を説明せよ。
    3 族番号と、その族の元素から形成されるイオンの電荷との関係性を説明せよ。
    4 電子配置の観点から、周期表の同一族にある元素の化学的性質の類似点を説明せよ。
    5 周期表における元素の位置を用いてその性質を予測する方法を説明する 6 元素に関する情報に基づき、族における傾向を同定する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The Periodic Table 周期表 arranges all the elements 元素 in order of increasing proton number 质子数 (the atomic number 原子序数). The horizontal rows are called periods 周期 and the vertical columns are called groups 族.

    A few key patterns:

    • Across a period, the elements change from metals 金属 on the left to non-metals 非金属 on the right.
    • The group number tells you the charge of the ions 离子 that the elements form. Group I forms $+1$ ions, Group II forms $+2$ ions, and Group VII forms $-1$ ions.
    • Elements in the same group have similar chemical properties. This is because they have the same number of outer-shell electrons 电子 (the same outer electronic configuration 电子排布).

    Because of these patterns, you can use an element's position to predict its properties.

    日本語
    A periodic table display with element samples
    The periodic table arranges the elements in order of atomic number.

    The Periodic Table 周期表 arranges all the elements 元素 in order of increasing proton number 质子数 (the atomic number 原子序数). The horizontal rows are called periods 周期 and the vertical columns are called groups 族.

    A few key patterns:

    • Across a period, the elements change from metals 金属 on the left to non-metals 非金属 on the right.
    • The group number tells you the charge of the ions 离子 that the elements form. Group I forms $+1$ ions, Group II forms $+2$ ions, and Group VII forms $-1$ ions.
    • Elements in the same group have similar chemical properties. This is because they have the same number of outer-shell electrons 电子 (the same outer electronic configuration 电子排布).

    Because of these patterns, you can use an element's position to predict its properties.

    A schematic periodic table: groups as numbered columns and periods as rows, with metals shaded on the left and non-metals on the right, split by a staircase line
    Groups are the columns and periods are the rows; metals lie to the left of the staircase, non-metals to the right
    Explore · ⁨探索⁩

    Spot a trend across a period

    Reading across a period, properties change in a regular, repeating pattern. Switch the trend to watch atomic radius, ionisation energy or melting point change element by element.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Periodic Table/ˌpɪərɪˈɒdɪk ˈteɪbl/ 元素周期表
    elements/ˈelɪmənts/ 元素
    proton number/ˈprəʊtɒn ˈnʌmbə/ 原子番号
    atomic number/əˈtɒmɪk ˈnʌmbə/ 原子番号
    periods/ˈpɪərɪədz/ 族(横行)
    groups/ɡruːps/ グループ
    metals/ˈmetlz/ 金属
    non-metals/nɒn ˈmetlz/ 非金属
    ions/ˈaɪɒnz/ イオン
    electrons/ɪˈlektrɒnz/ 電子
    electronic configuration/ɪlekˈtrɒnɪk kənˌfɪɡjəˈreɪʃn/ 電子配置
    8.2

    Group I — the alkali metals

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the Group I alkali metals, lithium, sodium and potassium, as relatively soft metals with general trends down the group, limited to: (a) decreasing melting point (b) increasing density (c) increasing reactivity
    2 Predict the properties of other elements in Group I, given information about the elements
    日本語
    コア サプリメント
    1 リチウム、ナトリウムおよびカリウムといったI族のアルカリ金属について、比較的柔らかい金属であり、族を下るにつれて以下の一般的な傾向があることを説明せよ:(a) 融点の低下 (b) 密度の増加 (c) 反応性の増加。
    2 元素に関する情報に基づき、I族の他の元素の性質を予測せよ。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Group I elements are the alkali metals 碱金属: lithium 锂, sodium 钠 and potassium 钾. They are soft 柔软 metals (you can cut them with a knife).

    Going down the group, there are clear trends 趋势:

    Property Trend going down the group
    melting point 熔点 decreases
    density 密度 increases
    reactivity 活泼性 increases

    You can use these trends to predict the properties of other Group I elements. For example, rubidium (below potassium) would be even more reactive and have an even lower melting point.

    Reaction with water, and what you see

    This is the reaction the exam asks about most. Every Group I metal reacts with water to give an alkali (the metal hydroxide) plus hydrogen - which is exactly why they are called the alkali metals:

    $$2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$$

    "Describe what you see" is worth marks of its own, so learn the list for sodium on water:

    • it floats - the metal is less dense than water.
    • it melts into a ball - the reaction gives out heat, and the metal's melting point is low.
    • it moves about on the surface - the hydrogen jetting off pushes it around.
    • there is effervescence 泡腾 (fizzing) - the hydrogen gas escaping.
    • it gets smaller and disappears as it is used up.
    • the solution turns alkaline, so universal indicator goes purple (it is now $\text{NaOH}$).

    The trend shows up in how violent this is: lithium fizzes steadily but does not melt; sodium melts into a ball and darts around; potassium is fast enough to ignite its own hydrogen, burning with a lilac flame. Two answers to avoid: "hydrogen is given off" is not an observation (you cannot see it - say effervescence), and the metal melts because the reaction is exothermic, not because "water is hot".

    日本語

    Group I elements are the alkali metals 碱金属: lithium 锂, sodium 钠 and potassium 钾. They are soft 柔软 metals (you can cut them with a knife).

    A small piece of potassium burns with a bright burst of sparks and a lilac flame on the surface of a dish of water
    Potassium reacts violently with water, giving off hydrogen that burns with a lilac flame — reactivity increases down Group I
    Electron-shell diagrams of lithium, sodium and potassium, each with a single highlighted outer electron
    Every Group I atom has one electron in its outer shell, which is why they react in similar ways

    Going down the group, there are clear trends 趋势:

    Property Trend going down the group
    melting point 熔点 decreases
    density 密度 increases
    reactivity 活泼性 increases

    You can use these trends to predict the properties of other Group I elements. For example, rubidium (below potassium) would be even more reactive and have an even lower melting point.

    Reaction with water, and what you see

    This is the reaction the exam asks about most. Every Group I metal reacts with water to give an alkali (the metal hydroxide) plus hydrogen - which is exactly why they are called the alkali metals:

    $$2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$$

    "Describe what you see" is worth marks of its own, so learn the list for sodium on water:

    • it floats - the metal is less dense than water.
    • it melts into a ball - the reaction gives out heat, and the metal's melting point is low.
    • it moves about on the surface - the hydrogen jetting off pushes it around.
    • there is effervescence 泡腾 (fizzing) - the hydrogen gas escaping.
    • it gets smaller and disappears as it is used up.
    • the solution turns alkaline, so universal indicator goes purple (it is now $\text{NaOH}$).

    The trend shows up in how violent this is: lithium fizzes steadily but does not melt; sodium melts into a ball and darts around; potassium is fast enough to ignite its own hydrogen, burning with a lilac flame. Two answers to avoid: "hydrogen is given off" is not an observation (you cannot see it - say effervescence), and the metal melts because the reaction is exothermic, not because "water is hot".

    Explore · ⁨探索⁩

    Group I trend lab

    Follow Group I metals down the group and watch reactivity increase.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    alkali metals/ˈælkəlaɪ ˈmetlz/ アルカリ金属
    lithium/ˈlɪθɪəm/ リチウム
    sodium/ˈsəʊdɪəm/ ナトリウム
    potassium/pəˈtæsɪəm/ カリウム
    soft/sɒft/ ソフト
    trends/trendz/ トレンド
    melting point/ˈmeltɪŋ pɔɪnt/ 融点
    density/ˈdensɪti/ 密度
    reactivity/rɪəkˈtɪvɪti/ 反応性
    effervescence/ˌefəˈvesəns/ 泡立ち
    8.3

    Group VII — the halogens

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the Group VII halogens, chlorine, bromine and iodine, as diatomic non-metals with general trends down the group, limited to: (a) increasing density (b) decreasing reactivity
    2 State the appearance of the halogens at r.t.p. as: (a) chlorine, a pale yellow-green gas (b) bromine, a red-brown liquid (c) iodine, a grey-black solid
    3 Describe and explain the displacement reactions of halogens with other halide ions
    4 Predict the properties of other elements in Group VII, given information about the elements
    日本語
    コア サプリメント
    1 クロリン、ブロミンおよびヨウ素といったVII族のハロゲンについて、二原子性の非金属であり、族を下るにつれて以下の一般的な傾向があることを説明せよ:(a) 密度の増加 (b) 反応性の低下。
    2 常温常圧におけるハロゲンの外観を述べる:(a) クロリン、淡黄緑色の気体 (b) ブロミン、赤茶色の液体 (c) ヨウ素、灰黒色の固体。
    3 ハロゲンと他のハライドイオンとの置換反応を説明し、その理由を述べる。
    4 元素に関する情報に基づき、VII族の他の元素の性質を予測せよ。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Group VII elements are the halogens 卤素: chlorine 氯气, bromine 溴 and iodine 碘. They are diatomic 双原子 non-metals (each molecule is made of two atoms, such as $\text{Cl}_2$).

    Their appearance at room temperature and pressure:

    Halogen Appearance
    chlorine a pale yellow-green gas
    bromine a red-brown liquid
    iodine a grey-black solid

    Going down the group, the density increases but the reactivity decreases (the opposite trend to Group I).

    Displacement reactions

    A more reactive halogen pushes out (displaces) a less reactive halide 卤化物 ion from its solution. This is a displacement reaction 置换反应.

    For example, chlorine is more reactive than bromine, so chlorine displaces bromine from potassium bromide:

    $$\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2$$

    Worked example. Aqueous chlorine is added to potassium iodide solution; then aqueous iodine is added to potassium bromide solution. What happens in each? Reactivity falls down Group VII: chlorine, then bromine, then iodine. (a) Chlorine is above iodine, so it is more reactive and displaces it: $\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2$, and the solution turns red-brown as iodine is released. (b) Iodine is below bromine, so it is less reactive and cannot displace it - there is no reaction and no colour change. "No reaction" is a complete answer worth marks: check the order in the group before you write any equation.

    日本語

    Group VII elements are the halogens 卤素: chlorine 氯气, bromine 溴 and iodine 碘. They are diatomic 双原子 non-metals (each molecule is made of two atoms, such as $\text{Cl}_2$).

    Their appearance at room temperature and pressure:

    Halogen Appearance
    chlorine a pale yellow-green gas
    bromine a red-brown liquid
    iodine a grey-black solid

    Going down the group, the density increases but the reactivity decreases (the opposite trend to Group I).

    Group I with a downward reactivity arrow next to Group VII with an upward reactivity arrow
    Group I gets more reactive down the group, while Group VII gets less reactive down — opposite trends

    Displacement reactions

    A more reactive halogen pushes out (displaces) a less reactive halide 卤化物 ion from its solution. This is a displacement reaction 置换反应.

    For example, chlorine is more reactive than bromine, so chlorine displaces bromine from potassium bromide:

    $$\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2$$
    Colourless potassium bromide solution turning orange when chlorine is added
    Chlorine is more reactive, so it displaces bromine from the solution; the released bromine turns the colourless solution orange

    Worked example. Aqueous chlorine is added to potassium iodide solution; then aqueous iodine is added to potassium bromide solution. What happens in each? Reactivity falls down Group VII: chlorine, then bromine, then iodine. (a) Chlorine is above iodine, so it is more reactive and displaces it: $\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2$, and the solution turns red-brown as iodine is released. (b) Iodine is below bromine, so it is less reactive and cannot displace it - there is no reaction and no colour change. "No reaction" is a complete answer worth marks: check the order in the group before you write any equation.

    Explore · ⁨探索⁩

    Group VII trend lab

    Follow halogens down the group and watch reactivity decrease.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    halogens/ˈhælədʒnz/ ハロゲン
    chlorine/ˈklɔːriːn/ 塩素
    bromine/ˈbrəʊmaɪn/ 臭素
    iodine/ˈaɪədiːn/ ヨウ素
    diatomic/ˌdaɪəˈtɒmɪk/ 二原子性(_diatomic_)
    halide/ˈhælaɪd/ ハロゲン化物
    displacement reaction/dɪˈspleɪsmənt rɪˈækʃn/ 置換反応
    8.4

    Transition elements

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the transition elements as metals that: (a) have high densities (b) have high melting points (c) form coloured compounds (d) often act as catalysts as elements and in compounds 2 Describe transition elements as having ions with variable oxidation numbers, including iron(II) and iron(III)
    日本語
    コア サプリメント
    1 遷移金属として以下の特徴を持つ金属であることを記述する:(a) 密度が高い (b) 融点が高い (c) 有色化合物を形成する (d) 元素および化合物としてしばしば触媒として機能する 2 遷移金属が可変酸化数のイオンを持ち、鉄(II)および鉄(III)を含むことを記述する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The transition elements 过渡元素 are the block of metals in the middle of the Periodic Table. Compared with Group I metals, they:

    • have high densities;
    • have high melting points;
    • form coloured 有色 compounds 化合物;
    • often act as catalysts, both as elements and in compounds.

    They can also have variable 可变 oxidation numbers 氧化数. For example, iron 铁 forms both iron(II) and iron(III) compounds.

    日本語

    The transition elements 过渡元素 are the block of metals in the middle of the Periodic Table. Compared with Group I metals, they:

    • have high densities;
    • have high melting points;
    • form coloured 有色 compounds 化合物;
    • often act as catalysts, both as elements and in compounds.

    They can also have variable 可变 oxidation numbers 氧化数. For example, iron 铁 forms both iron(II) and iron(III) compounds.

    Six beakers of brightly coloured solutions — red, orange, yellow, green, blue and purple
    Transition metals form coloured compounds: each of these solutions contains a different transition-metal ion
    Explore · ⁨探索⁩

    Transition element lab

    Match transition-metal properties to real examples.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    transition elements/trænˈsɪʃn ˈelɪmənts/ 遷移元素
    coloured/ˈkʌləd/ 有色
    compounds/ˈkɒmpaʊndz/ 化合物
    variable/ˈveərɪəbl/ 変数
    oxidation numbers/ˌɒksɪˈdeɪʃn ˈnʌmbəz/ 酸化数
    iron/ˈaɪən/ 鉄
    8.5

    Group VIII — the noble gases

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the Group VIII noble gases as unreactive, monatomic gases and explain this in terms of electronic configuration
    日本語
    コア サプリメント
    1 VIII族の希ガスを不活性で単原子性の気体であり、その理由を電子配置の観点から説明せよ。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The Group VIII noble gases 稀有气体 are unreactive 不活泼, monatomic 单原子 gases (they exist as single atoms, not as molecules).

    They are unreactive because they already have a full outer shell of electrons. This makes them stable, so they do not need to gain, lose or share electrons.

    日本語

    The Group VIII noble gases 稀有气体 are unreactive 不活泼, monatomic 单原子 gases (they exist as single atoms, not as molecules).

    They are unreactive because they already have a full outer shell of electrons. This makes them stable, so they do not need to gain, lose or share electrons.

    Five discharge tubes labelled He, Ne, Ar, Kr and Xe, each glowing a different colour
    Passing electricity through each noble gas makes it glow a characteristic colour — this is how neon signs and other lights work
    Explore · ⁨探索⁩

    Noble gas lab

    Classify noble gas uses by the property that makes them useful.

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    noble gases/ˈnəʊbl ˈɡæsɪz/ 希ガス
    unreactive/ʌnrɪˈæktɪv/ 不活性である
    monatomic/ˌmɒnəˈtɒmɪk/ 単原子性
    8.5

    Exam tips

    • The Periodic Table is ordered by proton number. Elements in the same group (column) have the same number of outer electrons, so they react in similar ways; periods are the rows.
    • Group I (alkali metals) get more reactive going down; Group VII (halogens) get less reactive going down — opposite trends.
    • A more reactive halogen displaces a less reactive one from a solution of its salt — for example chlorine displaces bromine.
    • Noble gases are unreactive because they have a full outer shell. Transition elements are hard and dense, form coloured compounds, act as catalysts and have variable oxidation numbers.
  • 9

    Metals

    Watch lesson · ⁨レッスンを視聴⁩
    9.1

    Properties of metals · ⁨金属の性質⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Compare the general physical properties of metals and non-metals, including: (a) thermal conductivity (b) electrical conductivity (c) malleability and ductility (d) melting points and boiling points
    2 Describe the general chemical properties of metals, limited to their reactions with: (a) dilute acids (b) cold water and steam (c) oxygen
    日本語
    コア サプリメント
    1 金属および非金属の一般的な物理的性質を、以下を含めて比較せよ:(a) 熱伝導率 (b) 電気伝導率 (c) 延性および伸性 (d) 融点および沸点。
    2 金属の一般的な化学的性質について、以下の反応に限って説明せよ:(a) 希酸 (b) 冷水および蒸気 (c) 酸素。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Physical properties

    Metals 金属 and non-metals 非金属 behave very differently:

    Property Metals Non-metals
    thermal conductivity 导热性 (conducting heat) good poor
    electrical conductivity 导电性 good poor (except graphite)
    malleability 展性 and ductility 延性 malleable and ductile brittle 易碎 (they snap)
    melting point 熔点 and boiling point 沸点 usually high usually low

    Chemical properties

    Metals react in three main ways:

    • with dilute acids 酸 → a salt 盐 + hydrogen 氢气
    • with cold water or steam 蒸汽 → a metal hydroxide (or oxide) + hydrogen
    • with oxygen 氧气 → a metal oxide 氧化物
    日本語

    物理的性質

    金属と非金属は非常に異なる振る舞いを示す:

    性質 金属 非金属
    熱伝導率(熱を伝える能力) 良い 悪い
    電気伝導率 良い 悪い(グラファイトを除く)
    展性および延性 展性および延性がある 脆い(折れる)
    融点および沸点 通常高い 通常低い

    化学的性質

    金属は主に3つの方法で反応する:

    • 希酸 → 塩 + 水素
    • 冷水または蒸気 → 金属水酸化物(または酸化物)+ 水素
    • 酸素 → 金属酸化物
    自然銅:金属は光沢があり、密度が高く、展性・延性があり、熱と電気の良質な伝導体である
    自然銅:金属は光沢があり、密度が高く、展性・延性があり、熱と電気の良質な伝導体である
    Explore · ⁨探索⁩

    Metallic bonding · ⁨金属結合⁩

    Positive ions sit in a sea of delocalised electrons — that is why metals conduct and bend. Push the layers and the bond holds. · ⁨正イオンは非局在電子の海の中に配置されている——これが金属が電気伝導性を持ち、曲げられる理由である。層を押し付けると結合は保たれる。⁩

    Explore · ⁨探索⁩

    Metal property lab · ⁨金属の性質実験⁩

    Link metal properties to the particle model. · ⁨金属の性質と粒子モデルを関連付ける。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    copper/ˈkɒpə/ 銅
    9.2

    Uses of metals · ⁨金属の用途⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the uses of metals in terms of their physical properties, including: (a) aluminium in the manufacture of aircraft because of its low density (b) aluminium in the manufacture of overhead electrical cables because of its low density and good electrical conductivity (c) aluminium in food containers because of its resistance to corrosion (d) copper in electrical wiring because of its good electrical conductivity and ductility
    日本語
    コア サプリメント
    1 金属の物理的性質に基づく利用を説明せよ:(a) 低密度のため航空機の製造におけるアルミニウム (b) 低密度および良好な電気伝導率のため架空送電線材の製造におけるアルミニウム (c) 腐食抵抗のため食品容器におけるアルミニウム (d) 良好な電気伝導率および伸性のため電気配線における銅。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A metal is chosen for a job because of its physical properties.

    • Aluminium 铝 is used to make aircraft because of its low density 密度 (it is light).
    • Aluminium is used for overhead electrical cables because of its low density and good electrical conductivity.
    • Aluminium is used for food containers because it resists corrosion 腐蚀.
    • Copper 铜 is used for electrical wiring because of its good electrical conductivity and its ductility (it can be drawn into wires).
    日本語

    金属は物理的性質に基づいて特定の作業に適しているために選ばれる。

    • アルミニウムは低密度(軽量)のため航空機の製造に使われる。
    • アルミニウムは低密度と良好な電気伝導率のため、架線電線に使われる。
    • アルミニウムは腐食に強い食品容器に使われる。
    • 銅は良好な電気伝導率と延性(線に引き伸ばせる)のため、電気配線に使われる。
    空港の滑走路に駐機している大型旅客機、背後には丘が見える
    航空機の胴体には多くのアルミニウムが使われている:強度はあるが密度が低いため、飛行機は軽量を保つ
    Explore · ⁨探索⁩

    Metal use lab · ⁨金属の用途実験⁩

    Choose the property that explains each metal use. · ⁨各金属の用途を説明する性質を選んでください。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    metals/ˈmetlz/ 金属
    non-metals/nɒn ˈmetlz/ 非金属
    thermal conductivity/ˈθɜːml kɒndəkˈtɪvɪti/ 熱伝導率
    electrical conductivity/ɪˈlektrɪkl kɒndəkˈtɪvɪti/ 電気伝導性
    malleability/ˌmæləˈbɪlɪti/ 延性
    ductility/dʌkˈtɪlɪti/ 延性
    brittle/ˈbrɪtl/ 脆い
    melting point/ˈmeltɪŋ pɔɪnt/ 融点
    boiling point/ˈbɔɪlɪŋ pɔɪnt/ 沸点
    acids/ˈæsɪdz/ 酸
    salt/sɒlt/ 塩
    hydrogen/ˈhaɪdrədʒn/ 水素
    steam/stiːm/ 蒸気
    oxygen/ˈɒksɪdʒn/ 酸素
    oxide/ˈɒksaɪd/ 酸化物
    aluminium/ˌæljʊˈmɪnɪəm/ アルミニウム
    density/ˈdensɪti/ 密度
    9.3

    Alloys · ⁨合金⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe an alloy as a mixture of a metal with other elements, including: (a) brass as a mixture of copper and zinc (b) stainless steel as a mixture of iron and other elements such as chromium, nickel and carbon
    2 State that alloys can be harder and stronger than the pure metals and are more useful 5 Explain in terms of structure how alloys can be harder and stronger than the pure metals because the different sized atoms in alloys mean the layers can no longer slide over each other
    3 Describe the uses of alloys in terms of their physical properties, including stainless steel in cutlery because of its hardness and resistance to rusting
    4 Identify representations of alloys from diagrams of structure
    日本語
    コア サプリメント
    1 合金とは、金属と他の元素との混合物であることを説明せよ:(a) 真鍮は銅と亜鉛の混合物 (b) ステンレス鋼は鉄とクロム、ニッケル、炭素などの他の元素との混合物。
    2 合金は純金属よりも硬く強靭であり、より有用であることを述べる 5 構造の観点から、合金中の異なるサイズの原子により層が互いに滑り込めなくなるため、合金が純金属より硬く強靭になる理由を説明する
    3 合金の物理的性質に基づく利用を説明せよ:硬さと錆びにくさのため食器器具におけるステンレス鋼など。
    4 構造図から合金の表現を特定せよ。

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    An alloy 合金 is a mixture 混合物 of a metal with one or more other elements.

    • Brass 黄铜 is a mixture of copper and zinc 锌.
    • Stainless steel 不锈钢 is a mixture of iron 铁 with other elements such as chromium 铬, nickel 镍 and carbon 碳.

    Alloys are usually harder and stronger than the pure metals, which makes them more useful. For example, stainless steel is used for cutlery because it is hard and does not rust.

    Why alloys are harder. In a pure metal, the atoms 原子 are all the same size, so the layers 层 can slide over each other easily. In an alloy, the different-sized atoms stop the layers from sliding, so the alloy is harder and stronger.

    日本語

    合金とは、金属と他の要素が1つ以上混ざり合った混合物である。

    • 真鍮は銅と亜鉛の混合物である。
    • ステンレス鋼は鉄とクロム、ニッケル、炭素などの他の元素の混合物である。

    合金は純金属よりも通常硬く強固であるため、より実用的である。例えば、ステンレス鋼は硬く錆びないため、カトラリーに使われる。

    **合金が硬くなる理由。**純金属では、原子の大きさがすべて同じであるため、層は互いに滑らかにすべり合うことができる。一方、合金では異なる大きさの原子が層のすべりを妨げるため、合金は純金属よりも硬く強くなる。

    同じ大きさの原子が整然と並び、層が滑りやすい様子と、大きな原子によって列が歪みすべりが阻害される合金の様子
    合金では、異なる大きさの原子が層のすべりを防ぐため、純金属より硬い
    Explore · ⁨探索⁩

    Alloy property lab · ⁨合金の性質実験⁩

    See why mixing atoms changes metal properties. · ⁨原子を混ぜることが金属の性質を変化させる理由を見てみましょう。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    alloy/ˈælɔɪ/ 合金 (alloy)
    mixture/ˈmɪkstʃə/ 混合物
    brass/bræs/ 真鍮
    zinc/zɪŋk/ 亜鉛
    stainless steel/ˈsteɪnləs stiːl/ ステンレス鋼
    9.4

    The reactivity series · ⁨反応性系列⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State the order of the reactivity series as: potassium, sodium, calcium, magnesium, aluminium, carbon, zinc, iron, hydrogen, copper, silver, gold 4 Describe the relative reactivities of metals in terms of their tendency to form positive ions, by displacement reactions, if any, with the aqueous ions of magnesium, zinc, iron, copper and silver
    2 Describe the reactions, if any, of: (a) potassium, sodium and calcium with cold water (b) magnesium with steam (c) magnesium, zinc, iron, copper, silver and gold with dilute hydrochloric acid and explain these reactions in terms of the position of the metals in the reactivity series
    5 Explain the apparent unreactivity of aluminium in terms of its oxide layer
    3 Deduce an order of reactivity from a given set of experimental results
    日本語
    コア サプリメント
    1 反応性の順序を以下のように記述せよ:カリウム、ナトリウム、カルシウム、マグネシウム、アルミニウム、炭素、亜鉛、鉄、水素、銅、銀、金 4 金属の相対的な反応性を、正イオン形成への傾向、および水溶液中のマグネシウム、亜鉛、鉄、銅、銀のイオンとの置換反応(あれば)によって説明せよ
    2 以下の反応(あれば)を記述せよ:(a) カリウム、ナトリウム、カルシウムと冷水 (b) マグネシウムと蒸気 (c) マグネシウム、亜鉛、鉄、銅、銀、金と希塩酸。これらの反応を反応性順序における金属の位置に基づいて説明せよ
    5 アルミニウムの見かけ上の不活性を、その酸化層によって説明せよ
    3 与えられた実験結果から反応性の順序を推論せよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    The reactivity series 金属活动性顺序 lists metals in order of how reactive they are. Carbon and hydrogen are included for comparison:

    potassium 钾, sodium 钠, calcium 钙, magnesium 镁, aluminium, carbon, zinc, iron, hydrogen, copper, silver, gold

    (most reactive at the top, least reactive at the bottom)

    The higher a metal is, the more easily it forms positive ions 离子. This explains its reactions:

    • potassium, sodium and calcium react with cold water.
    • magnesium reacts with steam (but only very slowly with cold water).
    • magnesium, zinc and iron react with dilute hydrochloric acid; copper, silver and gold do not.

    Displacement reactions

    A more reactive metal will displace a less reactive metal from a solution of its ions (a displacement reaction 置换反应). For example, zinc displaces copper from copper(II) sulfate solution, because zinc is more reactive than copper:

    $$\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$$

    Be ready to say what you would see, because those are separate marks:

    • the blue colour of the solution fades - the blue $\text{Cu}^{2+}$ ions are being used up (zinc sulfate is colourless).
    • a pink-brown solid coats the zinc - that is the displaced copper.
    • the grey zinc dissolves and gets smaller.
    • the mixture warms up - displacement is exothermic.

    Metal-and-acid reactions have their own list: effervescence 泡腾 (the hydrogen escaping) and the solid dissolves. With copper(II) oxide or carbonate and acid, add that the solution turns blue. Never write "hydrogen is given off" as an observation - you cannot see the gas, only the bubbles, so write effervescence.

    Worked example. Zinc and aluminium are both found as oxides. Which can be extracted by heating with carbon, and why? Find each metal's place relative to carbon in the reactivity series. Zinc sits below carbon, so carbon is the more reactive of the two and can take the oxygen away from it: zinc oxide is reduced by heating with carbon. Aluminium sits above carbon, so carbon is not reactive enough to displace it, and aluminium must be extracted by electrolysis instead - which is why aluminium costs far more to extract. The line in the series that decides the method is carbon's, not the top of the list.

    The special case of aluminium

    Aluminium seems less reactive than its position suggests. This is because it is covered by a thin, strong oxide layer 氧化层 that stops other substances reaching the metal underneath.

    日本語

    反応性系列は、金属を反応性の順に並べたものである。比較のために炭素と水素も含まれている:

    カリウム, ナトリウム, カルシウム, マグネシウム, アルミニウム, 炭素, 亜鉛, 鉄, 水素, 銅, 銀, 金

    (上部が最も反応性が高く、下部が最も反応性が低い)

    カリウムから金までの反応性系列で、炭素と水素を示し、炭素の左右でそれぞれ異なる抽出方法を示した図
    炭素より上の金属は電解法で抽出され、炭素より下の金属は炭素と加熱することで抽出できる

    金属が反応性系列で上位にあるほど、正のイオンになりやすく、これがその反応性を説明する:

    • カリウム、ナトリウム、カルシウムは冷水と反応する。
    • マグネシウムは蒸気と反応する(冷水とは非常にゆっくりしか反応しない)。
    • マグネシウム、亜鉛、鉄は希塩酸と反応する;銅、銀、金は反応しない。

    置換反応

    より反応性の高い金属は、そのイオンの溶液からより反応性の低い金属を置換する(置換反応)。例えば、亜鉛は銅より反応性が高いため、硫酸銅(II)溶液から銅を置換する:

    $$\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$$

    何が見えるかを説明できるよう準備しておくこと。これらは別々の採点ポイントだからだ:

    • 溶液の青色が薄れていく - 青色の$\text{Cu}^{2+}$イオンが使われている(硫酸亜鉛は無色)。
    • ピンク茶色の固体が亜鉛の表面に付着する - それが置換された銅である。
    • 灰色の亜鉛が溶解して小さくなる。
    • 混合物が温まる - 置換反応は発熱反応である。

    金属と酸の反応には独自の観察項目がある:泡立ち(水素の放出)と固体の溶解。酸化銅(II)や炭酸銅(II)と酸の反応については、溶液が青色になることも加える。「水素が出た」という記述は絶対に避けること。ガス自体は見えないので、泡だけが見える。したがって泡立ちと書くべきである。

    ** worked example. 亜鉛とアルミニウムはどちらも酸化物として存在する。どちらを炭素と加熱して抽出できるか、またなぜか? 反応性系列において各金属の炭素に対する位置を確認する。亜鉛は炭素の下にあるため、炭素の方がこの2つの中で反応性が高く、酸素を取り除くことができる:炭素と加熱することで酸化亜鉛は還元される。アルミニウムは炭素の上にあるため、炭素はそれを置換するだけの反応性を持たず、代わりにアルミニウムは電解法**で抽出されなければならない——これがアルミニウムの抽出コストが非常に高い理由である。方法を決定するのは系列の最上段ではなく、炭素のラインである。

    アルミニウムの特別なケース

    アルミニウムは反応性系列での位置よりも反応性が低く見える。これは、他の物質が下にある金属に到達するのを防ぐ、薄く強靭な酸化物層に覆われているためである。

    Explore · ⁨探索⁩

    Displacement and the reactivity series · ⁨置換反応と反応性系列⁩

    Step through a displacement. A more reactive metal pushes a less reactive one out of its compound — which is exactly what the reactivity series predicts. · ⁨置換反応の流れを確認しましょう。より反応性の高い金属は、より反応性の低い金属をその化合物から押し出します——これはまさに反応性系列が予測することです。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    reactivity series/rɪəkˈtɪvɪti ˈsɪəriːz/ 金属活動度順列
    potassium/pəˈtæsɪəm/ カリウム
    sodium/ˈsəʊdɪəm/ ナトリウム
    calcium/ˈkælsɪəm/ カルシウム
    magnesium/mæɡˈniːzɪəm/ マグネシウム
    ions/ˈaɪɒnz/ イオン
    displacement reaction/dɪˈspleɪsmənt rɪˈækʃn/ 置換反応
    oxide layer/ˈɒksaɪd ˈleɪə/ 酸化被膜
    Watch lesson · ⁨レッスンを視聴⁩
    9.5

    Corrosion of metals · ⁨金属の腐食⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State the conditions required for the rusting of iron and steel to form hydrated iron(III) oxide
    2 State some common barrier methods, including painting, greasing and coating with plastic 4 Describe the use of zinc in galvanising as an example of a barrier method and sacrificial protection
    3 Describe how barrier methods prevent rusting by excluding oxygen or water 5 Explain sacrificial protection in terms of the reactivity series and in terms of electron loss
    日本語
    コア サプリメント
    1 鉄および鋼が水化第三酸化鉄となるために必要な条件を記述せよ
    2 塗装、脂塗り、プラスチックコーティングなどの一般的なバリア方法を列挙する 4 亜鉛を用いたメッキをバリア方法および犠牲的保護の例として記述する
    3 バリア方法が酸素または水を排除することで錆びを防ぐ仕組みを記述する 5 反応性の順序および電子の損失の観点から、犠牲的保護を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Rusting 生锈 is the corrosion of iron and steel. Two things are needed for rusting: oxygen (from the air) and water. The rust 铁锈 formed is hydrated iron(III) oxide.

    Stopping rust

    Barrier methods 隔离法 keep oxygen and water away from the iron:

    • painting, greasing (covering with oil), and coating with plastic.

    Galvanising 镀锌 means coating iron with a layer of zinc. This works in two ways:

    • it is a barrier (the zinc keeps out air and water);
    • it gives sacrificial protection 牺牲保护. Because zinc is more reactive than iron, the zinc loses electrons 电子 and corrodes instead of the iron — even if the surface is scratched.
    日本語
    重度に腐食した金属の柱
    腐食は保護されていない金属を徐々に侵食していく。

    錆びとは、鉄および鋼の腐食のことである。錆びには2つの要素が必要である:酸素(空気中から)と水。生成される錆は水和酸化鉄(III)である。

    海岸の岩の上に置かれた重鉄鎖で、オレンジ茶色の錆が深く付着している写真
    海水と空気がこの鉄鎖を深刻に錆びさせた — その錆は水和酸化鉄(III)である
    試験管3本:水と空気の両方がある中の鉄釘、油で封じられた煮沸水中の鉄釘、乾燥剤入りの乾燥空気中の鉄釘
    錆びには水と酸素の両方が必要 — 鉄釘は両方を含む試験管内のみで錆びる

    錆びを防ぐ方法

    バリア方式により、鉄への酸素と水の接近を防ぐ:

    • 塗料塗布、油脂塗布(油で覆う)、プラスチックコーティング。

    亜鉛メッキとは、鉄の表面に亜鉛層を被覆することである。この方法は2つの働きをする:

    • バリアとなる(亜鉛が空気と水を通さない)。
    • 犠牲的保護を提供する。亜鉛は鉄より反応性が高いため、亜鉛が電子を失って腐食し、鉄の代わりに腐食する — 表面が傷ついても同様である。
    鉄棒に取り付けられた亜鉛のブロック、亜鉛から鉄へ電子が流れている様子
    より反応性の高い亜鉛が腐食し、電子を鉄に与えるため、表面が傷ついている箇所でも鉄は保護されます。
    Explore · ⁨探索⁩

    Why iron rusts — and how to stop it · ⁨鉄が錆びる理由と防止方法⁩

    Step through rusting. Iron needs BOTH water and oxygen, and once it starts the rust flakes off to expose fresh metal — which is why we protect iron. · ⁨錆びる流れを確認しましょう。鉄には水と酸素の両方が必要であり、一度錆び始めると錆が剥がれて新鮮な金属が露出するため、鉄を保護するのです。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    corrosion/kəˈrəʊʒn/ 腐食
    iron/ˈaɪən/ 鉄
    chromium/ˈkrəʊmɪəm/ クロム
    nickel/ˈnɪkl/ ニッケル
    carbon/ˈkɑːbən/ 炭素
    atoms/ˈætəmz/ 原子
    layers/ˈleɪəz/ レイヤー
    rusting/ˈrʌstɪŋ/ 錆び (rusting)
    rust/rʌst/ 錆
    barrier methods/ˈbærɪə ˈmeθədz/ バリア方式
    galvanising/ˈɡælvənaɪzɪŋ/ 亜鉛メッキ
    sacrificial protection/ˌsækrɪˈfɪʃl prəˈtekʃn/ 犠牲アノード保護
    electrons/ɪˈlektrɒnz/ 電子
    ore/ɔː/ 鉱石 (ore)
    electrolysis/ɪlekˈtrɒləsɪs/ 電気分解
    hematite/ˈhemətaɪt/ 赤鉄鉱
    9.6

    Extraction of metals · ⁨金属の抽出⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the ease in obtaining metals from their ores, related to the position of the metal in the reactivity series
    2 Describe the extraction of iron from hematite in the blast furnace, limited to: (a) the burning of carbon (coke) to provide heat and produce carbon dioxide (b) the reduction of carbon dioxide to carbon monoxide (c) the reduction of iron(III) oxide by carbon monoxide (d) the thermal decomposition of calcium carbonate/limestone to produce calcium oxide (e) the formation of slag Symbol equations are not required 4 State the symbol equations for the extraction of iron from hematite (a) $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ (b) $\text{C} + \text{CO}_2 \rightarrow 2\text{CO}$ (c) $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ (d) $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ (e) $\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3$
    3 State that the main ore of aluminium is bauxite and that aluminium is extracted by electrolysis 5 Describe the extraction of aluminium from purified bauxite/aluminium oxide, including: (a) the role of cryolite (b) why the carbon anodes need to be regularly replaced (c) the reactions at the electrodes, including ionic half-equations Details of the purification of bauxite are not required
    日本語
    コア サプリメント
    1 金属の反応性順序における位置と関連付け、鉱石からの金属の得やすさを記述せよ
    2 高炉におけるヘマタイトからの鉄の精製について、以下の範囲に限定して記述する:(a) 熱と二酸化炭素を生じるために炭素(コークス)を燃焼させる (b) 二酸化炭素を一酸化炭素に還元する (c) 一酸化炭素による三酸化二鉄の還元 (d) 二酸化カルシウム/石灰石の熱分解による酸化カルシウムの生成 (e) スラグの形成 化学方程式は不要である 4 ヘマタイトからの鉄の精製に対する化学方程式を述べる (a) $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ (b) $\text{C} + \text{CO}_2 \rightarrow 2\text{CO}$ (c) $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ (d) $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ (e) $\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3$
    3 アルミニウムの主要な鉱石はボクサイトであり、電解によって精製されることを述べる 5 精製されたボクサイト/酸化アルミニウムからのアルミニウムの精製について、以下の点を含めて記述する:(a) クリオリトの役割 (b) 炭素アノードが定期的な交換を必要とする理由 (c) 極での反応、イオン半反応式を含む ボクサイトの精製詳細は不要である

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    How a metal is taken from its ore 矿石 depends on its place in the reactivity series. Metals below carbon can be extracted by heating with carbon (which removes the oxygen). Metals above carbon are too reactive for this and must be extracted by electrolysis 电解.

    Iron from the blast furnace

    Iron is extracted from its ore hematite 赤铁矿 (iron(III) oxide) in a blast furnace 高炉:

    • Carbon (coke) burns in hot air to give carbon dioxide and lots of heat.
    • This carbon dioxide 二氧化碳 reacts with more carbon to form carbon monoxide 一氧化碳.
    • The carbon monoxide reduces the iron(III) oxide to iron (this is reduction 还原).
    • Limestone 石灰石 (calcium carbonate) breaks down in the heat to form calcium oxide 氧化钙.
    • The calcium oxide reacts with sandy impurities to form slag 炉渣, which is removed.

    Aluminium by electrolysis

    Aluminium is extracted from its ore bauxite 铝土矿 (purified to aluminium oxide) by electrolysis:

    • The aluminium oxide is dissolved in molten cryolite 冰晶石 to lower its melting point and save energy.
    • At the cathode 阴极, aluminium ions gain electrons to form aluminium metal.
    • At the anode 阳极, oxygen is formed. This oxygen reacts with the hot carbon anodes and burns them away, so they must be replaced regularly.
    日本語

    金属を鉱石から取り出す方法は、その金属が反応性系列における位置による。炭素より下の金属は炭素と加熱することで抽出できる(炭素が酸素を取り除く)。炭素より上の金属はこの方法では反応性が高すぎるため、電解法で抽出しなければならない。

    炭素より上の金属は電解法で、炭素より下の金属は炭素と加熱して抽出されることを示した図
    炭素より上位の金属は電気分解を要し、下位の金属は炭素とともに加熱する

    高炉からの鉄

    鉄は、鉱石であるヘマタイト(酸化鉄(III))を高炉で抽出する:

    • 炭素(コークス)が熱い空気中で燃焼し、二酸化炭素と大量の熱を生じる。
    • この二酸化炭素がさらに炭素と反応して一酸化炭素を生成する。
    • 一酸化炭素が酸化鉄(III)を鉄に還元する(これは還元反応である)。
    • 石灰石(炭酸カルシウム)が熱で分解して酸化カルシウムを生成する。
    • 酸化カルシウムが砂のような不純物と反応してスラグを生成し、これを除去する。
    ラベル付きの高炉:原料は上部から投入され、熱い空気は下部から吹き込まれ、溶融鉄とスラグは底部から排出される
    高炉では、一酸化炭素が鉄鉱石を還元し、石灰石が砂のような不純物をスラグとして除去する

    電解法によるアルミニウム

    アルミニウムは、その鉱石であるボーキサイト(精製して酸化アルミニウムとする)を電気分解によって抽出します:

    • 酸化アルミニウムの融点を下げてエネルギーを節約するため、溶融したクリオライトに溶解させます。
    • 陰極で、アルミニウムイオンが電子を受け取って金属アルミニウムとなります。
    • 陽極では酸素が発生します。この酸素は熱い炭素陽極と反応して燃やしてしまうため、定期的に交換する必要があります。
    Explore · ⁨探索⁩

    Extracting iron in the blast furnace · ⁨高炉による鉄の抽出⁩

    Iron is below carbon in reactivity, so carbon can reduce its ore to the metal. · ⁨鉄は炭素より反応性が低いため、炭素で鉱石を還元して金属を得ることができます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    blast furnace/blæst ˈfɜːnɪs/ 高炉
    carbon dioxide/ˈkɑːbən daɪˈɒksaɪd/ 二酸化炭素
    carbon monoxide/ˈkɑːbən mʌˈnɒksaɪd/ 一酸化炭素
    reduction/rɪˈdʌkʃn/ 還元
    limestone/ˈlaɪmstəʊn/ 石灰石
    calcium oxide/ˈkælsɪəm ˈɒksaɪd/ 酸化カルシウム
    slag/slæɡ/ スラグ
    bauxite/ˈbɔːksaɪt/ ボーキサイト
    cryolite/ˈkraɪəlaɪt/ クリオライト
    cathode/ˈkæθəʊd/ カソード
    anode/ˈænəʊd/ アノード
    effervescence/ˌefəˈvesəns/ 泡立ち
    9.6

    Exam tips · ⁨試験対策⁩

    English
    • Learn the reactivity series (potassium → gold). A more reactive metal displaces a less reactive one from a solution of its ions.
    • Metals above carbon are extracted by electrolysis; those below carbon (like iron) can be extracted by heating with carbon.
    • Rusting needs both water and oxygen. Barrier methods (paint, oil) keep them out; galvanising also gives sacrificial protection, because the more reactive zinc corrodes instead of the iron.
    • Alloys are harder than pure metals because the different-sized atoms stop the layers of ions sliding over each other.
    日本語
    • 活性度系列(カリウム → ールド)を学びましょう。より活性度の高い金属は、溶液から活性度の低い金属のイオンを置換します。
    • 炭素上にある金属は電気分解によって抽出され、炭素下にある金属(鉄など)は炭素と共に加熱することで抽出できます。
    • 錆びるには水と酸素の両方が必要です。バリア法(塗料、油)はこれらを遮断しますが、亜鉛メッキも犠牲的保護を提供します。これはより活性度の高い亜鉛が鉄の代わりに腐食するためです。
    • 合金は純金属よりも硬いです。異なるサイズの原子がイオンの層同士が滑り合うのを妨げるためです。
  • 10

    Chemistry of the environment

    Watch lesson · ⁨レッスンを視聴⁩
    10.1

    Water · ⁨水⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe chemical tests for the presence of water using anhydrous cobalt(II) chloride and anhydrous copper(II) sulfate
    2 Describe how to test for the purity of water using melting point and boiling point
    3 Explain that distilled water is used in practical chemistry rather than tap water because it contains fewer chemical impurities
    4 State that water from natural sources may contain substances, including: (a) dissolved oxygen (b) metal compounds (c) plastics (d) sewage (e) harmful microbes (f) nitrates from fertilisers (g) phosphates from fertilisers and detergents
    5 State that some of these substances are beneficial, including: (a) dissolved oxygen for aquatic life (b) some metal compounds provide essential minerals for life
    6 State that some of these substances are potentially harmful, including: (a) some metal compounds are toxic (b) some plastics harm aquatic life (c) sewage contains harmful microbes which cause disease (d) nitrates and phosphates lead to deoxygenation of water and damage to aquatic life Details of the eutrophication process are not required
    7 Describe the treatment of the domestic water supply in terms of: (a) sedimentation and filtration to remove solids (b) use of carbon to remove tastes and odours (c) chlorination to kill microbes
    日本語
    コア サプリメント
    1 無水コバルト(II)塩化物および無水銅(II)硫酸を用いた、水の存在を確認するための化学的試験を記述する
    2 融点および沸点を用いて、水の純度をテストする方法を記述する
    3 蒸留水が水道水よりも実験室で用いられるのは、化学的不純物が少ないためであることを説明する
    4 天然由来の水には以下の物質が含まれる可能性があることを述べる: (a) 溶解酸素 (b) 金属化合物 (c) プラスチック (d) 生活排水 (e) 有害微生物 (f) 肥料由来の硝酸塩 (g) 肥料および洗剤由来の燐酸塩
    5 これらの物質の一部は有益であり、以下を含むことを述べる: (a) 水生生物のための溶解酸素 (b) 一部の金属化合物は生命に必要なミネラルを提供する
    6 これらの物質の一部は潜在的に有害であり、以下を含むことを述べる: (a) 一部の金属化合物は毒性がある (b) 一部のプラスチックは水生生物を害する (c) 生活排水には疾病を引き起こす有害微生物が含まれる (d) 硝酸塩および燐酸塩は水中の脱酸素化と水生生物への被害をもたらす。富栄養化のプロセスの詳細は不要である
    7 家庭用給水処理を以下の観点から記述する: (a) 沈殿およびろ過による固形物の除去 (b) 味や臭いを除去するための炭素の使用 (c) 微生物を殺菌するための塩素添加

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Testing for water

    Two chemical tests show that water is present:

    • Anhydrous 无水 cobalt(II) chloride turns from blue to pink when water is added.
    • Anhydrous copper(II) sulfate turns from white to blue when water is added.

    These tests only show that water is there. To show that water is pure, you test its melting point 熔点 and boiling point 沸点: pure water melts at exactly $0\,{}^{\circ}\text{C}$ and boils at exactly $100\,{}^{\circ}\text{C}$. Any dissolved substance changes these values.

    This is why distilled water 蒸馏水 is used in chemistry instead of tap water — it has far fewer chemical impurities 杂质.

    What is in natural water

    Water from rivers, lakes and the sea is not pure. It may contain dissolved oxygen 氧气, metal compounds, plastics, sewage 污水, harmful microbes 微生物, and nitrates 硝酸盐 and phosphates 磷酸盐 (which come from fertilisers and detergents).

    Some of these are helpful:

    • dissolved oxygen lets aquatic life 水生生物 (fish and plants) breathe;
    • some metal compounds give essential minerals 矿物质.

    Others are harmful:

    • some metal compounds are toxic 有毒 (poisonous);
    • some plastics harm aquatic life;
    • sewage carries microbes that cause disease;
    • nitrates and phosphates cause deoxygenation 缺氧 (loss of oxygen) in the water, which harms aquatic life.

    Treating drinking water

    To make water safe to drink, the water supply is treated in steps:

    • sedimentation 沉降 and filtration 过滤 remove solid bits;
    • passing it through carbon 碳 removes bad tastes and smells;
    • chlorination 氯消毒 (adding chlorine) kills harmful microbes.
    日本語

    水の検出

    2つの化学試験により水が存在することを確認できます:

    • 無水コバルト(II)クロリドに水を加えると、青色から桃色に変化します。
    • 無水銅(II)硫酸塩に水を加えると、白色から青色に変化します。

    これらの試験は水があることを示すだけです。水が純粋であることを確認するには、融点と沸点を測定します:純水は正確に$0\,{}^{\circ}\text{C}$で融け、正確に$100\,{}^{\circ}\text{C}$で沸騰します。溶解している物質があればこれらの値は変化します。

    2つの色変化試験により水が存在することが示されます:コバルトクロリドは青から桃色へ、硫酸銅は白から青へ;固定された融点と沸点(0と100度)により純水であることが示される
    色変化試験は水が存在することを示し、固定された融点と沸点は純水であることを示します

    これが、化学実験において水道水ではなく蒸留水が使われる理由です——化学成分の不純物が非常に少ないためです。

    天然水中に含まれるもの

    河川、湖、海の水は純粋ではありません。溶解した酸素、金属化合物、プラスチック、下水、有害な微生物、そして窒素酸化物とリン酸塩(肥料や洗剤由来)を含んでいることがあります。

    これらのうち一部は有益です:

    • 溶解した酸素は水生生物(魚や植物)の呼吸を可能にします;
    • 一部の金属化合物は必須のミネラルを与えます。

    他のものは有害です:

    • 一部の金属化合物は有毒(毒物)です;
    • 一部のプラスチックは水生生物を害します;
    • 下水は疾病を引き起こす微生物を運びます;
    • 窒素酸化物とリン酸塩は水中の脱酸素(酸素の喪失)を引き起こし、水生生物を害します。

    飲料水の処理

    安全な飲料水にするために、給水システムは以下の手順で処理されます:

    • 沈殿とろ過により固形分を取り除きます;
    • 炭素を通すことで悪臭や異味を取り除きます;
    • 塩素添加(塩素添加)により有害な微生物を殺菌します。
    フローチャート:沈殿→ろ過→炭素通過→塩素添加による安全な飲料水の生成過程
    飲料水を安全にする手順:大きな粒子を沈殿させ、ろ過し、炭素で異味を取り除き、最後に塩素添加を行う
    Explore · ⁨探索⁩

    Water treatment lab · ⁨水処理実験⁩

    Classify water processes by what they remove or change. · ⁨除去または変換される物質に基づいて水処理プロセスを分類してください。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    anhydrous/ænˈhaɪdrəs/ 無水物(_anhydrous_)
    melting point/ˈmeltɪŋ pɔɪnt/ 融点
    boiling point/ˈbɔɪlɪŋ pɔɪnt/ 沸点
    distilled water/dɪˈstɪld ˈwɔːtə/ 蒸留水
    impurities/ɪmˈpjʊərɪtiz/ 不純物
    oxygen/ˈɒksɪdʒn/ 酸素
    sewage/ˈsuːɪdʒ/ 下水
    microbes/ˈmaɪkrəʊbz/ 微生物
    nitrates/ˈnaɪtreɪts/ 硝酸塩
    phosphates/ˈfɒsfeɪts/ リン酸塩
    aquatic life/əˈkwætɪk laɪf/ 水生生物
    minerals/ˈmɪnərəlz/ 無機塩類
    toxic/ˈtɒksɪk/ 有毒の
    deoxygenation/diːˌɒksɪdʒəˈneɪʃn/ 脱酸素化
    sedimentation/ˌsedɪmənˈteɪʃn/ 沈殿
    filtration/fɪlˈtreɪʃn/ ろ過
    carbon/ˈkɑːbən/ 炭素
    chlorination/ˌklɔːrɪˈneɪʃn/ 塩素殺菌
    fertilisers/ˈfɜːtəlaɪzəz/ 肥料
    nitrogen/ˈnaɪtrədʒn/ 窒素
    phosphorus/ˈfɒsfɔːrəs/ リン
    potassium/pəˈtæsɪəm/ カリウム
    noble gases/ˈnəʊbl ˈɡæsɪz/ 希ガス
    complete combustion/kəmˈpliːt kəmˈbʌstʃn/ 完全燃焼
    carbon monoxide/ˈkɑːbən mʌˈnɒksaɪd/ 一酸化炭素
    particulates/pəˈtɪkjʊleɪts/ 微粒子
    incomplete combustion/ɪŋkəmˈpliːt kəmˈbʌstʃn/ 不完全燃焼
    methane/ˈmiːθeɪn/ メタン
    oxides of nitrogen/ˈɒksaɪdz ɒv ˈnaɪtrədʒn/ 窒素酸化物
    fossil fuels/ˈfɒsl ˈfjuːəlz/ 化石燃料
    sulfur/ˈsʌlfɜː/ 硫黄
    carbon dioxide/ˈkɑːbən daɪˈɒksaɪd/ 二酸化炭素
    greenhouse gases/ˈɡriːnhaʊs ˈɡæsɪz/ 温室効果ガス
    global warming/ˈɡləʊbl ˈwɔːmɪŋ/ 地球温暖化
    climate change/ˈklaɪmət tʃeɪndʒ/ 気候変動
    respiratory/rɪˈspɪrətəri/ 呼吸器系
    cancer/ˈkænsə/ 癌
    acid rain/ˈæsɪd reɪn/ 酸性雨
    photochemical smog/ˌfəʊtəʊˈkemɪkl smɒɡ/ 光化学スモッグ
    renewable energy/rɪˈnjuːəbl ˈenədʒi/ 再生可能エネルギー
    catalytic converters/ˌkætəˈlɪtɪk kənˈvɜːtəz/ 触媒コンバーター
    flue gas desulfurisation/fluː ɡæs diːˌsʌlfjʊəraɪˈzeɪʃn/ 排ガス脱硫
    calcium oxide/ˈkælsɪəm ˈɒksaɪd/ 酸化カルシウム
    thermal energy/ˈθɜːml ˈenədʒi/ 熱エネルギーに変換する
    photosynthesis/ˌfəʊtəʊˈsɪnθəsɪs/ 光合成
    chlorophyll/ˌklɔːrəʊˈfɪl/ クロロフィル
    glucose/ˈɡluːkəʊs/ ブドウ糖
    10.2

    Fertilisers · ⁨肥料⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State that ammonium salts and nitrates are used as fertilisers
    2 Describe the use of NPK fertilisers to provide the elements nitrogen, phosphorus and potassium for improved plant growth
    日本語
    コア サプリメント
    1 アモニア塩類および硝酸塩が肥料として用いられることを述べる
    2 NPK肥料が窒素、リン、カリウムの元素を提供して植物の成長を向上させるために使用されることを記述する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Fertilisers 肥料 are added to soil to help plants grow. Ammonium salts and nitrates are common fertilisers.

    NPK fertilisers contain the three elements plants need most: nitrogen 氮气, phosphorus 磷 and potassium 钾.

    日本語

    肥料は土壌に追加され、植物の生育を助けます。アンモニウム塩と硝酸塩は一般的な肥料です。

    NPK肥料は植物が最も必要とする3つの要素を含む:窒素、リン、カリウム。

    NPKの3枚のカード:Nは葉と茎の成長のための窒素、Pは強い根のためのリン、Kは花と果実のためのカリウム
    NPK肥料は植物が最も必要とする3つの要素(窒素、リン、カリウム)を供給する
    Explore · ⁨探索⁩

    Fertiliser route lab · ⁨肥料ルート実験⁩

    Follow nitrogen from raw materials to crop growth. · ⁨原料からの窒素から作物の成長までの経路を追跡する。⁩

    10.3

    Air quality and climate · ⁨空気質と気候⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State the composition of clean, dry air as approximately 78% nitrogen, $\text{N}_2$, 21% oxygen, $\text{O}_2$ and the remainder as a mixture of noble gases and carbon dioxide, $\text{CO}_2$
    2 State the source of each of these air pollutants, limited to: (a) carbon dioxide from the complete combustion of carbon-containing fuels (b) carbon monoxide and particulates from the incomplete combustion of carbon-containing fuels (c) methane from the decomposition of vegetation and waste gases from digestion in animals (d) oxides of nitrogen from car engines (e) sulfur dioxide from the combustion of fossil fuels which contain sulfur compounds
    3 State the adverse effect of these air pollutants, limited to: (a) carbon dioxide: higher levels of carbon dioxide leading to increased global warming, which leads to climate change (b) carbon monoxide: toxic gas (c) particulates: increased risk of respiratory problems and cancer (d) methane: higher levels of methane leading to increased global warming, which leads to climate change (e) oxides of nitrogen: acid rain, photochemical smog and respiratory problems (f) sulfur dioxide: acid rain 7 Describe how the greenhouse gases carbon dioxide and methane cause global warming, limited to: (a) the absorption, reflection and emission of thermal energy (b) reducing thermal energy loss to space
    4 State and explain strategies to reduce the effects of these environmental issues, limited to: (a) climate change: planting trees, reduction in livestock farming, decreasing use of fossil fuels, increasing use of hydrogen and renewable energy, e.g. wind, solar (b) acid rain: use of catalytic converters in vehicles, reducing emissions of sulfur dioxide by using low-sulfur fuels and flue gas desulfurisation with calcium oxide 8 Explain how oxides of nitrogen form in car engines and describe their removal by catalytic converters, e.g. $2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2$
    5 Describe photosynthesis as the reaction between carbon dioxide and water to produce glucose and oxygen in the presence of chlorophyll and using energy from light
    6 State the word equation for photosynthesis, carbon dioxide + water $\rightarrow$ glucose + oxygen 9 State the symbol equation for photosynthesis, $6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$
    日本語
    コア サプリメント
    1 清浄で乾燥した空気の組成が、約78%の窒素、$\text{N}_2$、21%の酸素、$\text{O}_2$、残りが希ガスおよび二酸化炭素の混合物である$\text{CO}_2$ことを述べる
    2 以下の大気汚染物質の各々の起源を述べる(ただし範囲は限定する): (a) 炭素含有燃料の完全燃焼からの二酸化炭素 (b) 炭素含有燃料の不完全燃焼からの一酸化炭素および微粒子 (c) 植物および廃棄物の分解、動物の消化からのメタン (d) 自動車エンジンからの窒素酸化物 (e) 硫黄化合物を含む化石燃料の燃焼からの二酸化硫黄
    3 以下の大気汚染物質の悪影響を述べる(ただし範囲は限定する): (a) 二酸化炭素:高濃度の二酸化炭素が地球温暖化を促進し、結果として気候変動をもたらす (b) 一酸化炭素:有毒ガス (c) 微粒子:呼吸器系疾患およびがんのリスク増加 (d) メタン:高濃度のメタンが地球温暖化を促進し、結果として気候変動をもたらす (e) 窒素酸化物:酸雨、光化学スモッグ、呼吸器系疾患 (f) 二酸化硫黄:酸雨 7 温室効果ガスである二酸化炭素およびメタンが地球温暖化を引き起こす仕組みを説明する(ただし範囲は限定する): (a) 熱エネルギーの吸収、反射、放射 (b) 宇宙への熱エネルギー損失の減少
    4 環境問題の影響を軽減するための戦略を、以下の範囲に限定して述べよ:(a) 気候変動:樹木植栽、家畜飼育の削減、化石燃料の使用減少、水素および再生可能エネルギー(例:風力、太陽光)の利用増加 (b) 酸性雨:車両における触媒コンバーターの使用、低硫黄燃料および酸化カルシウムを用いた排ガス脱硫による二酸化硫黄の排出削減 8 窒素酸化物が自動車エンジン内で生成される仕組みと、触媒コンバーターによって除去されるプロセスを説明せよ。例: $2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2$
    5 光合成を、葉緑素の存在下で光エネルギーを用い、二酸化炭素と水からブドウ糖と酸素を生成する反応として記述せよ
    6 光合成の語式方程式を述べよ。二酸化炭素 + 水 $\rightarrow$ ブドウ糖 + 酸素 9 光合成の化学式方程式を述べよ。 $6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Clean, dry air is approximately:

    • 78% nitrogen ($\text{N}_2$)
    • 21% oxygen ($\text{O}_2$)
    • the rest is a mixture of noble gases 稀有气体 and carbon dioxide ($\text{CO}_2$).

    Air pollutants and their sources

    Pollutant Main source
    carbon dioxide ($\text{CO}_2$) complete combustion 完全燃烧 of carbon-containing fuels
    carbon monoxide 一氧化碳 and particulates 颗粒物 incomplete combustion 不完全燃烧 of carbon-containing fuels
    methane 甲烷 rotting plants and waste gases from animal digestion
    oxides of nitrogen 氮氧化物 car engines
    sulfur dioxide burning fossil fuels 化石燃料 that contain sulfur 硫

    Effects of these pollutants

    • Carbon dioxide 二氧化碳 and methane are greenhouse gases 温室气体: more of them causes global warming 全球变暖, which leads to climate change 气候变化.
    • Carbon monoxide is a toxic gas.
    • Particulates increase the risk of respiratory 呼吸 (breathing) problems and cancer 癌症.
    • Oxides of nitrogen cause acid rain 酸雨, photochemical smog 光化学烟雾 and breathing problems.
    • Sulfur dioxide causes acid rain.

    Worked example. A power station burns coal containing sulfur, and a lake downwind slowly turns acidic. Name the pollutant, explain the link, and give one way to stop it. Burning a fossil fuel that contains sulfur releases sulfur dioxide, which causes acid rain, so the rain falling on the lake is acidic. To stop it at the source, use flue gas desulfurisation, where calcium oxide removes the sulfur dioxide from the waste gases before they leave the chimney, or switch to a low-sulfur fuel. Match the fix to the pollutant: a catalytic converter treats a car's oxides of nitrogen and would do nothing about a coal-fired power station's sulfur dioxide.

    Reducing these problems

    To slow climate change: plant trees, farm fewer animals, burn fewer fossil fuels, and use more hydrogen and renewable energy 可再生能源 such as wind and solar power.

    To reduce acid rain: fit catalytic converters 催化转化器 in cars, use low-sulfur fuels, and use flue gas desulfurisation 烟气脱硫 with calcium oxide 氧化钙 to remove sulfur dioxide from waste gases.

    How greenhouse gases warm the Earth

    Greenhouse gases such as carbon dioxide and methane let sunlight through, but they absorb the thermal energy 热能 given off by the warm Earth, and send some of it back down. This reduces the thermal energy lost to space, so the Earth gets warmer.

    In a car engine, the high temperature makes nitrogen and oxygen from the air react to form oxides of nitrogen. A catalytic converter removes them, for example:

    $$2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2$$

    Photosynthesis

    Photosynthesis 光合作用 is the opposite of combustion — it removes carbon dioxide from the air. Plants use light energy and chlorophyll 叶绿素 to turn carbon dioxide and water into glucose 葡萄糖 and oxygen:

    $$6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$$
    日本語
    排気を出す発電所
    化石燃料の燃焼は大気を汚染し、気候を変化させます。

    清浄で乾燥した空気のおよそ組成は:

    • 窒素 78% ($\text{N}_2$)
    • 酸素 21% ($\text{O}_2$)
    • 残りは希ガスと二酸化炭素の混合物 ($\text{CO}_2$)。
    清浄乾燥空気の円グラフ:大きな窒素の切り口(78%)、酸素の切り口(21%)、残りの薄い切り口(1%)
    清浄で乾燥した空気は約78%の窒素、21%の酸素、1%の他のガス(希ガスと二酸化炭素)で構成される

    大気汚染物質とその発生源

    汚染物質 主な発生源
    二酸化炭素 ($\text{CO}_2$) 炭素を含む燃料の完全燃焼
    一酸化炭素と微粒子 炭素を含む燃料の不完全燃焼
    メタン 腐敗した植物と動物の消化による廃気
    窒素酸化物 自動車エンジン
    二酸化硫黄 硫黄を含む化石燃料の燃焼

    これらの汚染物質の影響

    • 二酸化炭素とメタンは温室効果ガスです:それらが増えると地球温暖化が進行し、気候変動を引き起こします。
    • 一酸化炭素は有毒なガスです。
    • 微粒子は呼吸器系(呼吸)の問題およびがんのリスクを増加させます。
    • 窒素酸化物は酸性雨、光化学スモッグおよび呼吸器系の問題を引き起こします。
    • 二酸化硫黄は酸性雨を引き起こします。

    ** worked example.** 硫黄を含む石炭を燃やす発電所があり、下風側の湖がゆっくりと酸性化している。汚染物質の名前を答え、その因果関係を説明し、対策を一つ挙げよ。硫黄を含む化石燃料を燃やすと二酸化硫黄が放出され、酸雨の原因となるため、湖に降る雨は酸性になる。発生源で防ぐには、煙突から排出される前に排ガス中の二酸化硫黄を除去する**排ガス脱硫(フリューガス脱硫)**を使用するか、低硫黄燃料に切り替える。対策と汚染物質を結びつけよ: 触媒コンバーターは自動車の窒素酸化物を処理するため、石炭火力発電所の二酸化硫黄には全く効果がない。

    これらの問題の軽減

    気候変動を緩やかにするには: 木を植える、家畜の飼育数を減らす、化石燃料の燃焼を減らす、そして風力や太陽光などの水素および再生可能エネルギーをより多く利用する。

    酸雨を減少させるには: 自動車に触媒コンバーターを取り付け、低硫黄燃料を使用し、排ガス中の二酸化硫黄を除去するために二酸化カルシウムを用いた排ガス脱硫を行う。

    触媒コンバーターから取り出したセラミックハニカムコア(数千もの微細なチャンネルが見える)
    触媒コンバーター内部には触媒金属でコーティングされたハニカムがあり、多くの微細なチャンネルにより巨大な表面積が得られている

    温室効果ガスが地球を暖める仕組み

    二酸化炭素やメタンなどの温室効果ガスは日光を通すが、暖まった地球から放射される熱エネルギーを吸収し、その一部を下方へ反射する。これにより宇宙へ失われる熱エネルギーが減少するため、地球は暖まる。

    太陽光が大気圏を通過して地球を温め、地球から熱が放出されます;温室効果ガスはその熱の一部を吸収して下方へ反射します
    温室効果ガスは日光を通しますが、地球から放出される熱の一部を閉じ込めるため、地球は暖まります

    自動車エンジンでは、高温により空気中の窒素と酸素が反応して窒素酸化物を形成します。触媒コンバーターはこれを除去します。例示如下:

    $$2\text{CO} + 2\text{NO} \rightarrow 2\text{CO}_2 + \text{N}_2$$

    光合成

    光合成は燃焼の逆プロセスであり、大気中の二酸化炭素を除去します。植物は光エネルギーとクロロフィルを使って、二酸化炭素と水をブドウ糖と酸素に変換します:

    $$6\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2$$
    Explore · ⁨探索⁩

    Air pollutant lab · ⁨大気汚染物質実験⁩

    Classify air gases by source and environmental effect. · ⁨大気ガスを発生源と環境への影響に基づいて分類する。⁩

    Explore · ⁨探索⁩

    Climate and air quality route · ⁨気候と大気品質ルート⁩

    Trace emissions from combustion to atmospheric effects. · ⁨燃焼からの微量排出を大気への影響まで追跡する。⁩

    Watch lesson · ⁨レッスンを視聴⁩
    10.3

    Exam tips · ⁨試験対策⁩

    English
    • Two tests show water is present (cobalt(II) chloride blue → pink; copper(II) sulfate white → blue), but only fixed melting and boiling points (0 °C and 100 °C) show it is pure.
    • Clean, dry air is about 78% nitrogen, 21% oxygen and 1% other gases (mostly noble gases and carbon dioxide).
    • Match each pollutant to its source and effect: carbon monoxide (incomplete combustion; toxic); sulfur dioxide and oxides of nitrogen (acid rain); carbon dioxide and methane (greenhouse gases → global warming).
    • Greenhouse gases let sunlight through but absorb the heat the warm Earth gives off and send some back down, so the Earth warms up.
    日本語
    • 2つの試験で水が存在することを確認できます(コバルト(II)塩化物:青→ピンク;硫酸銅(II):白→青)、ただし純度を示すのは固定された融点・沸点のみ(0 °C および 100 °C)です。
    • 清浄で乾燥した空気は約窒素78%、酸素21%、および他のガス1%(主に希ガスと二酸化炭素)から構成されています。
    • 各汚染物質をその発生源と影響に合わせます:一酸化炭素(不完全燃焼;有毒);二酸化硫黄および窒素酸化物(酸雨);二酸化炭素およびメタン(温室効果ガス → 温暖化)。
    • 温室効果ガスは太陽光を通すが、暖まった地球から放射される熱を吸収し、一部を地表に戻すため、地球は暖まる。
  • 11

    Organic chemistry

    Watch lesson · ⁨レッスンを視聴⁩

    Organic chemistry is the chemistry of carbon compounds. Carbon is special because it can join to other carbon atoms to make long chains and rings.

    11.1

    Formulae and key words

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Draw and interpret the displayed formula of a molecule to show all the atoms and all the bonds
    2 Write and interpret general formulae of compounds in the same homologous series, limited to: (a) alkanes, $\text{C}_n\text{H}_{2n+2}$ (b) alkenes, $\text{C}_n\text{H}_{2n}$ (c) alcohols, $\text{C}_n\text{H}_{2n+1}\text{OH}$ (d) carboxylic acids, $\text{C}_n\text{H}_{2n+1}\text{COOH}$
    3 Identify a functional group as an atom or group of atoms that determine the chemical properties of a homologous series
    7 State that a structural formula is an unambiguous description of the way the atoms in a molecule are arranged, including $\text{CH}_2=\text{CH}_2$, $\text{CH}_3\text{CH}_2\text{OH}$, $\text{CH}_3\text{COOCH}_3$
    8 Define structural isomers as compounds with the same molecular formula, but different structural formulae, including $\text{C}_4\text{H}_{10}$ as $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ and $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_3$ and $\text{C}_4\text{H}_8$ as $\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2$ and $\text{CH}_3\text{CH}=\text{CHCH}_3$
    4 State that a homologous series is a family of similar compounds with similar chemical properties due to the presence of the same functional group 9 Describe the general characteristics of a homologous series as: (a) having the same functional group (b) having the same general formula (c) differing from one member to the next by a –CH2– unit (d) displaying a trend in physical properties (e) sharing similar chemical properties
    5 State that a saturated compound has molecules in which all carbon–carbon bonds are single bonds
    6 State that an unsaturated compound has molecules in which one or more carbon–carbon bonds are not single bonds
    日本語
    コア サプリメント
    1 分子の表示式を描き、すべての原子とすべての結合を示すように解釈せよ
    2 同族系列にある化合物の一般式を書き、解釈せよ。ただし、以下のものに限る:(a) アルカンの場合、$\text{C}_n\text{H}_{2n+2}$ (b) アルケンの場合、$\text{C}_n\text{H}_{2n}$ (c) アルコールの場合、$\text{C}_n\text{H}_{2n+1}\text{OH}$ (d) カルボン酸の場合、$\text{C}_n\text{H}_{2n+1}\text{COOH}$
    3 官能基とは、同族系列の化学的性質を決定する原子または原子の集団であることを特定せよ
    7 構造式とは、分子内の原子の配置の曖昧でない記述であり、$\text{CH}_2=\text{CH}_2$、$\text{CH}_3\text{CH}_2\text{OH}$、$\text{CH}_3\text{COOCH}_3$ を含むことを述べよ
    8 異性体(構造異性体)とは、同じ分子式を持ちながら異なる構造式を持つ化合物であること定義せよ。具体的には、$\text{C}_4\text{H}_{10}$ が $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ および $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_3$ であり、$\text{C}_4\text{H}_8$ が $\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2$ および $\text{CH}_3\text{CH}=\text{CHCH}_3$ となることを含む
    4 同族系列とは、同じ官能基が存在するため、化学的性質が類似した化合物のファミリーであること述べよ 9 同族系列の一般的な特徴として以下を説明せよ:(a) 同じ官能基を持つ (b) 同じ一般式を持つ (c) 各成分同士は –CH2– 単位で異なる (d) 物理的性質に傾向が見られる (e) 類似した化学的性質を共有する
    5 飽和化合物とは、炭素-炭素結合がすべて単結合である分子を持つ化合物であること述べよ
    6 不飽和化合物とは、炭素-炭素結合の少なくとも一つが単結合ではない分子を持つ化合物であること述べよ

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    There are several ways to write an organic molecule:

    • The molecular formula 分子式 shows how many of each atom there are, for example $\text{C}_2\text{H}_6$.
    • The displayed formula 结构式 shows every atom and every bond drawn out in full.
    • The structural formula 结构简式 shows how the atoms are arranged without drawing every bond, for example $\text{CH}_3\text{CH}_2\text{OH}$.
    • The general formula 通式 works for a whole family, for example $\text{C}_n\text{H}_{2n+2}$ for alkanes.

    A homologous series 同系物 is a family of compounds with the same functional group 官能团 — the atom or group of atoms that gives the family its chemical properties. Members of a series have the same general formula, differ by a $\text{CH}_2$ unit each step, and have similar chemical properties with a gradual change in physical properties.

    Series Functional group General formula
    alkanes C–C single bonds only $\text{C}_n\text{H}_{2n+2}$
    alkenes C=C double bond $\text{C}_n\text{H}_{2n}$
    alcohols –OH $\text{C}_n\text{H}_{2n+1}\text{OH}$
    carboxylic acids –COOH $\text{C}_n\text{H}_{2n+1}\text{COOH}$

    A saturated 饱和 compound has only single carbon–carbon bonds. An unsaturated 不饱和 compound has one or more carbon–carbon bonds that are not single (such as a C=C double bond).

    Structural isomers 同分异构体 are compounds with the same molecular formula but different structural formulae. For example, $\text{C}_4\text{H}_{10}$ can be a straight chain or a branched chain.

    日本語

    There are several ways to write an organic molecule:

    • The molecular formula 分子式 shows how many of each atom there are, for example $\text{C}_2\text{H}_6$.
    • The displayed formula 结构式 shows every atom and every bond drawn out in full.
    • The structural formula 结构简式 shows how the atoms are arranged without drawing every bond, for example $\text{CH}_3\text{CH}_2\text{OH}$.
    • The general formula 通式 works for a whole family, for example $\text{C}_n\text{H}_{2n+2}$ for alkanes.

    A homologous series 同系物 is a family of compounds with the same functional group 官能团 — the atom or group of atoms that gives the family its chemical properties. Members of a series have the same general formula, differ by a $\text{CH}_2$ unit each step, and have similar chemical properties with a gradual change in physical properties.

    Series Functional group General formula
    alkanes C–C single bonds only $\text{C}_n\text{H}_{2n+2}$
    alkenes C=C double bond $\text{C}_n\text{H}_{2n}$
    alcohols –OH $\text{C}_n\text{H}_{2n+1}\text{OH}$
    carboxylic acids –COOH $\text{C}_n\text{H}_{2n+1}\text{COOH}$
    Displayed formulae of methane, ethane and propane, each one CH2 unit longer than the last
    The alkanes are a homologous series: each member has one more $\text{CH}_2$ unit (general formula $\text{C}_n\text{H}_{2n+2}$)

    A saturated 饱和 compound has only single carbon–carbon bonds. An unsaturated 不饱和 compound has one or more carbon–carbon bonds that are not single (such as a C=C double bond).

    Ethane with a single carbon-carbon bond next to ethene with a carbon-carbon double bond
    A saturated compound has only single C–C bonds; an unsaturated one has a C=C double bond

    Structural isomers 同分异构体 are compounds with the same molecular formula but different structural formulae. For example, $\text{C}_4\text{H}_{10}$ can be a straight chain or a branched chain.

    Butane drawn as a straight chain of four carbons beside 2-methylpropane drawn as a branched structure
    Structural isomers: butane and 2-methylpropane share the formula $\text{C}_4\text{H}_{10}$ but have different structures
    Explore · ⁨探索⁩

    Functional group lab · ⁨官能基実験室⁩

    Classify organic molecules by the group that controls their reactions. · ⁨有機分子を、その反応を支配する官能基に基づいて分類せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    molecular formula/məˈlekjʊlə ˈfɔːmjʊlə/ 分子式
    displayed formula/dɪˈspleɪd ˈfɔːmjʊlə/ 展開式
    structural formula/ˈstrʌktʃərəl ˈfɔːmjʊlə/ 構造式
    general formula/ˈdʒenərəl ˈfɔːmjʊlə/ 一般式
    homologous series/həˈmɒləɡəs ˈsɪəriːz/ writings系列
    functional group/ˈfʌŋkʃənl ɡruːp/ 官能基
    saturated/ˈsætʃəreɪtɪd/ 飽和溶液
    unsaturated/ʌnˈsætʃəreɪtɪd/ 不飽和溶液
    Structural isomers/ˈstrʌktʃərəl ˈaɪsəməz/ 構造異性体
    11.2

    Naming organic compounds

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Name and draw the displayed formulae of: (a) methane and ethane (b) ethene (c) ethanol (d) ethanoic acid (e) the products of the reactions stated in sections 11.4–11.7 3 Name and draw the structural and displayed formulae of unbranched: (a) alkanes (b) alkenes, including but-1-ene and but-2-ene (c) alcohols, including propan-1-ol, propan-2-ol, butan-1-ol and butan-2-ol (d) carboxylic acids containing up to four carbon atoms per molecule
    2 State the type of compound present, given a chemical name ending in -ane, -ene, -ol, or -oic acid or from a molecular formula or displayed formula 4 Name and draw the displayed formulae of the unbranched esters which can be made from unbranched alcohols and carboxylic acids, each containing up to four carbon atoms
    日本語
    コア サプリメント
    1 名称を述べ、表示式を描け:(a) メタンおよびエタン (b) エチレン (c) エタノール (d) エタノ酸 (e) 11.4〜11.7節で示された反応の生成物 3 直鎖(分岐のない)化合物の名称を述べ、構造式および表示式を描け:(a) アルカン (b) アルケン(ブテン-1-エンおよびブテン-2-エンを含む) (c) アルコール(プロパン-1-オール、プロパン-2-オール、ブタン-1-オールおよびブタン-2-オールを含む) (d) 分子あたりの炭素数が最大4個のカルボン酸
    2 化学名が -ane, -ene, -ol, または -oic acid で終わっている場合、または分子式・表示式から与えられた場合、存在する化合物の種類を述べよ 4 直鎖アルコールおよびカルボン酸から合成可能な直鎖エステル(それぞれ分子あたりの炭素数は最大4個)の名称を述べ、表示式を描け

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    The end of the name tells you the family:

    Ending Family Example
    -ane alkane methane, ethane
    -ene alkene ethene
    -ol alcohol ethanol
    -oic acid carboxylic acid ethanoic acid

    The start of the name tells you the number of carbon atoms: meth- = 1, eth- = 2, prop- = 3, but- = 4. For longer alkenes and alcohols, a number shows where the functional group is, for example but-1-ene and but-2-ene, or propan-1-ol and propan-2-ol.

    Explore · ⁨探索⁩

    Organic naming route · ⁨有機命名法の手順⁩

    Follow the route from structure to systematic name. · ⁨構造から系統名への手順に従う。⁩

    11.3

    Fuels

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Name the fossil fuels: coal, natural gas and petroleum
    2 Name methane as the main constituent of natural gas
    3 State that hydrocarbons are compounds that contain hydrogen and carbon only
    4 State that petroleum is a mixture of hydrocarbons
    5 Describe the separation of petroleum into useful fractions by fractional distillation
    6 Describe how the properties of fractions obtained from petroleum change from the bottom to the top of the fractionating column, limited to: (a) decreasing chain length (b) higher volatility (c) lower boiling points (d) lower viscosity
    7 Name the uses of the fractions as: (a) refinery gas fraction for gas used in heating and cooking (b) gasoline/petrol fraction for fuel used in cars (c) naphtha fraction as a chemical feedstock (d) kerosene/paraffin fraction for jet fuel (e) diesel oil/gas oil fraction for fuel used in diesel engines (f) fuel oil fraction for fuel used in ships and home heating systems (g) lubricating oil fraction for lubricants, waxes and polishes (h) bitumen fraction for making roads
    日本語
    コア サプリメント
    1 化石燃料の名称を述べよ:石炭、天然ガス、石油
    2 天然ガスの主な構成成分であるメタンの名称を述べよ
    3 炭化水素とは、水素と炭素のみからなる化合物であること述べよ
    4 石油は炭化水素の混合物であること述べよ
    5 石油を有用な画分に分ける蒸留分離(分餾)について記述せよ
    6 分餾塔の下から上へ向かって得られる画分の性質がどのように変化するかを、以下の点に限定して記述せよ:(a) 炭素鎖の長さの減少 (b) 揮発性の向上 (c) 沸点の低下 (d) 粘度の低下
    7 画分の用途を以下のように述べよ:(a) 精製ガス画分:暖房や調理用ガスとして (b) ガソリン画分:自動車の燃料として (c) ナフサ画分:化学原料として (d) ケロシン画分:ジェット燃料として (e) ディーゼル油画分:ディーゼルエンジンの燃料として (f) 燃料油画分:船舶および家庭用暖房システムの燃料として (g) 潤滑油画分:潤滑剤、、ポリッシュとして (h) ビトumen画分:道路舗装材として

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Fractional distillation of crude oil

    The three fossil fuels 化石燃料 are coal 煤, natural gas 天然气 and petroleum 石油 (crude oil). Methane 甲烷 is the main part of natural gas.

    A hydrocarbon 碳氢化合物 is a compound made of hydrogen and carbon only. Petroleum is a mixture of many different hydrocarbons.

    Fractional distillation

    Petroleum is separated into useful fractions 馏分 by fractional distillation 分馏. The mixture is heated, and the different hydrocarbons turn to gas and rise up a tall fractionating column 分馏塔. The column is hot at the bottom and cool at the top, so each fraction turns back to liquid at a different height.

    Going from the bottom to the top of the column, the fractions have:

    • shorter chain length 链长 (smaller molecules);
    • higher volatility 挥发性 (they turn to gas more easily);
    • lower boiling points;
    • lower viscosity 黏度 (they flow more easily).
    Fraction Use
    refinery gas gas for heating and cooking
    gasoline / petrol fuel for cars
    naphtha raw material for making chemicals
    kerosene / paraffin jet fuel
    diesel oil fuel for diesel engines
    fuel oil fuel for ships and home heating
    lubricating oil lubricants, waxes and polishes
    bitumen 沥青 making roads
    日本語
    Fractional distillation of crude oil
    An oil refinery at dusk
    An oil refinery separates crude oil into useful fuels by fractional distillation.

    The three fossil fuels 化石燃料 are coal 煤, natural gas 天然气 and petroleum 石油 (crude oil). Methane 甲烷 is the main part of natural gas.

    A hydrocarbon 碳氢化合物 is a compound made of hydrogen and carbon only. Petroleum is a mixture of many different hydrocarbons.

    Fractional distillation

    Petroleum is separated into useful fractions 馏分 by fractional distillation 分馏. The mixture is heated, and the different hydrocarbons turn to gas and rise up a tall fractionating column 分馏塔. The column is hot at the bottom and cool at the top, so each fraction turns back to liquid at a different height.

    A fractionating column with fractions drawn off at different heights, hottest at the bottom and coolest at the top
    Fractions separate by boiling point: small molecules leave the cool top, thick bitumen stays at the hot bottom

    Going from the bottom to the top of the column, the fractions have:

    • shorter chain length 链长 (smaller molecules);
    • higher volatility 挥发性 (they turn to gas more easily);
    • lower boiling points;
    • lower viscosity 黏度 (they flow more easily).
    Fraction Use
    refinery gas gas for heating and cooking
    gasoline / petrol fuel for cars
    naphtha raw material for making chemicals
    kerosene / paraffin jet fuel
    diesel oil fuel for diesel engines
    fuel oil fuel for ships and home heating
    lubricating oil lubricants, waxes and polishes
    bitumen 沥青 making roads
    Explore · ⁨探索⁩

    Fuel chain lab · ⁨燃料鎖実験⁩

    Follow a fuel from source to combustion products. · ⁨原料から燃焼生成物まで燃料を追跡せよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    fossil fuels/ˈfɒsl ˈfjuːəlz/ 化石燃料
    coal/kəʊl/ 石炭
    natural gas/ˈnætʃərəl ɡæs/ 天然ガス
    petroleum/pəˈtrəʊliːəm/ 石油
    Methane/ˈmiːθeɪn/ メタン
    hydrocarbon/ˈhaɪdrəkɑːbən/ 炭化水素
    fractions/ˈfrækʃnz/ 画分
    fractional distillation/ˈfrækʃənl dɪstɪˈleɪʃn/ 分餾
    fractionating column/ˈfrækʃənətɪŋ ˈkɒlʌm/ 蒸留塔
    chain length/tʃeɪn leŋθ/ 鎖長
    volatility/ˌvɒləˈtɪlɪti/ 揮発性
    viscosity/vɪˈskɒsɪti/ 粘性
    bitumen/ˈbɪtʃuːmən/ ビチューメン
    Watch lesson · ⁨レッスンを視聴⁩
    11.4

    Alkanes

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State that the bonding in alkanes is single covalent and that alkanes are saturated hydrocarbons
    2 Describe the properties of alkanes as being generally unreactive, except in terms of combustion and substitution by chlorine 3 State that in a substitution reaction one atom or group of atoms is replaced by another atom or group of atoms
    4 Describe the substitution reaction of alkanes with chlorine as a photochemical reaction, with ultraviolet light providing the activation energy, $E_a$, and draw the structural or displayed formulae of the products, limited to monosubstitution
    日本語
    コア サプリメント
    1 アルカンの結合は単結合共有結合であり、アルカンは飽和炭化水素であること述べよ
    2 アルカンの性質として、燃焼および塩素による置換反応を除き一般的に不活性であることを記述せよ 3 置換反応では、ある原子または原子の集団が別の原子または原子の集団に置き換わることを述べよ
    4 アルカンの塩素との置換反応を光化学反応として記述し、紫外線が活性化エネルギーを提供すること、$E_a$、そして生成物の構造式または表示式を描くこと。ただし、一置換反応に限定する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Alkanes 烷烃 have only single covalent bonds, so they are saturated hydrocarbons. They are generally unreactive. Their two important reactions are:

    • Combustion 燃烧: they burn in plenty of oxygen to give carbon dioxide and water.
    • Substitution with chlorine.

    In a substitution reaction 取代反应, one atom (or group of atoms) is replaced by another. Alkanes react with chlorine only in ultraviolet light 紫外线 — this is a photochemical reaction 光化学反应, where the light provides the activation energy 活化能. For example:

    $$\text{CH}_4 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{HCl}$$
    日本語

    Alkanes 烷烃 have only single covalent bonds, so they are saturated hydrocarbons. They are generally unreactive. Their two important reactions are:

    • Combustion 燃烧: they burn in plenty of oxygen to give carbon dioxide and water.
    • Substitution with chlorine.

    In a substitution reaction 取代反应, one atom (or group of atoms) is replaced by another. Alkanes react with chlorine only in ultraviolet light 紫外线 — this is a photochemical reaction 光化学反应, where the light provides the activation energy 活化能. For example:

    $$\text{CH}_4 + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{HCl}$$
    Explore · ⁨探索⁩

    Burning an alkane · ⁨アルカンの燃焼⁩

    Step through combustion. With plenty of oxygen an alkane burns cleanly; starve it of oxygen and you get toxic carbon monoxide and soot instead. · ⁨燃焼プロセスを確認する。十分な酸素があればアルカンはきれいに燃焼しますが、酸素を不足させると有毒な一酸化炭素と黒煙が代わりの生成物として得られます。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Alkanes/ˈælkeɪnz/ アルカン
    Combustion/kəmˈbʌstʃn/ 燃焼反応
    substitution reaction/ˌsʌbstɪˈtjuːʃn rɪˈækʃn/ 置換反応
    ultraviolet light/ˌʊltrəˈvaɪəlɪt laɪt/ 紫外線
    photochemical reaction/ˌfəʊtəʊˈkemɪkl rɪˈækʃn/ 光化学反応
    activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ 活性化エネルギー
    11.5

    Alkenes

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 State that the bonding in alkenes includes a double carbon–carbon covalent bond and that alkenes are unsaturated hydrocarbons
    2 Describe the manufacture of alkenes and hydrogen by the cracking of larger alkane molecules using a high temperature and a catalyst
    3 Describe the reasons for the cracking of larger alkane molecules
    5 State that in an addition reaction only one product is formed
    4 Describe the test to distinguish between saturated and unsaturated hydrocarbons by their reaction with aqueous bromine 6 Describe the properties of alkenes in terms of addition reactions with: (a) bromine or aqueous bromine (b) hydrogen in the presence of a nickel catalyst (c) steam in the presence of an acid catalyst and draw the structural or displayed formulae of the products
    日本語
    コア サプリメント
    1 アルケンの結合には炭素-炭素の二重共有結合が含まれており、アルケンは不飽和炭化水素であること述べよ
    2 高温および触媒を用いて大きなアルカン分子をクラックすることで、アルケンおよび水素を製造するプロセスを記述せよ
    3 大きなアルカン分子をクラックする理由を記述せよ
    5 付加反応では、一つの生成物だけが形成されることを述べよ
    4 水溶液臭素との反応により、飽和炭化水素と不飽和炭化水素を区別するための試験を述べよ 6 付加反応におけるアルケンの性質を次のように記述せよ:(a) 臭素または水溶液臭素 (b) ニッケル触媒存在下での水素 (c) 酸触媒存在下での蒸気。生成物の構造式または展開式を描け

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Alkenes 烯烃 have a carbon–carbon double bond (C=C), so they are unsaturated hydrocarbons.

    Cracking

    Large alkane molecules are not very useful. Cracking 裂化 breaks them into smaller, more useful molecules — smaller alkanes and alkenes — using a high temperature and a catalyst 催化剂. Cracking also makes hydrogen 氢气 and provides alkenes for making plastics.

    Worked example. Cracking one molecule of decane, $\text{C}_{10}\text{H}_{22}$, gives octane, $\text{C}_8\text{H}_{18}$, and one other product. Identify it, and say how you would tell the two products apart. Atoms are conserved, so subtract: carbon $10 - 8 = 2$, hydrogen $22 - 18 = 4$. The other product is $\text{C}_2\text{H}_4$, ethene - an alkene, because it fits $\text{C}_n\text{H}_{2n}$. To tell them apart, shake each with orange bromine water: ethene has a C=C, so it decolourises the bromine water, while octane is saturated and leaves it orange. Balance a cracking equation by counting atoms on each side; the alkene is whatever is left over.

    Reactions of alkenes

    The C=C double bond makes alkenes reactive. They take part in addition reactions 加成反应, where two molecules join to form a single product.

    • Test for unsaturation: shake the compound with bromine 溴 water (which is orange). An alkene turns the bromine water colourless; an alkane does not change it.
    • With hydrogen and a nickel 镍 catalyst, an alkene becomes an alkane.
    • With steam 蒸汽 and an acid catalyst, an alkene becomes an alcohol.
    日本語

    Alkenes 烯烃 have a carbon–carbon double bond (C=C), so they are unsaturated hydrocarbons.

    Cracking

    Large alkane molecules are not very useful. Cracking 裂化 breaks them into smaller, more useful molecules — smaller alkanes and alkenes — using a high temperature and a catalyst 催化剂. Cracking also makes hydrogen 氢气 and provides alkenes for making plastics.

    A long alkane chain breaking into a shorter alkane and an alkene with a double bond
    Cracking breaks one large alkane into a smaller alkane plus an alkene (with a C=C), using heat and a catalyst

    Worked example. Cracking one molecule of decane, $\text{C}_{10}\text{H}_{22}$, gives octane, $\text{C}_8\text{H}_{18}$, and one other product. Identify it, and say how you would tell the two products apart. Atoms are conserved, so subtract: carbon $10 - 8 = 2$, hydrogen $22 - 18 = 4$. The other product is $\text{C}_2\text{H}_4$, ethene - an alkene, because it fits $\text{C}_n\text{H}_{2n}$. To tell them apart, shake each with orange bromine water: ethene has a C=C, so it decolourises the bromine water, while octane is saturated and leaves it orange. Balance a cracking equation by counting atoms on each side; the alkene is whatever is left over.

    Reactions of alkenes

    The C=C double bond makes alkenes reactive. They take part in addition reactions 加成反应, where two molecules join to form a single product.

    • Test for unsaturation: shake the compound with bromine 溴 water (which is orange). An alkene turns the bromine water colourless; an alkane does not change it.
    Two test tubes of orange bromine water: the alkene one has gone colourless, the alkane one is still orange
    The bromine water test: an alkene decolourises the orange bromine water (its C=C reacts), while an alkane leaves it orange
    • With hydrogen and a nickel 镍 catalyst, an alkene becomes an alkane.
    • With steam 蒸汽 and an acid catalyst, an alkene becomes an alcohol.
    Explore · ⁨探索⁩

    Addition across the C=C double bond · ⁨C=C二重結合への付加反応⁩

    Step through an addition reaction. The C=C double bond is an alkene's reactive spot — molecules add right across it, which also gives the bromine-water test. · ⁨付加反応のプロセスを確認する。C=C二重結合はアルケンの反応部位であり、分子がその両端に直接付加することで、ブロム水試験にもつながる。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Alkenes/ˈælkiːnz/ アルケン
    Cracking/ˈkrækɪŋ/ クラッキング
    catalyst/ˈkætəlɪst/ 触媒
    hydrogen/ˈhaɪdrədʒn/ 水素
    addition reactions/əˈdɪʃn rɪˈækʃnz/ 付加反応
    bromine/ˈbrəʊmaɪn/ 臭素
    nickel/ˈnɪkl/ ニッケル
    steam/stiːm/ 蒸気
    Watch lesson · ⁨レッスンを視聴⁩
    11.6

    Alcohols

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the manufacture of ethanol by: (a) fermentation of aqueous glucose at 25–35 °C in the presence of yeast and in the absence of oxygen (b) catalytic addition of steam to ethene at 300 °C and 6000 kPa / 60 atm in the presence of an acid catalyst 4 Describe the advantages and disadvantages of the manufacture of ethanol by: (a) fermentation (b) catalytic addition of steam to ethene
    2 Describe the combustion of ethanol
    3 State the uses of ethanol as: (a) a solvent (b) a fuel
    日本語
    コア サプリメント
    1 エタノール製造法を記述せよ:(a) 酵母の存在下、無酸素状態で 25–35 °C の水溶液グルコースの発酵 (b) 酸触媒存在下、300 °C および 6000 kPa / 60 atm でエチレンへの蒸気の触媒付加 4 エタノール製造法((a) 発酵、(b) エチレンへの蒸気の触媒付加)の利点と欠点を述べよ
    2 エタノールの燃焼を説明する
    3 エタノールの用途として次のように述べる:(a) 溶媒として (b) 燃料として

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Alcohols 醇 contain the –OH functional group. The most important one is ethanol 乙醇. There are two ways to make ethanol.

    Method Conditions Notes
    fermentation 发酵 of glucose yeast, 25–35 °C, no oxygen uses renewable sugar, but slow and gives impure ethanol
    addition of steam to ethene 乙烯 300 °C, 60 atm, acid catalyst fast and pure, but uses petroleum (non-renewable)

    In fermentation, yeast 酵母 turns glucose 葡萄糖 into ethanol and carbon dioxide.

    Ethanol burns well (combustion), so it is used as a fuel. It also dissolves many substances, so it is used as a solvent 溶剂.

    日本語

    Alcohols 醇 contain the –OH functional group. The most important one is ethanol 乙醇. There are two ways to make ethanol.

    Two routes to ethanol: fermentation of glucose, and hydration of ethene with steam
    Two routes to ethanol: fermentation, and hydration of ethene
    Method Conditions Notes
    fermentation 发酵 of glucose yeast, 25–35 °C, no oxygen uses renewable sugar, but slow and gives impure ethanol
    addition of steam to ethene 乙烯 300 °C, 60 atm, acid catalyst fast and pure, but uses petroleum (non-renewable)

    In fermentation, yeast 酵母 turns glucose 葡萄糖 into ethanol and carbon dioxide.

    Ethanol burns well (combustion), so it is used as a fuel. It also dissolves many substances, so it is used as a solvent 溶剂.

    Explore · ⁨探索⁩

    Making ethanol by fermentation · ⁨発酵によるエタノール製造⁩

    Step through fermentation. Yeast turns sugar into ethanol with no air — the same alcohol can then be oxidised to vinegar. · ⁨発酵の過程。酵母は酸素なしで糖をエタノールに変える。このアルコール subsequently 酢酸(酢)に酸化される。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Alcohols/ˈælkəhɒlz/ アルコール類
    ethanol/ˈeθənɒl/ エタノール
    fermentation/fɜːmənˈteɪʃn/ 発酵
    ethene/ˈiːθiːn/ エテン
    yeast/jiːst/ 酵母
    glucose/ˈɡluːkəʊs/ ブドウ糖
    solvent/ˈsɒlvənt/ 溶媒
    11.7

    Carboxylic acids

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe the reaction of ethanoic acid with: (a) metals (b) bases (c) carbonates including names and formulae of the salts produced 2 Describe the formation of ethanoic acid by the oxidation of ethanol: (a) with acidified aqueous potassium manganate(VII) (b) by bacterial oxidation during vinegar production
    3 Describe the reaction of a carboxylic acid with an alcohol using an acid catalyst to form an ester
    日本語
    コア サプリメント
    1 酢酸の反応を記述せよ:(a) 金属 (b) 塩基 (c) カルボネート。生成される塩の名前と化学式を含める 2 エタノールの酸化による酢酸の生成を記述せよ:(a) 酸性水溶液マンガン(VII)酸カリウムを用いる場合 (b) 酢酸製造中の細菌酸化による場合
    3 酸触媒を用いてカルボン酸とアルコールの反応を記述し、エステルを生成させる

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Carboxylic acids 羧酸 contain the –COOH functional group. Ethanoic acid 乙酸 is the one to know. Like other acids, it reacts with:

    • metals 金属 → a salt + hydrogen;
    • bases 碱 → a salt 盐 + water;
    • carbonates 碳酸盐 → a salt + water + carbon dioxide.

    The salts formed are called ethanoates.

    Ethanoic acid can be made by the oxidation 氧化 of ethanol, either using acidified potassium manganate(VII) 高锰酸钾, or by bacteria during the making of vinegar 醋.

    When a carboxylic acid reacts with an alcohol (using an acid catalyst), it forms an ester 酯. For example, ethanol + ethanoic acid → ethyl ethanoate + water. The ester's functional group is $-\text{COO}-$, and you name it from its two parents: the alcohol gives the first part with an -yl ending (ethyl), the acid gives the second part with an -oate ending (ethanoate). Its displayed formula is $\text{CH}_3\text{-COO-CH}_2\text{-CH}_3$.

    日本語

    Carboxylic acids 羧酸 contain the –COOH functional group. Ethanoic acid 乙酸 is the one to know. Like other acids, it reacts with:

    Carboxylic acid plus alcohol reacting to form an ester plus water
    A carboxylic acid and an alcohol react to make an ester and water
    • metals 金属 → a salt + hydrogen;
    • bases 碱 → a salt 盐 + water;
    • carbonates 碳酸盐 → a salt + water + carbon dioxide.

    The salts formed are called ethanoates.

    Ethanoic acid can be made by the oxidation 氧化 of ethanol, either using acidified potassium manganate(VII) 高锰酸钾, or by bacteria during the making of vinegar 醋.

    When a carboxylic acid reacts with an alcohol (using an acid catalyst), it forms an ester 酯. For example, ethanol + ethanoic acid → ethyl ethanoate + water. The ester's functional group is $-\text{COO}-$, and you name it from its two parents: the alcohol gives the first part with an -yl ending (ethyl), the acid gives the second part with an -oate ending (ethanoate). Its displayed formula is $\text{CH}_3\text{-COO-CH}_2\text{-CH}_3$.

    Explore · ⁨探索⁩

    A weak acid's pH · ⁨弱酸のpH⁩

    Carboxylic acids are weak acids — they only partly ionise, so they sit just below pH 7. Slide the scale to see where they fall. · ⁨カルボン酸は弱酸であり、部分的に電離するため、pH 7より少し低い位置にある。スケールをスライドさせてその位置を確認する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    Carboxylic acids/ˌkɑːbəkˈsɪlɪk ˈæsɪdz/ カルボン酸
    Ethanoic acid/ˌeθəˈnəʊɪk ˈæsɪd/ 酢酸
    metals/ˈmetlz/ 金属
    bases/ˈbeɪsɪz/ 塩基
    salt/sɒlt/ 塩
    carbonates/ˈkɑːbəneɪts/ 炭酸塩
    oxidation/ˌɒksɪˈdeɪʃn/ 酸化反応
    potassium manganate(VII)/pəˈtæsɪəm ˈmæŋɡəneɪt/ 高マンガン酸カリウム
    vinegar/ˈvɪnɪɡə/ 酢
    ester/ˈestə/ エステル
    11.8

    Polymers

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Define polymers as large molecules built up from many smaller molecules called monomers 6 Identify the repeat units and/or linkages in addition polymers and in condensation polymers
    2 Describe the formation of poly(ethene) as an example of addition polymerisation using ethene monomers 7 Deduce the structure or repeat unit of an addition polymer from a given alkene and vice versa
    8 Deduce the structure or repeat unit of a condensation polymer from given monomers and vice versa, limited to: (a) polyamides from a dicarboxylic acid and a diamine (b) polyesters from a dicarboxylic acid and a diol
    9 Describe the differences between addition and condensation polymerisation
    10 Describe and draw the structure of: (a) nylon, a polyamide [image] (b) PET, a polyester [image] The full name for PET, polyethylene terephthalate, is not required
    3 State that plastics are made from polymers
    4 Describe how the properties of plastics have implications for their disposal 11 State that PET can be converted back into monomers and re-polymerised
    5 Describe the environmental challenges caused by plastics, limited to: (a) disposal in landfill sites (b) accumulation in oceans (c) formation of toxic gases from burning
    12 Describe proteins as natural polyamides and that they are formed from amino acid monomers with the general structure: [image] where R represents different types of side-chain
    13 Describe and draw the structure of proteins as: [image]
    日本語
    コア サプリメント
    1 ポリマーを、単量体と呼ばれる小さな分子の集合体から構成される大きな分子として定義せよ 6 付加重合ポリマーおよび縮合重合ポリマーの反復単位および/または結合部を特定せよ
    2 エチレン単量体を用いた付加重合の例としてポリエチレンの生成を記述せよ 7 与えられたアルケンから付加重合ポリマーの構造または反復単位を導き、逆にその構造または反復単位からアルケンを導け
    8 与えられた単量体から縮合重合ポリマーの構造または反復単位を導き、逆にその構造または反復単位から単量体を導け。対象は次に限定する:(a) 二カルボン酸およびジアミンからのポリアミド (b) 二カルボン酸およびジオールからのポリエステル
    9 付加重合と縮合重合の違いを記述せよ
    10 以下の構造を記述し描け:(a) ポリアミドであるナイロン [画像] (b) ポリエステルであるPET [画像] PETの完全名であるポリエチレンテレフタルレートは必要ない
    3 プラスチックがポリマーで作られていることを述べる
    4 プラスチックの性質が廃棄に与える影響を記述せよ 11 PETが単量体に戻され、再重合できることを述べよ
    5 プラスチックが引き起こす環境問題について、次の点に限定して説明する:(a) ごみ処分場への埋め立て (b) 海洋への蓄積 (c) 燃焼による有毒ガス生成
    12 プロテインを天然ポリアミドであり、一般構造 [画像] を持つアミノ酸単量体から形成されることを記述せよ。ここで R は異なる種類の側鎖を表す
    13 プロテインの構造を [画像] として記述し描け

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    A polymer 聚合物 is a very large molecule built from many small molecules called monomers 单体.

    Addition polymerisation

    In addition polymerisation 加聚, many unsaturated monomers join together with no other product. For example, many ethene monomers join to make poly(ethene). The small part that repeats along the chain is the repeat unit 重复单元.

    Condensation polymerisation

    In condensation polymerisation 缩聚, monomers join and a small molecule (usually water) is lost each time a bond forms. This makes two important types:

    • Polyamides 聚酰胺 are made from a dicarboxylic acid and a diamine. Nylon 尼龙 is a polyamide.
    • Polyesters 聚酯 are made from a dicarboxylic acid and a diol. PET is a polyester, and it can be broken back into its monomers and re-made.

    The new bond joining the monomers is the linkage: an amide linkage 酰胺键 (written $-\text{CO-NH}-$) in a polyamide such as nylon, and an ester linkage 酯键 (written $-\text{CO-O}-$) in a polyester such as PET. A repeat unit is drawn in brackets, e.g. nylon as $-[\text{CO-}(\text{CH}_2)_4\text{-CO-NH-}(\text{CH}_2)_6\text{-NH}]-$; in an exam you may be asked to circle the linkage in a drawn repeat unit.

    Plastics and the environment

    Plastics 塑料 are made from polymers. Because many plastics do not break down, getting rid of them is a problem:

    • they build up in landfill 填埋 sites;
    • they collect in the oceans and harm sea life;
    • burning them can make toxic gases.

    Proteins

    Proteins 蛋白质 are natural polyamides. They are made from amino acid 氨基酸 monomers joined in long chains.

    日本語

    A polymer 聚合物 is a very large molecule built from many small molecules called monomers 单体.

    Addition polymerisation

    In addition polymerisation 加聚, many unsaturated monomers join together with no other product. For example, many ethene monomers join to make poly(ethene). The small part that repeats along the chain is the repeat unit 重复单元.

    Many ethene monomers with C=C double bonds joining into a poly(ethene) chain with single bonds, shown as a repeat unit in brackets
    In addition polymerisation the C=C bonds open up and many ethene monomers join into a long poly(ethene) chain

    Condensation polymerisation

    In condensation polymerisation 缩聚, monomers join and a small molecule (usually water) is lost each time a bond forms. This makes two important types:

    • Polyamides 聚酰胺 are made from a dicarboxylic acid and a diamine. Nylon 尼龙 is a polyamide.
    • Polyesters 聚酯 are made from a dicarboxylic acid and a diol. PET is a polyester, and it can be broken back into its monomers and re-made.

    The new bond joining the monomers is the linkage: an amide linkage 酰胺键 (written $-\text{CO-NH}-$) in a polyamide such as nylon, and an ester linkage 酯键 (written $-\text{CO-O}-$) in a polyester such as PET. A repeat unit is drawn in brackets, e.g. nylon as $-[\text{CO-}(\text{CH}_2)_4\text{-CO-NH-}(\text{CH}_2)_6\text{-NH}]-$; in an exam you may be asked to circle the linkage in a drawn repeat unit.

    Plastics and the environment

    Plastics 塑料 are made from polymers. Because many plastics do not break down, getting rid of them is a problem:

    • they build up in landfill 填埋 sites;
    • they collect in the oceans and harm sea life;
    • burning them can make toxic gases.
    A pebble beach covered in washed-up plastic bottles, cups and broken plastic fragments
    Most plastics are non-biodegradable, so waste plastic collects on beaches and in the oceans for hundreds of years

    Proteins

    Proteins 蛋白质 are natural polyamides. They are made from amino acid 氨基酸 monomers joined in long chains.

    Explore · ⁨探索⁩

    Making a polymer · ⁨ポリマーの製造⁩

    Small monomers join into a long chain — the basis of all plastics. · ⁨小さなモノマーが長い鎖になる — これはすべてのプラスチックの基礎である。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    polymer/ˈpɒlɪmə/ ポリマー
    monomers/ˈmɒnəʊməz/ 単量体
    addition polymerisation/əˈdɪʃn ˌpɒlɪməraɪˈzeɪʃn/ 付加重合
    repeat unit/rɪˈpiːt ˈjuːnɪt/ 繰り返し単位
    condensation polymerisation/kɒndenˈseɪʃn ˌpɒlɪməraɪˈzeɪʃn/ 縮加重合
    Polyamides/ˌpɒlɪˈeɪmaɪdz/ ポリアミド
    Nylon/ˈnaɪlɒn/ ナイロン
    Polyesters/ˌpɒlɪˈestəz/ ポリエステル
    amide linkage/əˈmaɪd ˈlɪŋkɪdʒ/ アミド結合
    ester linkage/ˈestə ˈlɪŋkɪdʒ/ エステル結合
    Plastics/ˈplæstɪks/ プラスチック
    landfill/ˈlændfɪl/ 埋立地
    Proteins/ˈprəʊtiːnz/ タンパク質
    amino acid/əˈmiːnəʊ ˈæsɪd/ アミノ酸
    Watch lesson · ⁨レッスンを視聴⁩
    11.8

    Exam tips

    • Learn each homologous series by its functional group and general formula: alkanes $\text{C}_n\text{H}_{2n+2}$, alkenes $\text{C}_n\text{H}_{2n}$, alcohols $-\text{OH}$, carboxylic acids $-\text{COOH}$.
    • The bromine water test for unsaturation: an alkene (with its C=C) turns orange bromine water colourless; an alkane leaves it orange.
    • Cracking breaks large, less useful alkanes into smaller alkanes and alkenes, using heat and a catalyst.
    • Two routes to ethanol: fermentation of glucose (renewable but slow and impure) and adding steam to ethene (fast and pure, but from petroleum).
    • Addition polymerisation joins unsaturated monomers with no other product; condensation polymerisation joins monomers and loses a small molecule (usually water) each time.
  • 12

    Experimental techniques and chemical analysis

    Watch lesson · ⁨レッスンを視聴⁩
    12.1

    Experimental design · ⁨実験計画⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Name appropriate apparatus for the measurement of time, temperature, mass and volume, including: (a) stop-watches (b) thermometers (c) balances (d) burettes (e) volumetric pipettes (f) measuring cylinders (g) gas syringes
    2 Suggest advantages and disadvantages of experimental methods and apparatus
    3 Describe a: (a) solvent as a substance that dissolves a solute (b) solute as a substance that is dissolved in a solvent (c) solution as a mixture of one or more solutes dissolved in a solvent (d) saturated solution as a solution containing the maximum concentration of a solute dissolved in the solvent at a specified temperature (e) residue as a substance that remains after evaporation, distillation, filtration or any similar process (f) filtrate as a liquid or solution that has passed through a filter
    日本語
    コア サプリメント
    1 時間、温度、質量、体積の測定に適した器具を列挙する。以下を含む:(a) ストップウォッチ (b) 温度計 (c) 天秤 (d) メスシリンダー (e) メスピペット (f) 量筒 (g) ガスシリンジ
    2 実験手法および器具の利点と欠点を提案する
    3 次を説明する:(a) 溶媒:溶質を溶解する物質 (b) 溶質:溶媒に溶解する物質 (c) 溶液:溶媒に1つ以上の溶質が溶解した混合物 (d) 飽和溶液:指定された温度において溶媒に最大濃度の溶質が溶解した溶液 (e) 残渣:蒸発、蒸留、ろ过などの類似プロセス後に残る物質 (f) 濾液:フィルターを通過した液体または溶液

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Apparatus

    You should know which piece of apparatus to use for each measurement:

    Measurement Apparatus
    time stop-watch 秒表
    temperature thermometer 温度计
    mass balance 天平
    volume (accurate) burette 滴定管 or volumetric pipette 移液管
    volume (rough) measuring cylinder 量筒
    volume of a gas gas syringe 注射器

    Key words

    These words are used throughout practical chemistry:

    • A solvent 溶剂 is a substance that dissolves a solute.
    • A solute 溶质 is the substance that dissolves in the solvent.
    • A solution 溶液 is a mixture of one or more solutes dissolved in a solvent.
    • A saturated solution 饱和溶液 holds the maximum amount of solute that will dissolve at a given temperature.
    • A residue 残渣 is the substance left behind after evaporation, distillation or filtration.
    • A filtrate 滤液 is the liquid that has passed through a filter.
    日本語

    器具

    各測定に対応する器具を知っている必要があります:

    測定項目 器具
    時間 ストップウォッチ
    温度 温度計
    質量 天秤
    体積(精密) ビュレット または 容積ピペット
    体積(おおよそ) 量筒
    気体の体積 ガスピレット

    重要用語

    これらの用語は実習化学全体で使用されます:

    • 溶媒とは、溶質を溶解させる物質のことです。
    • 溶質とは、溶媒に溶解する物質のことです。
    • 溶液とは、溶媒に1つ以上の溶質が溶解した混合物のことです。
    • 飽和溶液とは、特定の温度において最大量の溶質が溶解している状態の溶液です。
    • 残渣とは、蒸発、蒸留またはろ過後に残る物質のことです。
    • 濾液とは、フィルターを通過した液体のことです。
    Explore · ⁨探索⁩

    Experiment design route · ⁨実験設計の手順⁩

    Follow a fair test from question to reliable conclusion. · ⁨質問から信頼できる結論まで、公平なテストに従う。⁩

    12.2

    Acid–base titrations · ⁨酸塩基滴定⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe an acid–base titration to include the use of a: (a) burette (b) volumetric pipette (c) suitable indicator
    2 Describe how to identify the end-point of a titration using an indicator
    日本語
    コア サプリメント
    1 酸塩基滴定を説明し、以下の使用を含める:(a) メスシリンダー (b) メスピペット (c) 適切な指示薬
    2 指示薬を用いて滴定の終了点を特定する方法を説明する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English
    Titration curve and equivalence point

    A titration 滴定 finds the exact volume of one solution that reacts with another.

    • Use a volumetric pipette to measure a fixed volume of one solution into a flask.
    • Add a few drops of a suitable indicator 指示剂.
    • Add the other solution from a burette, slowly, until the colour just changes.

    The end-point 终点 is the moment the indicator changes colour, which shows the reaction is exactly complete. You read the burette to find the volume added.

    日本語
    滴定曲線と等量点

    滴定法は、ある溶液と正確に反応するもう一つの溶液の体積を求める手法です。

    • 容積ピペットを使用して、固定された体積の溶液をフラスコに測り取ります。
    • 適当な指示薬の数滴を加えます。
    • ビュレットから別の溶液をゆっくりと加え、色がちょうど変わるまで続けます。

    指示薬の色が変わった瞬間を終点と呼び、これは反応が正確に完了したことを示します。ビュレットの目盛りを読んで加えた体積を確認します。

    白いタイルの上に置かれた溶液と指示薬が入ったコニカルフラスコの上部にクランプで固定された溶液入りビュレット
    溶液をビュレットからフラスコへ流し、指示薬がちょうど色が変わる(終点)まで行います
    実際の滴定装置:オレンジ色の指示薬溶液を含むコニカルフラスコの上にあるビュレット
    実際の滴定:ビュレットからの溶液をフラスコに加えていくと、指示薬がオレンジ色に変わります
    Explore · ⁨探索⁩

    Acid–base titrations · ⁨酸塩基滴定⁩

    pH jumps at the equivalence point · ⁨等量点でpHが跳ね上がる⁩

    Track the pH as base is added — it leaps through neutral at the equivalence point. · ⁨塩基を添加するにつれてpHを追跡する — 等量点で中性を急激に通過する。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    titration/taɪˈtreɪʃn/ 滴定
    indicator/ˈɪndɪkeɪtə/ 指示薬
    end-point/end pɔɪnt/ 終了点
    Watch lesson · ⁨レッスンを視聴⁩
    12.3

    Chromatography · ⁨クロマトグラフィー⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe how paper chromatography is used to separate mixtures of soluble coloured substances, using a suitable solvent 3 Describe how paper chromatography is used to separate mixtures of soluble colourless substances, using a suitable solvent and a locating agent Knowledge of specific locating agents is not required
    2 Interpret simple chromatograms to identify: (a) unknown substances by comparison with known substances (b) pure and impure substances 4 State and use the equation for $R_{\text{f}}$:
    $$R_{\text{f}} = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}}$$
    日本語
    コア サプリメント
    1 適切な溶媒を用いて、紙クロマトグラフィーが可溶性有色物質の混合物を分離するためにどのように用いられるかを記述せよ 3 適切な溶媒および検出剤を用いて、紙クロマトグラフィーが可溶性無色物質の混合物を分離するためにどのように用いられるかを記述せよ。特定の検出剤に関する知識は不要
    2 単純なクロマトグラムを解釈して識別せよ:(a) 既知物質と比較して未知物質 (b) 純物質および不純物 4 $R_{\text{f}}$ の方程式を述べ、使用せよ:
    $$R_{\text{f}} = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}}$$

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Paper chromatography 纸色谱法 separates a mixture of soluble substances. You put a spot of the mixture near the bottom of the paper, then stand the paper in a solvent. As the solvent rises up the paper, the substances move different distances, so they separate.

    • For coloured substances you can see the spots directly.
    • For colourless substances you must spray a locating agent 显色剂 to make the spots show up.

    The finished paper is a chromatogram 色谱图. You can use it to:

    • identify an unknown substance by comparing it with known substances;
    • tell if a substance is pure (a pure substance gives only one spot).

    You can also calculate the $R_{\text{f}}$ value of a spot:

    $$R_{\text{f}} = \frac{\text{distance moved by the substance}}{\text{distance moved by the solvent}}$$

    Worked example. On a chromatogram the solvent front has risen 8.0 cm from the baseline, and a spot has moved 6.0 cm. Find its $R_{\text{f}}$ value.

    $$R_{\text{f}} = \frac{6.0}{8.0} = 0.75$$

    Both distances are measured from the baseline (the pencil line), and the spot's distance is taken to the centre of the spot. An $R_{\text{f}}$ has no unit, and it is always less than 1, because a spot can never travel further than the solvent carrying it - an answer above 1 means the two distances have been divided the wrong way round.

    日本語

    紙クロマトグラフィーは、可溶性物質の混合物を分離するために用いられます。紙の下部付近に混合物のスポットを置き、それを溶媒に浸します。溶媒が紙を上昇するにつれて、各物質が異なる距離を移動するため、分離されます。

    溶媒に浸されたクロマトグラフィー紙:ベースライン上の混合物が溶媒前線より下で異なる高さにスポットとして分離しています
    溶媒が上昇するにつれ、物質は異なる距離を進んで分離します;$R_f$はスポット距離($a$)を溶媒移動距離($b$)で割った値です
    実際のクロマトグラフィー紙:各色スポットがストリップの異なる高さに上がっています
    実際のクロマトグラムでは、混合物中の各物質が異なる距離だけ上昇し、それぞれ独立した色のスポットを残します
    • 有色物質については、スポットを直接確認できます。
    • 無色物質については、スポットを可視化するために**現像剤(ロケーティング・エージェント)**をスプレーする必要があります。

    完成した紙はクロマトグラムです。これを使って:

    • 既知の物質と比較して、未知の物質を同定する;
    • 物質が純物かどうかを調べる(純物は1つのスポットのみを示す)。

    スポットの $R_{\text{f}}$ 値も計算できます:

    $$R_{\text{f}} = \frac{\text{distance moved by the substance}}{\text{distance moved by the solvent}}$$

    ** worked example.** クロマトグラムにおいて、溶媒前線がベースラインから8.0 cm上昇し、あるスポットが6.0 cm移動しました。その $R_{\text{f}}$ 値を求めよ。

    $$R_{\text{f}} = \frac{6.0}{8.0} = 0.75$$

    両方の距離はベースライン(鉛筆で引いた線)から測り、スポットの距離はスポットの中心までとする。$R_{\text{f}}$ に単位はなく、常に1未満である。なぜなら、スポットはそれを運ぶ溶媒より遠くへ移動できないからであり、答えが1を超えれば、2つの距離の割り算の順序が逆になっていることを意味する。

    Explore · ⁨探索⁩

    Chromatography route · ⁨クロマトグラフィーの手順⁩

    Watch a mixture separate as the solvent front moves. · ⁨溶媒前線が移動するにつれて混合物が分離する様子を見る。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    stop-watch/stɒp wɒtʃ/ ストップウォッチ
    thermometer/θɜːˈmɒmɪtə/ 温度計
    balance/ˈbæləns/ バランス
    burette/bjʊəˈret/ ビュレット
    volumetric pipette/ˌvɒljuːˈmetrɪk pɪˈpet/ 量管
    measuring cylinder/ˈmeʒərɪŋ ˈsɪlɪndə/ メジャーリングカップ
    gas syringe/ɡæs sɪˈrɪndʒ/ ガスシリンジ
    solvent/ˈsɒlvənt/ 溶媒
    solute/ˈsɒljuːt/ 溶質
    solution/səˈluːʃn/ 溶液
    saturated solution/ˈsætʃəreɪtɪd səˈluːʃn/ 飽和溶液
    residue/ˈresɪdjuː/ 残渣
    filtrate/ˈfɪltreɪt/ 濾過液
    paper chromatography/ˈpeɪpə krəʊməˈtɒɡrəfi/ 紙クロマトグラフィー
    locating agent/ləʊˈkeɪtɪŋ ˈeɪdʒənt/ 検出剤
    chromatogram/krəʊˈmætəɡræm/ クロマトグラム
    Watch lesson · ⁨レッスンを視聴⁩
    12.4

    Separation and purification · ⁨分離と精製⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe and explain methods of separation and purification using: (a) a suitable solvent (b) filtration (c) crystallisation (d) simple distillation (e) fractional distillation
    2 Suggest suitable separation and purification techniques, given information about the substances involved
    3 Identify substances and assess their purity using melting point and boiling point information
    日本語
    コア サプリメント
    1 以下の方法を用いた分離および精製の方法を説明・解説する:(a) 適切な溶媒 (b) ろ过 (c) 結晶化 (d) 単一蒸留 (e) 分餾
    2 関わる物質に関する情報を与え、適切な分離および精製技術を提案する
    3 融点および沸点の情報を用いて物質を同定し、純度を評価する

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    You choose a method based on the mixture:

    Method Used to separate
    dissolving in a suitable solvent, then filtration 过滤 an insoluble solid from a liquid
    crystallisation 结晶 a soluble solid from its solution
    simple distillation 蒸馏 a solvent (the liquid) from a solution
    fractional distillation 分馏 two or more liquids with different boiling points

    You can check the purity of a substance using its melting point 熔点 and boiling point 沸点: a pure substance melts and boils at sharp, fixed temperatures, while impurities lower the melting point and raise the boiling point.

    日本語

    混合物に基づいて方法を選択する:

    方法 分離対象
    適切な溶媒での溶解後、ろ過 液体からの不溶性固体
    結晶化 溶液からの可溶性固体
    蒸留 溶液からの溶媒(液体)
    分餾 異なる沸点を持つ2つ以上の液体
    フィルターペーパーで内張られたろ紙漏斗が不溶性残渣を捕集し、ろ液が下のビーカーに滴り落ちる様子
    ろ過: 不溶性固体はフィルターペーパー上に残り(残渣)、液体は通過する(ろ液)
    溶液が入った加熱フラスコ、温度計、水冷凝縮管、および蒸留液を収容するビーカー
    蒸留: 溶媒が蒸発し、凝縮管によって再び液体に戻され、純粋な蒸留液が収容される
    蒸留と同様だが、ガラスビーズ入りの分餾カラムとカラム上部の温度計があるもの
    分餾には分餾カラムを使用するため、沸点の異なる液体が明確に分離されます

    物質の純度は融点や沸点を用いて確認できる:純物は鋭利で一定の温度で融解・沸騰するのに対し、不純物は融点を下げ、沸点を上げる。

    Explore · ⁨探索⁩

    Separation method lab · ⁨分離手法の実験⁩

    Choose the method that matches the mixture. · ⁨混合物に合う方法を選んでください。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    filtration/fɪlˈtreɪʃn/ ろ過
    crystallisation/ˌkrɪstəlaɪˈzeɪʃn/ 結晶化
    simple distillation/ˈsɪmpl dɪstɪˈleɪʃn/ 単一蒸留
    fractional distillation/ˈfrækʃənl dɪstɪˈleɪʃn/ 分餾
    melting point/ˈmeltɪŋ pɔɪnt/ 融点
    boiling point/ˈbɔɪlɪŋ pɔɪnt/ 沸点
    anion/ˈænaɪən/ アニオン
    carbonate/ˈkɑːbəneɪt/ 炭酸塩
    carbon dioxide/ˈkɑːbən daɪˈɒksaɪd/ 二酸化炭素
    chloride/ˈklɔːraɪd/ 塩化物
    nitric acid/ˈnaɪtrɪk ˈæsɪd/ 硝酸
    silver nitrate/ˈsɪlvə ˈnaɪtreɪt/ 硝酸銀
    bromide/ˈbrəʊmaɪd/ 臭化物
    iodide/ˈaɪədaɪd/ ヨウ化物
    nitrate/ˈnaɪtreɪt/ 窒素酸
    aluminium/ˌæljʊˈmɪnɪəm/ アルミニウム
    sodium hydroxide/ˈsəʊdɪəm haɪˈdrɒksaɪd/ 水酸化ナトリウム
    ammonia/æˈməʊnɪə/ アンモニア
    sulfate/ˈsʌlfeɪt/ 硫酸塩
    barium nitrate/ˈbeərɪəm ˈnaɪtreɪt/ 硝酸バリウム
    sulfite/ˈsʌlfaɪt/ 亜硫酸塩
    potassium manganate(VII)/pəˈtæsɪəm ˈmæŋɡəneɪt/ 高マンガン酸カリウム
    cation/ˈkætaɪən/ カチオン
    12.5

    Identifying ions and gases · ⁨イオンとガスの同定⁩

    Syllabus · ⁨シラバス⁩
    English
    Core Supplement
    1 Describe tests to identify the anions: (a) carbonate, $\text{CO}_3^{2-}$, by reaction with dilute acid and then testing for carbon dioxide gas (b) chloride, $\text{Cl}^-$, bromide, $\text{Br}^-$, and iodide, $\text{I}^-$, by acidifying with dilute nitric acid then adding aqueous silver nitrate (c) nitrate, $\text{NO}_3^-$, reduction with aluminium foil and aqueous sodium hydroxide and then testing for ammonia gas (d) sulfate, $\text{SO}_4^{2-}$, by acidifying with dilute nitric acid and then adding aqueous barium nitrate (e) sulfite, $\text{SO}_3^{2-}$, by reaction with acidified aqueous potassium manganate(VII)
    2 Describe tests using aqueous sodium hydroxide and aqueous ammonia to identify the aqueous cations: (a) aluminium, $Al^{3+}$ (b) ammonium, $NH_4^+$ (c) calcium, $Ca^{2+}$ (d) chromium(III), $Cr^{3+}$ (e) copper(II), $Cu^{2+}$ (f) iron(II), $Fe^{2+}$ (g) iron(III), $Fe^{3+}$ (h) zinc, $Zn^{2+}$
    3 Describe tests to identify the gases: (a) ammonia, $NH_3$, using damp red litmus paper (b) carbon dioxide, $CO_2$, using limewater (c) chlorine, $Cl_2$, using damp litmus paper (d) hydrogen, $H_2$, using a lighted splint (e) oxygen, $O_2$, using a glowing splint (f) sulfur dioxide, $SO_2$, using acidified aqueous potassium manganate(VII)
    4 Describe the use of a flame test to identify the cations: (a) lithium, $Li^+$ (b) sodium, $Na^+$ (c) potassium, $K^+$ (d) calcium, $Ca^{2+}$ (e) barium, $Ba^{2+}$ (f) copper(II), $Cu^{2+}$
    日本語
    コア サプリメント
    1 陰イオンの識別試験を記述せよ:(a) カルボネート、$\text{CO}_3^{2-}$、希酸との反応後、二酸化炭素ガスのテスト (b) クロライド、$\text{Cl}^-$、ブロマイド、$\text{Br}^-$、ヨウ化物、$\text{I}^-$、希硝酸で酸性化した後、水溶液硝酸銀の添加 (c) ナイトレート、$\text{NO}_3^-$、アルミウム箔および水溶液水酸化ナトリウムによる還元後、アンモニアガスのテスト (d) スルフェート、$\text{SO}_4^{2-}$、希硝酸で酸性化した後、水溶液硝酸バリウムの添加 (e) スルファイト、$\text{SO}_3^{2-}$、酸性水溶液マンガン(VII)酸カリウムとの反応
    2 水溶液水酸化ナトリウムおよび水溶液アンモニアを用いて水溶液陽イオンの識別試験を記述せよ:(a) アルミニウム、$Al^{3+}$ (b) アンモニウム、$NH_4^+$ (c) カルシウム、$Ca^{2+}$ (d) クロム(III)、$Cr^{3+}$ (e) 銅(II)、$Cu^{2+}$ (f) 鉄(II)、$Fe^{2+}$ (g) 鉄(III)、$Fe^{3+}$ (h) 亜鉛、$Zn^{2+}$
    3 以下のガスを同定するためのテストを説明せよ: (a) アンモニア, $NH_3$(湿潤赤リトマス紙を用いる) (b) 二酸化炭素, $CO_2$(石灰水を用いる) (c)塩素, $Cl_2$(湿潤リトマス紙を用いる) (d) 水素, $H_2$(燃えさし棒を用いる) (e) 酸素, $O_2$(発光する燃えさし棒を用いる) (f) 二酸化硫黄, $SO_2$(酸性化された高マンガン酸カリウム(VII)水溶液を用いる)
    4 炎色反応を用いて以下の陽イオンを同定する方法を説明せよ: (a) リチウム, $Li^+$ (b) ナトリウム, $Na^+$ (c) カリウム, $K^+$ (d) カルシウム, $Ca^{2+}$ (e) バリウム, $Ba^{2+}$ (f) 銅(II), $Cu^{2+}$

    Source: Cambridge International syllabus · ⁨出典: Cambridge International シラバス⁩

    English

    Tests for anions

    An anion 阴离子 is a negative ion.

    Anion Test Result
    carbonate 碳酸盐 ($\text{CO}_3^{2-}$) add dilute acid fizzes; the gas turns limewater milky (carbon dioxide 二氧化碳)
    chloride 氯化物 ($\text{Cl}^-$) add dilute nitric acid 硝酸, then silver nitrate 硝酸银 white precipitate
    bromide 溴化物 ($\text{Br}^-$) add dilute nitric acid, then silver nitrate cream precipitate
    iodide 碘化物 ($\text{I}^-$) add dilute nitric acid, then silver nitrate yellow precipitate
    nitrate 硝酸盐 ($\text{NO}_3^-$) add aluminium 铝 foil and sodium hydroxide 氢氧化钠, warm ammonia 氨气 gas given off
    sulfate 硫酸盐 ($\text{SO}_4^{2-}$) add dilute nitric acid, then barium nitrate 硝酸钡 white precipitate
    sulfite 亚硫酸盐 ($\text{SO}_3^{2-}$) add acidified potassium manganate(VII) 高锰酸钾 purple colour fades

    Tests for cations

    A cation 阳离子 is a positive ion. Add aqueous sodium hydroxide, or aqueous ammonia, and look at the precipitate 沉淀 formed.

    Cation With sodium hydroxide With aqueous ammonia
    aluminium ($\text{Al}^{3+}$) white, dissolves in excess white, stays
    ammonium 铵 ($\text{NH}_4^+$) ammonia gas when warmed —
    calcium 钙 ($\text{Ca}^{2+}$) white, stays no precipitate
    chromium(III) 铬 ($\text{Cr}^{3+}$) green, dissolves in excess green, stays
    copper(II) 铜 ($\text{Cu}^{2+}$) light blue, stays light blue, dissolves to deep blue
    iron(II) 铁 ($\text{Fe}^{2+}$) green, stays green, stays
    iron(III) ($\text{Fe}^{3+}$) red-brown, stays red-brown, stays
    zinc 锌 ($\text{Zn}^{2+}$) white, dissolves in excess white, dissolves in excess

    Tests for gases

    Gas Test Result
    ammonia ($\text{NH}_3$) damp red litmus 石蕊 paper turns blue
    carbon dioxide ($\text{CO}_2$) bubble through limewater 石灰水 turns milky
    chlorine 氯气 ($\text{Cl}_2$) damp litmus paper bleached white
    hydrogen 氢气 ($\text{H}_2$) a lighted splint burns with a squeaky pop
    oxygen 氧气 ($\text{O}_2$) a glowing splint relights
    sulfur dioxide ($\text{SO}_2$) acidified potassium manganate(VII) purple colour fades

    Flame tests

    A flame test 焰色试验 identifies some metal cations by the colour they give to a flame:

    Cation Flame colour
    lithium 锂 ($\text{Li}^+$) red
    sodium 钠 ($\text{Na}^+$) yellow
    potassium 钾 ($\text{K}^+$) lilac (purple)
    calcium ($\text{Ca}^{2+}$) orange-red
    barium 钡 ($\text{Ba}^{2+}$) light green
    copper(II) ($\text{Cu}^{2+}$) blue-green
    日本語

    陰イオンの検出

    陰イオンとは負の電荷を持つイオンである。

    陰イオン 試験 結果
    炭酸塩 ($\text{CO}_3^{2-}$) 希酸を加える 泡立ち;気体が石灰水を白濁させる (二酸化炭素)
    塩化物 ($\text{Cl}^-$) 希硝酸を加えた後、硝酸銀を加える 白色沈殿
    臭化物 ($\text{Br}^-$) 希硝酸を加えた後、硝酸銀を加える 乳白色沈殿
    ヨウ化物 ($\text{I}^-$) 希硝酸を加えた後、硝酸銀を加える 黄色沈殿
    硝酸塩 ($\text{NO}_3^-$) アルミニウム箔と水酸化ナトリウムを加え、温める アンモニアガスが発生
    硫酸塩 ($\text{SO}_4^{2-}$) 希硝酸を加えた後、硝酸バリウムを加える 白色沈殿
    亜硫酸塩 ($\text{SO}_3^{2-}$) 酸性化したパーガンガン酸カリウム(VII) を加える 紫色が消える

    陽イオンの検出

    陽イオンとは正の電荷を持つイオンである。水溶液の水酸化ナトリウムまたは水溶液のアンモニアを加え、生成した沈殿を見る。

    滴管が試験管内の溶液に水酸化ナトリウムを加え、底部に有色沈殿を形成する様子
    水酸化ナトリウムを加えると、一部の金属イオンで有色沈殿ができます
    陽イオン 水酸化ナトリウム添加時 水溶液のアンモニア添加時
    アルミウム ($\text{Al}^{3+}$) 白色、過剰量で溶解 白色、残る
    アンモニウム ($\text{NH}_4^+$) 温めるとアンモニアガス —
    カルシウム ($\text{Ca}^{2+}$) 白色、残る 沈殿なし
    クロム(III) ($\text{Cr}^{3+}$) 緑色、過剰量で溶解 緑色、残る
    銅(II) ($\text{Cu}^{2+}$) 薄青、残る 薄青、溶解して深青色になる
    鉄(II) ($\text{Fe}^{2+}$) 緑色、残る 緑色、残る
    鉄(III) ($\text{Fe}^{3+}$) 赤茶、残る 赤茶、残る
    亜鉛 ($\text{Zn}^{2+}$) 白色、過剰量で溶解 白色、過剰量で溶解

    ガスの検出

    ガス 試験 結果
    アンモニア ($\text{NH}_3$) 湿らせた赤リトマス紙 青色に変化する
    二酸化炭素 ($\text{CO}_2$) 石灰水に通す 白濁する
    塩素 ($\text{Cl}_2$) 湿らせたリトマス紙 漂白されて白色になる
    水素 ($\text{H}_2$) 点火したマッチ棒 「ピッ」という音を立てて燃焼する
    酸素 ($\text{O}_2$) 余熱のあるマッチ棒 再燃する
    二酸化硫黄 ($\text{SO}_2$) 酸性化したパーガンガン酸カリウム(VII) 紫色が消える
    2本の試験管: 水素には点火したマッチ棒で「ピッ」という音がし、酸素には余熱のあるマッチ棒が再燃する様子
    ガスの試験:水素はピッと鳴り、酸素は燃えかかった木片を再点火させます

    炎色反応

    炎色反応は、金属陽イオンが炎に与える色によって特定のものを同定する:

    ブーセンバーナーの炎で試料を持つ清浄なワイヤーが、有色の炎を生む様子
    炎色反応:金属イオンが炎に色を与えます
    陽イオン 炎の色
    リチウム ($\text{Li}^+$) 赤
    ナトリウム ($\text{Na}^+$) 黄色
    カリウム ($\text{K}^+$) ラベンダー(紫)
    カルシウム ($\text{Ca}^{2+}$) 橙赤
    バリウム ($\text{Ba}^{2+}$) 薄緑
    銅(II) ($\text{Cu}^{2+}$) 青緑
    Explore · ⁨探索⁩

    Ion and gas test lab · ⁨イオンとガス試験の実験⁩

    Match test observations to the ion or gas present. · ⁨試験の結果观察を存在するイオンまたはガスに一致させよ。⁩

    Vocabulary · ⁨語彙⁩ Train · ⁨練習する⁩
    English 日本語
    precipitate/prɪˈsɪpɪteɪt/ 沈殿物
    ammonium/æˈməʊnɪəm/ アンモニウム
    calcium/ˈkælsɪəm/ カルシウム
    chromium/ˈkrəʊmɪəm/ クロム
    copper/ˈkɒpə/ 銅
    iron/ˈaɪən/ 鉄
    zinc/zɪŋk/ 亜鉛
    litmus/ˈlɪtməs/ リトマス試薬
    limewater/ˈlaɪmwɔːtə/ 石灰水
    chlorine/ˈklɔːriːn/ 塩素
    hydrogen/ˈhaɪdrədʒn/ 水素
    oxygen/ˈɒksɪdʒn/ 酸素
    flame test/fleɪm test/ 炎色反応
    lithium/ˈlɪθɪəm/ リチウム
    sodium/ˈsəʊdɪəm/ ナトリウム
    potassium/pəˈtæsɪəm/ カリウム
    barium/ˈbeərɪəm/ バリウム
    12.5

    Exam tips · ⁨試験対策⁩

    English
    • Pick apparatus by the measurement: a burette or pipette for accurate volumes, a measuring cylinder for a rough volume, a gas syringe for a gas volume, a balance for mass.
    • $R_f = \dfrac{\text{distance moved by the substance}}{\text{distance moved by the solvent}}$, so it is always less than 1. A pure substance gives a single spot.
    • Learn the anion tests: a carbonate fizzes and the gas turns limewater milky; halides with silver nitrate give a chloride white, bromide cream, iodide yellow precipitate; a sulfate gives a white precipitate with barium nitrate.
    • Learn the gas tests: hydrogen gives a squeaky pop, oxygen relights a glowing splint, carbon dioxide turns limewater milky, chlorine bleaches damp litmus, ammonia turns damp red litmus blue.
    • For cations, add sodium hydroxide and read the precipitate colour: copper(II) light blue, iron(II) green, iron(III) red-brown.
    日本語
    • 測定に応じて器具を選ぶ: 正確な体積にはメレットやピペット、おおよその体積には目盛り付きコップ、気体の体積にはガスシリンジ、質量には天秤。
    • $R_f = \dfrac{\text{distance moved by the substance}}{\text{distance moved by the solvent}}$であるため、常に1より小さいです。純粋な物質は単一のスポットを与えます。
    • 陰イオンの検出法を覚える: 炭酸塩は泡立ち、気体が石灰水を白濁させる;ハロゲン化物は硝酸銀で、塩化物白色、臭化物乳白色、ヨウ化物黄色の沈殿を与える;硫酸塩は硝酸バリウムで白色沈殿を与える。
    • 気体テストを学ぼう:水素はピシッという音を立て、酸素は燃えかかった木片を再点火させ、二酸化炭素は石灰水を白く濁らせ、塩素は湿ったリトマス紙を漂白し、アンモニアは湿った赤色リトマス紙を青色に変える。
    • カチオンについては、水酸化ナトリウムを加えて沈殿の色を確認する:銅(II)は薄い青、鉄(II)は緑、鉄(III)は赤茶色。

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