Pearson Edexcel · International GCSE · Chemistry

The language of chemistry

Everything in this course is written in one language: particles, formulae and equations. Section 1 teaches it — how particles arrange themselves in solids, liquids and gases; how atoms are built and lined up in the Periodic Table; how to turn a sentence about a reaction into a balanced equation and a mass calculation; and how the three kinds of bonding give every substance its properties.

Edexcel International GCSE Chemistry 4CH1, Section 1 (statements 1.1–1.60C) is the largest section and the foundation for the other three. C-suffixed statements are the separate-science (Chemistry-only) extension.

States of matter and solutions

Solid: particles vibrate about fixed positions in a regular pattern. Liquid: particles touch but move past each other at random. Gas: particles far apart, fast, random. Interconversions: melting 熔化 / freezing 凝固, boiling 沸腾 / condensation 凝结, evaporation 蒸发 (surface only, below boiling), sublimation 升华 (solid → gas directly). Energy is absorbed to separate particles and released as they bond again.

A solvent 溶剂 dissolves the solute 溶质 to make a solution 溶液; a saturated solution 饱和溶液 holds all the solute it can at that temperature. (C) Solubility is the mass dissolved per 100 g of solvent at a stated temperature, read from a solubility curve 溶解度曲线 — points on the line are saturated; below it, unsaturated.

Dilution of coloured solutions and gas diffusion (bromine spreading through air) both show particles moving from where they are many to where they are few.

Elements, compounds, mixtures and separation

An element 元素 contains one type of atom; a compound 化合物 holds different atoms chemically joined; a mixture 混合物 is not chemically joined. A pure substance has a sharp melting/boiling point; mixtures melt and boil over a range — the test for purity.

Separation techniques, matched to the job: filtration 过滤 (insoluble solid from liquid); crystallisation 结晶 (evaporate a solution to leave crystals); simple distillation 蒸馏 (liquid from dissolved solid — evaporate then condense); fractional distillation 分馏 (liquids with close boiling points, using a fractionating column); paper chromatography 纸层析 (solutes that dissolve differently in a solvent).

Rf = distance moved by the spot ÷ distance moved by the solvent. The same substance always has the same Rf in the same solvent — comparing with known values identifies the components of the mixture.

Atoms and the Periodic Table

Atomic number 质子数 = protons; mass number 质量数 = protons + neutrons. Isotopes 同位素 have the same protons but different neutrons. Relative atomic mass ($A_r$) is the weighted mean over the isotopes.

Worked check. Chlorine: 75 % Cl-35, 25 % Cl-37 → $A_r = (75 \times 35 + 25 \times 37)/100 = 35.5$.

The Periodic Table lists elements by atomic number into groups 族 (columns, same outer electrons) and periods 周期 (rows). The first 20 electronic configurations follow 2,8,8,2 shells (e.g. sodium 2.8.1). Metals sit left/centre: their oxides are basic and they conduct; non-metal oxides are acidic. Group members behave alike because they have the same number of outer electrons; Group 0 noble gases have full outer shells and do not readily react.

Formulae, equations and calculations

Balance equations with state symbols 状态符号: (s), (l), (g), (aq). $M_r$ sums the $A_r$ values. The mole 摩尔 is the amount of substance: $n = m/M_r$ (mass in grams).

Worked check (reacting masses). What mass of magnesium oxide forms from 6.0 g of magnesium? $2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}$: $n(\mathrm{Mg}) = 6.0/24 = 0.25$ mol → $n(\mathrm{MgO}) = 0.25$ mol → mass $= 0.25 \times 40 = 10$ g.

Percentage yield 产率 $=$ actual ÷ theoretical × 100 %. Empirical formula 实验式 is the simplest whole-number ratio (divide each mass by its $A_r$, scale to integers); the molecular formula is the true count ($M_r$ ÷ empirical mass gives the multiplier).

Worked check. A hydrocarbon is 85.7 % C, 14.3 % H: C 85.7/12 = 7.14, H 14.3/1 = 14.3 → ratio 1:2 → empirical CH₂. With $M_r = 56$: 56/14 = 4 → molecular C₄H₈.

(C) Solutions: $n = c \times V$ with $c$ in mol/dm³ and $V$ in dm³ (cm³ ÷ 1000). Gases: molar volume 24 dm³ at rtp, so $n = V/24$. (C-practical) A metal oxide's formula comes from combustion (weigh magnesium, burn, weigh the oxide) or reduction (weigh copper(II) oxide, reduce with hydrogen, weigh the copper).

Bonding

Ionic bonding 离子键: metals lose electrons to non-metals; the charged ions attract electrostatically in a giant lattice — dot-and-cross diagrams show the transfer (Groups 1–3 with 5–7). High melting points (strong attractions); conduct only when molten or dissolved (ions free to move). Learn the charges: Groups 1/2/3 → 1+/2+/3+; Groups 5/6/7 → 3−/2−/1−; Ag⁺, Cu²⁺, Fe²⁺, Fe³⁺, Pb²⁺, Zn²⁺; OH⁻, NH₄⁺, CO₃²⁻, NO₃⁻, SO₄²⁻. Formulae balance the charges: calcium nitrate Ca(NO₃)₂, ammonium sulfate (NH₄)₂SO₄.

Covalent bonding 共价键: shared pairs of electrons (dot-and-cross for H₂, O₂, N₂, halogens, HCl, H₂O, NH₃, CO₂ and small organics). Simple molecular substances have weak intermolecular forces — low melting points, rising with $M_r$; they do not conduct (no ions, no free electrons). Giant covalent: diamond (each C bonded to 4 — very hard, non-conductor), graphite (3 bonds + one free electron per C, layers with weak forces — soft, conducts, slippery), C₆₀ fullerene (simple molecular cage).

(C) Metallic: a lattice of positive ions in a sea of delocalised electrons — electrostatic attraction; the free electrons conduct electricity, and layers slide (malleable).

Electrolysis (C)

Electrolysis 电解 uses electricity to split molten or dissolved ionic compounds: the cathode 阴极 (negative) attracts cations 阳离子; the anode 阳极 (positive) attracts anions 阴离子. Molten PbBr₂ gives Pb at the cathode (Pb²⁺ + 2e⁻ → Pb) and Br₂ at the anode (2Br⁻ → Br₂ + 2e⁻). In aqueous solution, hydrogen forms at the cathode if the metal is more reactive than hydrogen; at the anode, halide ions give the halogen, otherwise oxygen (4OH⁻ → O₂ + 2H₂O + 4e⁻). Loss of electrons is oxidation 氧化, gain is reduction 还原.

Check yourself

1. Describe the particle arrangement and movement in a liquid.
2. Define saturated solution, and state the units of solubility.
3. Which technique separates two miscible liquids with close boiling points?
4. An isotope has 17 protons and 20 neutrons. Give its mass number and the numbers of each particle.
5. Calculate $A_r$ of an element with 80 % of mass 20 and 20 % of mass 22.
6. Balance with state symbols: magnesium + oxygen → magnesium oxide.
7. 8.4 g of the oxide of calcium forms from 6.0 g of calcium. Show the empirical formula of the oxide.
8. Write the formula of ammonium carbonate.
9. Explain why graphite conducts but diamond does not.
10. (C) What forms at each electrode in the electrolysis of molten potassium chloride?

Answers: 1 particles touch but are disordered, moving randomly past each other; 2 a solution containing the maximum mass of dissolved solute at that temperature; g per 100 g of solvent; 3 fractional distillation; 4 mass number 37; 17 protons, 20 neutrons, 17 electrons; 5 $(80 \times 20 + 20 \times 22)/100 = 20.4$; 6 2Mg(s) + O₂(g) → 2MgO(s); 7 Ca 6.0/40 = 0.15 mol; O mass 8.4 − 6.0 = 2.4 g → 2.4/16 = 0.15 mol; ratio 1:1 → empirical formula CaO; 8 (NH₄)₂CO₃; 9 graphite uses three bonds per carbon, leaving one delocalised electron per carbon free to move between layers — diamond uses all four in fixed bonds; 10 (C) potassium at the cathode (K⁺ + e⁻ → K) and chlorine at the anode (2Cl⁻ → Cl₂ + 2e⁻).

The reactive world of inorganic chemistry

Drop lithium in water and it fizzes gently; drop potassium and it ignites. This section follows the patterns behind those observations — the families of the Periodic Table, the air we breathe, the ladder of reactivity that decides how every metal is extracted and why iron rusts — and then the huge practical centre of the course: acids, alkalis, salt preparation and the identification tests.

Edexcel International GCSE Chemistry 4CH1, Section 2 (statements 2.1–2.50) supplies most of the recall and practical questions on both papers. C-suffixed statements are Chemistry-only.

Group 1 and Group 7

Group 1 (Li, Na, K) react with water to make the hydroxide and hydrogen — fizzing, floating, melting (Na), lilac flame (K). The reactions get faster and more vigorous down the group: a family with a trend. (C) Reactivity rises because the outer electron is further from the nucleus and lost more easily.

Group 7 (Cl₂ pale green gas, Br₂ orange-brown liquid, I₂ grey-black solid) darken down the group. A more reactive halogen displaces a less reactive one's salt: chlorine displaces bromine from potassium bromide (solution turns orange). (C) Reactivity falls down the group — the outer shell is further away, so gaining an electron is harder.

Air and gases

Dry air: about 78 % nitrogen, 21 % oxygen, 0.96 % argon, 0.04 % carbon dioxide (the four most abundant). The oxygen percentage can be measured by using it up — iron rusting in damp air over water, or phosphorus burning, so the water rises about one-fifth of the way. Elements burn in oxygen: magnesium gives a bright white flame and white MgO; hydrogen burns with a pop to steam; sulfur gives a blue flame and choking SO₂. Copper(II) carbonate decomposes on heating: CuCO₃ → CuO + CO₂ (limewater turns milky). CO₂ is a greenhouse gas 温室气体 whose rising level contributes to climate change.

The reactivity series and metals

Order to learn: potassium, sodium, lithium, calcium, magnesium, aluminium, (carbon), zinc, iron, (hydrogen), copper, silver, gold. Metals above hydrogen — everything down to iron — react with dilute HCl/H₂SO₄ giving hydrogen; the more reactive, the faster the fizz. A metal displaces any less reactive metal from its salt solution or oxide.

Rusting 生锈 needs oxygen and water; prevented by barrier methods (paint, oil, plastic), galvanising 镀锌 (zinc coat) and sacrificial protection 牺牲保护 (a more reactive metal such as zinc or magnesium corrodes instead). Oxidation 氧化 is loss of electrons or gain of oxygen; reduction 还原 the reverse; a redox reaction has both; the oxidising agent 氧化剂 is reduced, the reducing agent 还原剂 is oxidised.

(C) Extraction matches reactivity: unreactive metals (gold, silver) occur native; those below carbon (iron) are reduced by carbon in a blast furnace; those above (aluminium) need electrolysis of molten ore. Aluminium (low density, resists corrosion), copper (conducts), iron and steels (strong; stainless resists rust) are chosen for their properties. Alloys 合金 mix a metal with other elements and are harder because the different-sized atoms disturb the layers' sliding.

Acids, alkalis and salts

Acids provide H⁺(aq); alkalis provide OH⁻(aq). Indicators: litmus (red in acid, blue in alkali), phenolphthalein (colourless → pink), methyl orange (red → yellow). The pH scale pH值 0–14 classifies solutions (0–3 strongly acidic … 11–14 strongly alkaline); universal indicator measures approximate pH. Neutralisation: acid + alkali → salt + water. (C) Acids are proton donors, bases proton acceptors; alkalis are soluble bases. (C) Titration 滴定: fixed volume of alkali with indicator in a flask, acid from a burette dropwise to the colour change, repeat for concordant results.

Solubility rules: sodium, potassium and ammonium salts — soluble; all nitrates — soluble; chlorides — soluble except AgCl and PbCl₂; sulfates — soluble except BaSO₄, CaSO₄ and PbSO₄; carbonates — insoluble except Na⁺, K⁺, NH₄⁺; hydroxides — insoluble except Na⁺, K⁺ (Ca(OH)₂ slightly).

Acid + metal / base / carbonate gives a salt: e.g. Zn + H₂SO₄ → ZnSO₄ + H₂; CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O. Soluble salt from an insoluble reactant (CuO + warm dilute H₂SO₄): add excess base, filter, evaporate, crystallise. (C) From acid + alkali: titrate, then evaporate. (C) Insoluble salt: mix two solutions so a precipitate 沉淀 forms (Pb(NO₃)₂ + Na₂SO₄ → PbSO₄↓ + 2NaNO₃), filter, wash, dry.

Chemical tests

Gases: hydrogen — a lit splint pops; oxygen — relights a glowing splint; carbon dioxide — turns limewater milky; ammonia — turns damp red litmus blue; chlorine — bleaches damp litmus paper white. Cations: flame tests 焰色试验 (Li⁺ red, Na⁺ yellow, K⁺ lilac, Ca²⁺ orange-red, Cu²⁺ blue-green); NH₄⁺ warms with NaOH(aq) giving ammonia; Cu²⁺/Fe²⁺/Fe³⁺ give blue/green/rust-brown precipitates with NaOH(aq). Anions: Cl⁻/Br⁻/I⁻ give white/cream/pale-yellow precipitates with acidified silver nitrate; SO₄²⁻ gives a white precipitate with acidified barium chloride; CO₃²⁻ fizzes with dilute HCl giving CO₂. Water: turns anhydrous copper(II) sulfate white → blue; a pure water sample boils sharply at 100 °C.

Check yourself

1. Describe the trend in reactivity down Group 1 and give the (C) electron reason.
2. What is seen when chlorine water is added to potassium bromide, and why?
3. Name the four most abundant gases in dry air with approximate percentages.
4. State the two conditions needed for iron to rust, and two prevention methods.
5. Define oxidation and reduction in terms of electrons.
6. (C) Why is aluminium extracted by electrolysis but iron by carbon reduction?
7. Which chlorides are insoluble, and which sulfates?
8. Write the equation for copper(II) oxide reacting with sulfuric acid.
9. Describe the flame-test colours for sodium, potassium and copper.
10. Describe the test for sulfate ions and the positive result.

Answers: 1 reactivity increases down the group (B1); (C) the outer electron is in a shell further from the nucleus, so it is lost more easily; 2 the solution turns orange — chlorine displaces bromine because chlorine is the more reactive halogen; 3 nitrogen ≈78 %, oxygen ≈21 %, argon ≈0.96 %, carbon dioxide ≈0.04 %; 4 oxygen and water; barrier methods (paint/grease/plastic) and galvanising or sacrificial zinc; 5 oxidation is loss of electrons, reduction is gain; 6 (C) aluminium lies above carbon in the reactivity series, so carbon cannot reduce its oxide — electrolysis of molten alumina is needed; iron lies below carbon, so carbon displaces it from its oxide; 7 AgCl and PbCl₂; BaSO₄, CaSO₄ and PbSO₄; 8 CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l); 9 sodium yellow, potassium lilac, copper blue-green; 10 acidify with dilute hydrochloric acid, add barium chloride solution — a white precipitate of BaSO₄ confirms sulfate.

Energy, speed and balance

Three questions decide whether a reaction is useful: does it heat or cool, how fast is it, and does it finish? Physical chemistry answers them — calorimetry and Q = mcΔT, collision theory with its four levers, and the dynamic equilibrium a reversible reaction reaches in a closed vessel.

Edexcel International GCSE Chemistry 4CH1, Section 3 (statements 3.1–3.22C) is short but calculation-heavy. C-suffixed statements are Chemistry-only.

Energetics

Exothermic 放热 reactions give out heat (combustion, neutralisation, most displacements) — the mixture warms, ΔH is negative. Endothermic 吸热 reactions take heat in (thermal decomposition, some dissolving) — the mixture cools. Simple calorimetry 量热: measure the temperature change of a known mass (polystyrene cup for solutions; spirit burner under a water beaker for combustion).

Worked check. 50 g of water rises 24 °C in a spirit-burner trial: $Q = 50 \times 4.2 \times 24 = 5040$ J. If 0.010 mol of fuel burned: $\Delta H = -5040/0.010 = -504\,000\ \mathrm{J\,mol^{-1}} = -504\ \mathrm{kJ\,mol^{-1}}$.

(C) Bond breaking is endothermic; bond making is exothermic. ΔH = bonds broken − bonds formed. Worked check. H₂ + Cl₂ → 2HCl: broken 436 + 242 = 678; formed 2 × 431 = 862; ΔH = 678 − 862 = −184 kJ/mol (exothermic).

Rates of reaction

Experiments follow a reaction that is visible: gas volume (marble + acid), mass loss on a balance, or disappearing cross (sodium thiosulfate + acid). Four levers: surface area of a solid, concentration (or gas pressure), temperature, and a catalyst 催化剂.

Collision theory 碰撞理论 explains them all: reactions need collisions with at least the activation energy 活化能. Powder gives more exposed surface → more collisions per second; higher concentration or pressure packs particles closer → more frequent collisions; higher temperature makes particles faster → collisions both more frequent and more energetic — the only lever that raises the fraction of successful collisions. A catalyst provides an alternative pathway with lower activation energy — more collisions succeed — and is chemically unchanged at the end; it does not change the amount of product, only how fast it appears.

Reversible reactions and equilibria

A reversible reaction 可逆反应 (⇌) runs both ways: hydrated copper(II) sulfate ⇌ anhydrous salt + water (heat drives it forward, water reverses it); ammonium chloride ⇌ ammonia + hydrogen chloride on heating. (C) In a sealed container it reaches dynamic equilibrium 动态平衡: forward and reverse rates are equal, and all concentrations stay constant — both reactions still run.

(C) Temperature up → equilibrium shifts in the endothermic direction; pressure up → shifts towards the side with fewer moles of gas; a catalyst speeds both directions equally and leaves the position unchanged.

Check yourself

1. Classify as exo- or endothermic: combustion, thermal decomposition, neutralisation.
2. 100 g of water cools by 8.0 °C in a dissolving experiment. Calculate Q.
3. (C) State the sign of ΔH for bond breaking and bond making.
4. Use bond energies to find ΔH for 2H₂ + O₂ → 2H₂O (H–H 436, O=O 498, O–H 463 kJ/mol).
5. Why does powdered marble react faster with acid than one large chip?
6. Which rate lever increases the fraction of successful collisions, and why?
7. Define a catalyst.
8. (C) State the two characteristics of dynamic equilibrium.
9. (C) N₂ + 3H₂ ⇌ 2NH₃: what does raising the pressure do, and why?
10. (C) Why does a catalyst not change the position of equilibrium?

Answers: 1 exothermic (combustion), endothermic (thermal decomposition), exothermic (neutralisation); 2 $Q = 100 \times 4.2 \times 8.0 = 3360$ J taken in — the water cools, so the dissolving is endothermic; 3 (C) breaking positive (endothermic), making negative (exothermic); 4 broken: 2(436) + 498 = 1370; formed: 4(463) = 1852; ΔH = 1370 − 1852 = −482 kJ/mol (per 2 mol H₂O); 5 powder exposes far more surface, so more acid collisions per second; 6 temperature — faster particles collide more often AND more collisions exceed the activation energy; 7 a substance that increases the rate but is chemically unchanged at the end; 8 (C) forward and reverse rates equal; concentrations of all substances constant; 9 (C) shifts right — 4 moles of gas on the left, 2 on the right, so higher pressure favours the fewer-mole side.

Answers continued: 10 (C) it lowers the activation energy of forward and reverse equally, speeding both without changing their balance.

Семейство углерода

Нефть представляет собой смесь углеводородов — соединений, состоящих только из углерода и водорода, — от самых мелких молекул газовой фракции до битумных смол. В этом разделе они структурированы по семействам, прослеживается путь нефти от скважины до фракции, после чего последовательно рассматривается каждое семейство: инертные алканы, алкены с двойной связью и (C) спирты, кислоты, эфиры и полимеры, которые следуют за ними.

Edexcel International GCSE Chemistry 4CH1, Раздел 4 (утверждения 4.1–4.50C) завершает курс. Утверждения с суффиксом C относятся только к химии.

Введение и нефть

Углеводород содержит только водород и углерод. Молекулы изображаются с помощью эмпирической, молекулярной, общей, структурной и развернутой формул. Гомологический ряд характеризуется общей формулой и функциональной группой, члены ряда отличаются на CH₂; изомерия — изомеры имеют одинаковую молекулярную формулу, но различаются по структуре. Номенклатура IUPAC охватывает цепочку из шести атомов углерода.

Нефть разделяют методом фракционной перегонки: печь испаряет нефть, пары поднимаются вверх по колонне, охлаждаясь по мере подъёма — мелкие молекулы с низкими температурами кипения скапливаются сверху, крупные вязкие вещества — внизу.

Фракции темнеют и густеют вниз по колонне (повышение температуры кипения и вязкости). Полное горение даёт CO₂ + H₂O; неполное горение также образует угарный газ — ядовитый, поскольку он снижает способность крови переносить кислород — и сажу. В двигателях горячий воздух заставляет азот и кислород реагировать с образованием оксидов азота; примеси серы сгорают, образуя SO₂; оба газа растворяются в осадках, делая их кислыми (кислотные дожди повреждают листья, озёра и здания). Крекинг превращает длинноцепочечные алканы в алкены и короткоцепочечные алканы на диоксиде кремния/алюминии при 600–700 °C — потому что предложение длинных фракций на нефтеперерабатывающем заводе превышает спрос, тогда как короткие фракции требуются в большем объёме.

Алканы и алкены

Алканы ($C_nH_{2n+2}$) являются насыщенными — содержат только одинарные связи C–C. Метан, этан, пропан, бутан, пентан. С галогенами под действием ультрафиолетового света происходит замещение одного атома водорода: CH₄ + Cl₂ → CH₃Cl + HCl (замещение).

Алкены ($C_nH_{2n}$) содержат C=C — ненасыщенные. С бромом они присоединяются по двойной связи, образуя дибромалкан (присоединение): C₂H₄ + Br₂ → C₂H₄Br₂. Бромная вода позволяет их различить: обесцвечивается (оранжевый → бесцветный) алкеном, остаётся неизменной при взаимодействии с алканом.

Спирты, кислоты и эфиры (C)

Спирты содержат –OH: метанол, этанол, пропан-1-ол, бутан-1-ол. Этанол окисляется до уксусной кислоты путём полного сгорания, микробного окисления на воздухе или нагревания с дихроматом калия(VI) в разбавленной серной кислоте. Этанол получают реакцией этена + пара над фосфорнокислотным катализатором (≈300 °C, 60–70 атм) или ферментацией глюкозы с участием ферментов дрожжей в отсутствие воздуха при ≈30 °C — оптимальной температуре для ферментов, при отсутствии кислорода дрожжи производят этанол, а не воду.

Карбоновые кислоты содержат –COOH: метановая, уксусная (этановая), пропановая, бутановая. Их растворы реагируют с металлами (выделяя H₂) и карбонатами (выделяя CO₂), как и другие кислоты; уксус — это разбавленный раствор уксусной (этановой) кислоты.

Эфиры содержат –COO–: этанол + уксусная (этановая) кислота в присутствии кислотного катализатора дают уксусно-этиловый эфир (этилэтанат) и воду (эфирный синтез). Зная спирт и кислоту, соедините –COOH кислоты с –O– спирта. Эфиры летучи, имеют сладкий запах, применяются как ароматизаторы и отдушки.

Синтетические полимеры

Присоединительные полимеры образуются при соединении множества алкеновых мономеров с раскрытием их двойных связей C=C.

Нарисуйте повторяющееся звено полиэтилена, полипропилена, полихлорэтена, тефлона; по повторяющемуся звену закройте две связи обратно в C=C, чтобы восстановить мономер. Проблемы утилизации: полимеры инертны и не биоразлагаются, а при горении могут выделяться токсичные газы. (C) Конденсационная полимеризация: дикарбоновая кислота + диол образуют сложноэфирные связи с выделением воды — полиэстер; некоторые (биопolíэстеры) биоразлагаются.

Проверьте себя

1. Дайте определения углеводороду, гомологическому ряду и изомерии.
2. Назовите шесть основных фракций нефти в порядке повышения температуры кипения.
3. Запишите продукты полного и неполного горения углеводорода.
4. Приведите два загрязняющих газа, образующихся при сжигании топлива, и источник каждого из них.
5. Почему необходима крекинг?
6. Напишите уравнение реакции метана с хлором под действием УФ-света, указав тип реакции.
7. Как бромная вода различает алкен и алкан?
8. (C) Назовите два способа получения этанола, указав одно условие для каждого.
9. (C) Какой спирт и какая кислота образуют уксусно-этиловый эфир, и что еще образуется?
10. (C) Какие мономеры образуют полиэстер, и какая малая молекула выделяется?

Ответы: 1 соединение, состоящее только из водорода и углерода; семейство с общей формулой и функциональной группой, отличающееся на CH₂; молекулы с одинаковой молекулярной формулой, но разной структурой; 2 нефтегазовые газы, бензин, керосин, дизельное топливо, мазут, битум; 3 полное: углекислый газ + вода; неполное добавляет угарный газ (и сажу); 4 угарный газ — неполное горение; оксид серы — сгорание примесей серы; оксиды азота — горячий воздух в двигателе; 5 предложение длинноцепочечных фракций превышает спрос, тогда как короткие фракции и алкены в дефиците; 6 CH₄ + Cl₂ → CH₃Cl + HCl — замещение (фотохимическое, УФ); 7 алкен обесцвечивает оранжевую бромную воду; алкан оставляет её оранжевой; 8 (C) этилен + водяной пар, катализатор фосфорная кислота, ~300 °C, 60–70 атм; брожение глюкозы с дрожжами, без доступа воздуха, ~30 °C; 9 (C) этанол + уксусная (этановая) кислота — также образуется вода (кислотный катализатор); 10 (C) дикарбоновая кислота и диол; вода выделяется на каждой сложноэфирной связи.