主题 1：原子结构与元素周期表
AQA · GCSE · 化学 知识点 1 13:56 英语旁白 · 英中内嵌字幕
章节
文本转录
Begin with particles.
从粒子开始。
Proton number identifies the element; electron arrangement helps explain reaction patterns.
质子数决定元素种类，电子排布有助于解释反应规律。
We will use evidence to choose a separation, test the atomic model and predict displacement.
我们将依据证据选择分离方法、检验原子模型并预测置换反应。
Keep particle counting separate from reaction prediction.
粒子计数与反应预测要分开思考。
Oxygen molecules contain two oxygen atoms, but only one element.
氧分子含两个氧原子，但只含一种元素。
Water contains two different elements chemically joined in a fixed ratio.
水含两种元素，按固定比例化合。
Three water molecules contain six hydrogen atoms and three oxygen atoms.
三个水分子共有六个氢原子和三个氧原子。
A coefficient multiplies the entire formula.
化学式前的系数乘整个化学式。
The product is aluminium oxide, Al two O three.
产物是氧化铝，Al₂O₃。
Changing its subscripts changes the substance.
改下标就改变了物质。
Use three oxygen molecules and two product units to make six oxygen atoms on each side. Then use four aluminium atoms.
用三个氧分子和两个产物单位，使两边各有六个氧原子，再配四个铝原子。
Count each element again at the end.
最后逐一检查每种元素。
Sand, salt and water form a mixture.
沙、盐和水构成混合物。
Filter first: sand remains as residue while salt solution passes through.
先过滤，沙留作滤渣，盐溶液成为滤液。
Gently concentrate the filtrate, then cool to crystallise the salt. Filter and dry the crystals.
温和蒸发部分水，再冷却结晶，过滤并干燥晶体。
Simple distillation instead recovers water as condensate.
若要回收水，应采用简单蒸馏收集冷凝液。
All of these are physical separations.
这些都是物理分离。
Use fractional distillation for miscible liquids with different boiling points, such as ethanol and water.
对乙醇和水这类沸点不同、互溶的液体用分馏。
Repeated evaporation and condensation enrich the lower-boiling liquid in the collected fraction.
反复蒸发和冷凝，使收集馏分中较低沸点液体的比例增加。
Chromatography separates dissolved substances by their different movement between stationary and mobile phases.
色谱法利用溶质在固定相与流动相之间运动的差别来分离物质。
A single coloured spot is evidence, not automatic proof of purity.
一个色斑只是证据，不能自动证明纯度。
Follow the changing representation: an indivisible sphere, negative electrons in a positive background, then a tiny positive nucleus.
沿模型变化观察：不可分的小球、正电背景中的负电子，再到微小正电原子核。
Bohr introduced specific electron energy levels. Later evidence identified protons and neutrons.
玻尔提出特定电子能级，后续证据确认质子和中子。
The next frame connects the scattering observations to the nuclear model.
下一页把散射现象与核式模型联系起来。
Explain each observation before naming the replacement model.
先解释观察，再说替代模型。
Most alpha particles pass through, so most atomic volume is empty space.
大多数α粒子直穿，说明原子的大部分空间是空的。
A small fraction are strongly deflected or rebound: positive charge and most mass are concentrated in a tiny nucleus.
少量粒子大角度偏转或反弹，说明正电荷与大部分质量集中在微小原子核中。
This evidence replaced the spread-out positive charge of plum pudding.
证据否定了正电荷均匀分布的布丁模型。
Bohr later introduced specific electron energy levels.
玻尔随后提出特定电子能级。
Pause and calculate the three particle counts for the ion.
暂停，计算这个离子的三种粒子数。
Atomic number is twelve, mass number is twenty-six, and the charge is two plus.
原子序数为12，质量数为26，电荷为2+。
Explain which particle changed when the ion formed. Do not change the nucleus to account for charge.
说明成离子时改变的是哪种粒子，不能用改变原子核来解释电荷。
The correct option is A.
正确选项为A。
Twelve protons identify magnesium.
12个质子确定它是镁。
Neutrons equal twenty-six minus twelve, giving fourteen.
中子数是26减12，等于14。
A two-plus ion has two fewer electrons than protons, so ten.
2+离子比质子少两个电子，所以有10个。
Option B treats the mass number as neutron number; C changes the element; D gains electrons and would make a negative ion.
B把质量数当中子数；C改变了元素；D增加电子，形成的会是负离子。
Relative atomic mass is an abundance-weighted mean on the carbon-twelve scale.
相对原子质量是以碳12为基准的丰度加权平均。
Multiply each isotope mass by its percentage, add and divide by one hundred. The result is thirty-five point five, closer to the more common isotope.
每种同位素质量乘它的百分丰度，相加后除以100，得到35.5，更接近丰度较高的同位素。
Relative atomic mass has no unit; mass number is a count for one isotope.
相对原子质量没有单位；质量数是某一种同位素的粒子计数。
An atom has radius about zero point one nanometre, or ten to the minus ten metre.
原子半径约为0.1纳米，也就是10的负10次方米。
The nucleus radius is less than one ten-thousandth of the atom radius. Almost all mass lies in that nucleus.
原子核半径小于原子半径的一万分之一，而几乎全部质量都集中在原子核中。
A classroom shell diagram is not a scale drawing and must not be used to measure a real nucleus.
课堂电子层图并不按比例绘制，不能拿它测量真实原子核。
Sodium has eleven electrons arranged two, eight, one.
钠有11个电子，排布为2、8、1。
Three occupied shells place it in Period three, and one outer electron explains Group one.
三个占据的电子层说明它在第三周期，一个外层电子对应第一族。
Chlorine is two, eight, seven, in the same period but Group seven.
氯是2、8、7，在同一周期但位于第七族。
Helium is the important Group zero exception: its first shell is full with two electrons.
氦是零族的重要例外：第一层只有两个电子就已充满。
Mendeleev worked with incomplete information.
门捷列夫掌握的信息并不完整。
He left gaps rather than placing every known element in strict atomic-weight order. He predicted properties of missing elements.
他没有把所有已知元素硬塞进严格的原子量顺序，而是留下空位，预测未知元素的性质。
Later discoveries matching those predictions supported his arrangement.
后来发现的元素符合预测，支持了他的排列。
Modern proton-number order and isotope knowledge explain why atomic-weight order sometimes fails.
现代质子数顺序以及同位素知识解释了为什么原子量顺序有时失效。
Group one metals become more reactive down the group because increased distance and shielding make the outer electron easier to lose.
第一族向下反应性增强，因为距离与屏蔽增加，外层电子更容易失去。
Group seven becomes less reactive because an incoming electron is less strongly attracted down the group.
第七族向下吸引外来电子的能力减弱，反应性降低。
Group zero atoms have stable outer arrangements and are monatomic.
零族原子外层稳定，通常以单原子存在。
Their boiling points rise down the group as attractions between atoms increase.
向下原子间吸引增强，所以沸点上升。
Group one metals become more reactive down the group because increased distance and shielding make the outer electron easier to lose.
第一族向下反应性增强，因为距离与屏蔽增加，外层电子更容易失去。
Group seven becomes less reactive because an incoming electron is less strongly attracted down the group.
第七族向下吸引外来电子的能力减弱，反应性降低。
Group zero atoms have stable outer arrangements and are monatomic.
零族原子外层稳定，通常以单原子存在。
Their boiling points rise down the group as attractions between atoms increase.
向下原子间吸引增强，所以沸点上升。
Sodium reacts with water to produce sodium hydroxide solution and hydrogen.
钠与水反应生成氢氧化钠溶液和氢气。
Balance two sodium atoms with two water molecules.
用两个钠原子配两个水分子。
The solution is alkaline because hydroxide ions are present.
溶液因含氢氧根离子而显碱性。
Sodium also reacts with chlorine to form sodium chloride and with oxygen to form an oxide.
钠也与氯气反应生成氯化钠，与氧反应生成氧化物。
Compare the specified lithium, sodium and potassium observations using the same controlled teacher demonstration.
通过相同条件下的教师演示比较锂、钠、钾的指定现象。
Predict whether chlorine water reacts with potassium bromide solution, and whether iodine reacts with the same solution.
预测氯水与溴化钾溶液是否反应，以及碘与相同溶液是否反应。
Write an equation for any reaction.
为发生的反应写方程式。
Your explanation must use the reactivity order, not the memorised colour alone.
解释必须依据反应性顺序，不能只背颜色。
The correct option is B.
正确选项为B。
Chlorine is more reactive than bromine, so chlorine plus two potassium bromide gives two potassium chloride plus bromine.
氯比溴活泼，Cl₂与2KBr反应生成2KCl和Br₂。
Iodine is less reactive, so it cannot displace bromine.
碘比溴不活泼，不能置换出溴。
Halogens are diatomic, so write chlorine and bromine with subscript two.
卤素以双原子分子存在，因此氯和溴都要带下标2。
Bromine gives an orange aqueous solution; iodine colour alone would not prove displacement.
溴水呈橙色；仅看碘本身的颜色不能证明发生了置换。
This section belongs to separate GCSE Chemistry.
本节属于单独的GCSE化学。
Compared with Group one, the specified transition metals are generally stronger, harder, denser and less reactive with water, oxygen and halogens, with higher melting points.
与第一族相比，指定过渡金属通常更强、更硬、密度更大、熔点更高，对水、氧和卤素的反应性较弱。
Their compounds often have colours; iron forms two-plus and three-plus ions. Iron catalyses the Haber process.
它们的化合物常有颜色，铁可形成2+和3+离子，铁还催化哈伯法。
Treat these as typical properties of the specified examples, not absolute claims about every metal.
这些是指定实例的典型性质，不是适用于所有金属的绝对规律。
This is Higher Tier.
这是高阶内容。
Write the full reaction of aqueous silver nitrate and sodium chloride, then separate soluble substances into ions.
先写硝酸银溶液与氯化钠溶液的完整反应，再把可溶物拆成离子。
Cancel identical sodium and nitrate ions on both sides. Keep solid silver chloride intact.
消去两边相同的钠离子和硝酸根离子，保留固体氯化银。
The net equation conserves one silver atom, one chlorine atom and total charge zero.
净离子方程式守恒银原子、氯原子以及总电荷零。
In a half equation, place electrons on the side needed to conserve charge.
半反应中，电子应放在能使电荷守恒的一侧。
Use the supplied periodic table and learn the first twenty names and symbols.
使用题给元素周期表，熟悉前二十号元素的名称和符号。
Capitalisation matters: a second lowercase letter belongs to the same element symbol.
大小写很重要，第二个小写字母仍属于同一个元素符号。
A coefficient multiplies the entire chemical formula.
系数乘整个化学式。
Pause and choose a sequence using particle size, solubility and boiling point.
暂停，根据粒子大小、溶解性和沸点选择操作顺序。
State which product is retained and which passes through.
说明哪种产物被截留，哪种通过滤纸。
The filter separates an insoluble solid, not two mixed liquids.
过滤分离不溶固体，不能分开两种混合液体。
Repeated evaporation and condensation in the fractionating column improve liquid separation.
分馏柱中的反复蒸发和冷凝改善液体分离。
These are physical changes that do not make new substances.
这些都是不产生新物质的物理变化。
Pause and use diameter for the spacing of neighbouring atom centres.
暂停，相邻原子中心间距要用直径。
Use the specified less-than-one-ten-thousandth radius ratio for the enlarged nucleus.
放大后的原子核使用题给小于原子半径万分之一的关系。
The estimate assumes touching atoms and no gaps.
估计假设原子相接且没有空隙。
The radius ratio is an upper comparison, not an exact nucleus size.
半径比例是上限比较，不是精确核半径。
A classroom shell drawing is schematic and cannot supply a measured nuclear radius.
课堂电子层图是示意图，不能测量真实原子核。
Count the dots in every shell.
逐层数清电子点。
Calcium has four occupied shells and twenty electrons; chlorine has three and seventeen.
钙有四个占据电子层、二十个电子，氯有三个电子层、十七个电子。
These are neutral atoms. The shell sizes in the diagram are not to scale.
图示是中性原子，电子层大小不按真实比例。
Pause before the answer.
暂停作答。
Add the electrons to identify proton number in the neutral atom; use occupied shells for period and outer electrons for group.
中性原子的电子总数等于质子数；用占据层数判断周期，用外层电子判断主族。
The first-twenty model fills two, eight, eight, then begins the fourth shell.
前二十号元素模型按二、八、八填充，再进入第四层。
Positive charge does not mean extra protons.
正电荷不表示增加了质子。
Helium remains the full-first-shell exception in Group zero.
氦是零族中第一层两个电子即充满的例外。
An abundance-weighted isotope average need not increase strictly with proton number.
同位素丰度加权平均不一定随质子数严格增加。
Isotopes differ in neutron number, while proton number fixes element identity.
同位素的中子数不同，而质子数决定元素。
This explains why atomic-weight order could conflict with related properties.
这解释了为什么按原子量排列有时与性质分组冲突。
Use characteristic properties and electronic structure together.
结合典型性质与电子结构判断。
The periodic table places most metals towards the left and bottom.
周期表中金属主要位于左侧和下方。
A property such as conductivity has exceptions and is not by itself a universal definition.
导电性等性质存在例外，不能单独作为无例外的定义。
Lithium, sodium and potassium react with oxygen, chlorine and water.
锂、钠、钾与氧、氯和水反应。
Water gives hydroxide and hydrogen.
与水生成氢氧化物和氢气。
Their increasing water reactivity is seen in supervised demonstrations; use given oxygen-product formulas rather than assuming every condition gives one identical oxide type.
教师演示可比较与水反应性；氧化产物应使用题给化学式，不应假设各种条件都生成同一类氧化物。
At room temperature chlorine is a pale-green gas, bromine a red-brown liquid and iodine a grey-black solid.
室温下氯为浅绿色气体，溴为红棕色液体，碘为灰黑色固体。
A halogen can form ionic compounds with metals and covalent compounds with non-metals. Do not infer identical bonding from the presence of chlorine alone.
卤素与金属可形成离子化合物，与非金属可形成共价化合物，不能只因含氯就判断成键相同。
Pause and use independent observations.
暂停，使用三个独立观察。
Do not claim these three values identify the element or establish every property of all its compounds.
不能声称这些数值确定了元素身份，也不能推断它所有化合物的性质。
Typical properties guide a defensible prediction.
典型性质支持合理预测。
To show catalytic activity, compare reaction rates under otherwise identical conditions; colour alone does not demonstrate catalysis.
要证明催化作用，应在其他条件相同下比较反应速率；颜色本身不能证明催化。
The two compounds contain the same elements but different proportions and iron charges.
两种化合物元素相同，但比例和铁离子电荷不同。
Many transition elements have variable ion charges and coloured compounds.
许多过渡元素具有可变价离子和有色化合物。
Iron metal catalyses the Haber process; nickel catalyses hydrogenation.
铁金属催化哈伯法，镍催化加氢。
Attempt a final transfer task. A sample dissolves in water, and filtering leaves no residue.
尝试最后的迁移题：样品溶于水，过滤后没有滤渣。
Does this prove the sample is an element?
这能证明样品是元素吗？
Choose a method to recover the dissolved solid. Then explain which further information would distinguish element from compound.
选择回收溶解固体的方法，再说明需要什么进一步信息来区分元素与化合物。
Use the definitions and separation model, not a guessed substance name.
依据定义与分离模型，不要猜物质名称。
The answer is C.
答案为C。
Soluble elements and soluble compounds are both possible in principle, so the observation does not identify composition.
仅凭溶解现象无法确定组成。
Concentrate, cool, filter and dry crystals to recover the solute.
温和浓缩、冷却、过滤并干燥晶体可回收溶质。
To classify it as an element or compound, establish whether it contains one element or different elements chemically combined.
要判断它是元素还是化合物，需要确定它只含一种元素，还是含不同元素的化合。
Solubility alone is insufficient evidence.
单独的溶解性证据不足。