📚 Year 10 Edexcel Chemistry: Unit Test Mock Paper Analysis | Year 10 Edexcel 化学：单元测试模拟卷解析
This article provides a detailed analysis of a typical Year 10 Edexcel Chemistry unit test mock paper, covering key topics such as atomic structure, bonding, quantitative chemistry, acids and bases, and electrolysis. Each question is broken down with step-by-step reasoning and mark scheme guidance to help you improve exam technique and secure top grades.
本文详细解析了一份典型的 Year 10 Edexcel 化学单元测试模拟卷，涵盖原子结构、化学键、定量化学、酸碱与电解等核心主题。每一道题目都配有逐步推理和评分标准指导，帮助你提升应试技巧，争取高分。
1. Atomic Structure and Isotopes | 原子结构与同位素
Question 1: Give the number of protons, neutrons and electrons in an atom of ³⁵Cl. Then explain why ³⁵Cl and ³⁷Cl are isotopes. (4 marks)
问题1：指出一个 ³⁵Cl 原子中的质子、中子和电子数目。然后解释为什么 ³⁵Cl 和 ³⁷Cl 互为同位素。（4分）
sotopes are atoms of the same element with the same number of protons but different numbers of neutrons. Hence ³⁵Cl and ³⁷Cl are isotopes because both have 17 protons, but ³⁵Cl has 18 neutrons while ³⁷Cl has 20 neutrons.
氯原子的原子序数为17，因此含有17个质子和17个电子。³⁵Cl的质量数为35，中子数 = 35 - 17 = 18。同位素是同一种元素的原子，具有相同的质子数但中子数不同。因此 ³⁵Cl 和 ³⁷Cl 是同位素，因为它们都有17个质子，但 ³⁵Cl 有18个中子，³⁷Cl 有20个中子。
t the atomic number from the mass number. Also, a full mark answer must explicitly state that isotopes have the same proton number, not just 'different neutrons'.
常见错误：学生常常混淆质量数和原子序数，直接写出中子数 = 35。一定要用质量数减去原子序数。此外，得满分的答案必须明确说明同位素具有相同的质子数，而不仅仅是“中子数不同”。
2. Ionic Bonding and Ionic Compounds | 离子键与离子化合物
Question 2: Draw a dot-and-cross diagram to show the bonding in magnesium oxide, MgO. Explain why MgO has a very high melting point. (4 marks)
问题2：用点叉图画出氧化镁 MgO 的键合情况。解释为什么 MgO 的熔点非常高。（4分）
Magnesium (2,8,2) loses its two outer electrons to become Mg²⁺ with a full outer shell (2,8). Oxygen (2,6) gains these two electrons to become O²⁻ (2,8). The dot-and-cross diagram should show Mg with no electrons drawn around it, a squared bracket with eight crosses, and a 2+ charge; oxygen with eight dots and crosses combined inside a squared bracket with a 2- charge. The high melting point is due to the strong electrostatic forces of attraction between oppositely charged Mg²⁺ and O²⁻ ions in the giant ionic lattice; a large amount of energy is needed to overcome these forces.
镁（2,8,2）失去两个最外层电子，形成 Mg²⁺ 并获得满壳层（2,8）。氧（2,6）得到这两个电子变成 O²⁻（2,8）。点叉图中，Mg 周围不画电子，用方括号加上八个叉，并标2+电荷；氧周围有八个点和叉的组合，置于方括号内，标2-电荷。MgO 的高熔点是由于巨型离子晶格中带相反电荷的 Mg²⁺ 和 O²⁻ 离子之间强烈的静电吸引力；克服这些力需要大量的能量。
Examiner tip: Ensure brackets and charges are clearly shown; missing charges will lose marks. Explain melting point using the term "electrostatic forces" rather than just "strong bonds".
考官提示：务必清晰画出方括号和电荷；缺少电荷会被扣分。解释熔点时要用“静电力”这一术语，而不是仅说“强键”。
3. Covalent Bonding and Simple Molecules | 共价键与简单分子
Question 3: Draw a dot-and-cross diagram for methane, CH₄. Explain why methane has a low boiling point. (4 marks)
问题3：画出甲烷 CH₄ 的点叉图。解释为什么甲烷的沸点很低。（4分）
Carbon has four outer electrons, and each hydrogen has one. Carbon shares one electron with each of the four hydrogen atoms, forming four single covalent bonds. The dot-and-cross diagram shows carbon in the centre with four hydrogen atoms around it, overlapping circles each containing one dot from H and one cross from C (or vice versa). Methane has a low boiling point because it exists as simple molecules; the intermolecular forces between CH₄ molecules are weak van der Waals' forces, which require little energy to overcome. Melting and boiling do not break the strong covalent bonds within the molecule.
碳有四个最外层电子，每个氢有一个。碳与四个氢原子各共享一个电子，形成四个共价单键。点叉图中碳在中心，周围四个氢原子，重叠的圆圈中各自含有一个来自氢的点和一个来自碳的叉（或反之）。甲烷的沸点低是因为它以简单分子形式存在；CH₄ 分子之间的分子间作用力是微弱的范德华力，只需很少的能量就能被克服。熔化和沸腾并不会破坏分子内的强共价键。
Key distinction: Many students lose marks by saying "covalent bonds break" during boiling. Always state that only intermolecular forces are overcome, not the covalent bonds inside the molecule.
关键区别：许多学生因说“共价键断裂”而失分。一定要说明只有分子间作用力被克服，分子内部的共价键并没有断裂。
4. Giant Covalent Structures | 巨型共价结构
Question 4: Diamond and graphite are both allotropes of carbon. Explain why diamond is hard and does not conduct electricity, whereas graphite is soft and conducts electricity. (6 marks)
问题4：金刚石和石墨都是碳的同素异形体。解释为什么金刚石坚硬且不导电，而石墨柔软且能导电。（6分）
In diamond, each carbon atom forms four strong covalent bonds to four other carbon atoms in a tetrahedral arrangement, creating a rigid giant covalent lattice. This makes diamond extremely hard. All outer electrons are localised in covalent bonds, so there are no delocalised electrons or charged particles free to move; hence diamond does not conduct electricity. Graphite has a layered structure where each carbon is bonded to only three others, forming hexagons. The fourth outer electron becomes delocalised and can move between the layers, carrying charge, which allows electrical conductivity. The layers are held together by weak forces, so they can slide over each other easily, making graphite soft and slippery.
在金刚石中，每个碳原子与另外四个碳原子形成四个强共价键，呈四面体排列，构成坚硬的巨型共价晶格，因此金刚石极为坚硬。所有最外层电子都定域在共价键中，没有自由移动的离域电子或带电粒子，所以金刚石不导电。石墨具有层状结构，每个碳原子只与另外三个碳原子成键，形成六边形。第四个最外层电子变为离域电子，可以在层间移动，携带电荷，从而能够导电。石墨层与层之间依靠微弱的力结合，因此层间可以轻易相互滑动，使石墨柔软滑腻。
or. Graphite: 3 bonds + delocalised electrons + weak interlayer forces = soft + conductor.
视觉核对：确保你的答案将结构与性质联系起来。金刚石：4个键 + 无自由电子 = 坚硬 + 绝缘体。石墨：3个键 + 离域电子 + 弱层间力 = 柔软 + 导体。
5. States of Matter and Changes of State | 物质的状态变化
Question 5: When iodine crystals are heated gently, they change directly from solid to gas without melting. Name this process and explain what happens to the particles. (3 marks)
问题5：碘晶体被微热时，会不经熔化直接由固态变为气态。写出这一过程的名称，并从粒子角度加以解释。（3分）
The process is called sublimation. Iodine exists as simple molecules (I₂). In the solid state, I₂ molecules are held together in a regular lattice by weak intermolecular forces. When heated, the energy supplied breaks only the intermolecular forces between I₂ molecules, not the covalent bonds within the molecules. The molecules gain enough kinetic energy to move far apart and form a gas. Because the intermolecular forces are weak, iodine can change directly from solid to gas.
这一过程叫做升华。碘以简单分子（I₂）形式存在。在固态时，I₂ 分子通过微弱的分子间作用力按规则排列成晶格。加热时，提供的能量仅仅破坏了 I₂ 分子之间的分子间作用力，而分子内部的共价键并未断裂。分子获得了足够的动能，彼此远离形成气体。由于分子间力很弱，碘可以直接从固态变为气态。
Mark reminder: You must name "sublimation" and state that intermolecular forces are overcome, leaving the covalent bond within the I₂ molecule intact. Using the term "intermolecular forces" is essential.
评分提醒：你必须写出“升华”，并说明克服的是分子间作用力，而 I₂ 分子内的共价键保持完整。使用“分子间作用力”一词至关重要。
6. Chemical Equations and Conservation of Mass | 化学方程式与质量守恒
uation: __Mg + __O₂ → __MgO. Explain why mass is conserved during this reaction. (3 marks)
问题6：配平下列方程式：__Mg + __O₂ → __MgO。解释为什么该反应中质量守恒。（3分）
O. To balance, we need 2 Mg atoms on each side and 2 O atoms on each side. Mass is conserved because no atoms are created or destroyed in a chemical reaction; they are only rearranged. The total number and types of atoms in the reactants are exactly the same as in the products, so the total mass remains unchanged.
配平后的方程式为 2Mg + O₂ → 2MgO。配平时，我们需要左右两边各有2个 Mg 原子和2个 O 原子。质量守恒是因为化学反应中原子既不能被创造也不能被消灭，只是重新排列。反应物中原子的总数和种类与生成物中完全相同，因此总质量保持不变。
Common error: Some students think mass can change because gas is produced. Even if a gas escapes, if you collect all products, the total mass stays the same. State the rearrangement of atoms clearly.
常见错误：有些学生认为因为有气体产生，质量会改变。即使气体逸出，如果能收集所有产物，总质量仍然不变。答题时要清楚说明原子的重新排列。
7. Mole Calculations and Reacting Masses | 摩尔计算与反应质量
ed when 10 g of calcium carbonate decomposes completely. (Aᵣ: Ca=40, C=12, O=16) (4 marks)
问题7：碳酸钙加热分解：CaCO₃ → CaO + CO₂。计算10 g碳酸钙完全分解时生成的氧化钙（CaO）的质量。（相对原子质量：Ca=40, C=12, O=16）（4分）
0 mol. Step 3: From the balanced equation, 1 mol CaCO₃ produces 1 mol CaO, so moles of CaO = 0.10 mol. Step 4: Mass of CaO = moles × Mᵣ = 0.10 × 56 = 5.6 g.
第一步：计算 CaCO₃ 的相对式量：40 + 12 + (3 × 16) = 100。CaO 的相对式量：40 + 16 = 56。第二步：CaCO₃ 的物质的量 = 质量 ÷ 相对式量 = 10 ÷ 100 = 0.10 mol。第三步：由配平方程式可知，1 mol CaCO₃ 生成 1 mol CaO，所以 CaO 的物质的量 = 0.10 mol。第四步：CaO 的质量 = 物质的量 × 相对式量 = 0.10 × 56 = 5.6 g。
Method tip: Always show the mole ratio step clearly. Marks are awarded for correct Mᵣ values, mole calculation, ratio application and final mass. Include units throughout.
方法提示：一定要清晰地展示物质的量比例的步骤。评分点包括正确的相对式量、摩尔计算、比例应用和最终质量。整个过程中要写上单位。
8. Acids, Bases and Neutralisation | 酸碱与中和反应
Question 8: Write the balanced symbol equation for the reaction between hydrochloric acid and sodium hydroxide. Describe how you would obtain pure, dry crystals of sodium chloride from the resulting solution. (5 marks)
问题8：书写盐酸与氢氧化钠反应的配平符号方程式。描述你如何从所得溶液中获得纯净干燥的氯化钠晶体。（5分）
NaCl + H₂O. To obtain pure dry sodium chloride crystals: first pour the neutral solution into an evaporating basin and heat it gently over a water bath or a low flame to evaporate some of the water until the solution is saturated (or crystals start to form at the edge). Then leave the solution to cool slowly, allowing more crystals to form. Filter the mixture to collect the crystals and finally pat them dry with filter paper or leave them in a warm place to dry completely. Avoid heating to dryness to prevent the crystals from decomposing or splattering.
配平方程式为：HCl + NaOH → NaCl + H₂O。获取纯净干燥氯化钠晶体的方法：首先将中和后的溶液倒入蒸发皿中，在水浴或小火上缓慢加热，蒸发掉部分水分，直到溶液达到饱和（或边缘开始出现晶体）。然后让溶液慢慢冷却，析出更多晶体。过滤混合物，收集晶体，最后用滤纸吸干或在温暖处晾干。避免加热至完全干燥，以防晶体分解或迸溅。
Practical detail: During evaporation, do not boil dry; use a water bath for better control. Also mention testing for neutrality with litmus paper if the solution was not pre-neutralised, but in this case it is already neutral.
实验细节：蒸发时不要蒸干；使用水浴能更好地控制。如果溶液事先未经中和，要提到用石蕊试纸测试中性，但此处已经是中性溶液。
9. Electrolysis and Molten Electrolysis | 电解与熔融电解
Question 9: Molten lead(II) bromide, PbBr₂, is electrolysed using inert electrodes. Predict the products at the anode and cathode and write the half-equations. (5 marks)
问题9：使用惰性电极电解熔融的溴化铅 PbBr₂。预测阳极和阴极的产物，并写出半反应方程式。（5分）
trons and are reduced to form molten lead metal. The negative Br⁻ ions move to the positive anode, where they lose electrons and are oxidised to bromine gas. Cathode half-equation: Pb²⁺ + 2e⁻ → Pb. Anode half-equation: 2Br⁻ → Br₂ + 2e⁻. Observation: silvery liquid lead collects at the bottom, and red-brown bromine gas is released at the anode.
熔融溴化铅含有 Pb²⁺ 和 Br⁻ 离子。电解时，带正电的 Pb²⁺ 离子向阴极（负极）移动，在阴极得到电子被还原，生成熔融铅金属。带负电的 Br⁻ 离子向阳极（正极）移动，在阳极失去电子被氧化，生成溴气。阴极半反应：Pb²⁺ + 2e⁻ → Pb。阳极半反应：2Br⁻ → Br₂ + 2e⁻。可观察到的现象：银白色液态铅聚集在底部，阳极有红棕色溴气放出。
Electrode identification: Cathode is negative and attracts cations; anode is positive and attracts anions. Half-equations must balance charge and atoms. Use state symbols: Pb (l) and Br₂ (g).
电极识别：阴极是负极，吸引阳离子；阳极是正极，吸引阴离子。半反应必须同时配平电荷和原子。要标注状态符号：Pb (l) 和 Br₂ (g)。
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