📚 Year 11 Cambridge Chemistry: Unit Test Mock Paper Walkthrough | 剑桥Year 11化学：单元测试模拟卷解析
Mock papers are one of the most effective revision tools for Cambridge IGCSE Chemistry. By working through typical unit test questions and understanding the reasoning behind each answer, students can identify gaps, master core principles and build confidence for the real examination. This walkthrough analyses a representative mock paper covering atomic structure, bonding, stoichiometry, electrolysis, kinetics, equilibrium, acid–base calculations, organic chemistry, energetics and periodic trends. Each question is followed by a step‑by‑step bilingual explanation so that you can consolidate your learning in both English and Chinese. Use this article as a self‑study companion or a classroom discussion guide.
模拟卷是剑桥IGCSE化学最有效的复习工具之一。通过演练典型的单元测试题目并理解每道题背后的推理，学生可以发现知识盲区、掌握核心原理并为真正的大考建立信心。本文解析一份涵盖原子结构、化学键、化学计量、电解、动力学、平衡、酸碱计算、有机化学、能量学和周期规律的模拟卷。每道题后都提供中英双语的分步讲解，帮助你巩固学习。你可以将本文用作自学伴侣，也可作为课堂讨论的指南。
1. Multiple Choice on Atomic Structure | 原子结构选择题
Question: Which row correctly describes the relative charge and relative mass of a neutron?
题目：哪一行正确描述了中子的相对电荷和相对质量？
A. Relative charge +1, relative mass 1
A. 相对电荷 +1，相对质量 1
B. Relative charge 0, relative mass 1
B. 相对电荷 0，相对质量 1
C. Relative charge –1, relative mass 1/1840
C. 相对电荷 –1，相对质量 1/1840
D. Relative charge 0, relative mass 1/1840
D. 相对电荷 0，相对质量 1/1840
Answer and explanation: A neutron carries no electric charge and its relative mass is 1 (compared to the carbon‑12 scale). Protons have a relative charge of +1 and mass 1; electrons have a charge of –1 and a negligible mass of about 1/1840. Therefore, the correct option is B.
答案与解析：中子不带电，其相对质量为1（以碳‑12为标准）。质子相对电荷 +1、质量1；电子电荷 –1、质量可忽略，约为1/1840。因此正确答案为B。
2. Ionic Bonding and Formulae | 离子键与化学式
Question: Aluminium ions carry a 3+ charge and oxide ions carry a 2– charge. Write the formula for aluminium oxide.
题目：铝离子带3+电荷，氧离子带2–电荷。写出氧化铝的化学式。
The compound must be electrically neutral. The lowest common multiple of 3 and 2 is 6. To achieve a total positive charge of +6, two Al³⁺ ions are needed (2 × +3 = +6). To achieve a total negative charge of –6, three O²⁻ ions are needed (3 × –2 = –6). The formula is therefore Al₂O₃. Do not write the charges in the final formula.
化合物必须呈电中性。3和2的最小公倍数是6。为得到+6的总正电荷，需要两个Al³⁺离子（2 × +3 = +6）；为得到–6的总负电荷，需要三个O²⁻离子（3 × –2 = –6）。因此化学式为Al₂O₃。最终化学式中不写出电荷。
3. Balancing Equations and Mole Concept | 方程式配平与摩尔概念
Question: Propane (C₃H₈) burns completely in oxygen. Balance the equation and calculate how many moles of O₂ are required to combust 0.50 mol of propane.
题目：丙烷（C₃H₈）在氧气中完全燃烧。配平方程式并计算燃烧0.50 mol丙烷需要多少摩尔氧气。
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
The balanced equation shows a 1 : 5 mole ratio between propane and oxygen. For every 1 mol of C₃H₈, 5 mol of O₂ are needed. With 0.50 mol of propane, the required oxygen is 0.50 × 5 = 2.5 mol. Always use the coefficients in the balanced equation to convert between moles of different substances.
配平后的方程式显示丙烷与氧气的摩尔比为1 : 5。每1 mol C₃H₈需要5 mol O₂。现有0.50 mol丙烷，所需氧气为0.50 × 5 = 2.5 mol。始终利用配平方程中的系数进行不同物质之间的摩尔换算。
4. Electrolysis of Aqueous Solutions | 水溶液电解
Question: Concentrated aqueous sodium chloride is electrolysed using inert electrodes. State the product at the cathode and write the half‑equation for the anode reaction.
题目：用惰性电极电解浓氯化钠水溶液。指出阴极产物并写出阳极反应的半方程式。
At the cathode, hydrogen gas is formed because H⁺ ions from water are discharged in preference to Na⁺ ions. The half‑equation is:
2H⁺ + 2e⁻ → H₂
At the anode, chloride ions are discharged instead of hydroxide ions due to the high concentration of Cl⁻. The half‑equation is:
2Cl⁻ → Cl₂ + 2e⁻
Thus, the overall products are hydrogen at the cathode and chlorine at the anode. If the solution were dilute, oxygen might be released at the anode instead.
在阴极，氢气生成，因为水中的H⁺离子优先于Na⁺离子放电。半方程式为：2H⁺ + 2e⁻ → H₂。在阳极，由于Cl⁻浓度高，氯离子优先于氢氧根离子放电。半方程式为：2Cl⁻ → Cl₂ + 2e⁻。因此总产物为阴极氢气和阳极氯气。若为稀溶液，阳极可能放出氧气。
5. Rates of Reaction Graph Analysis | 反应速率图像分析
Question: A student investigates the effect of concentration on the rate of reaction between magnesium and excess hydrochloric acid. The graph shows the volume of hydrogen produced against time for two experiments: one with 1.0 mol/dm³ HCl and one with 2.0 mol/dm³ HCl. Explain why the curve for the higher concentration is steeper at the start but both curves eventually reach the same final volume of gas.
题目：某学生探究浓度对镁与过量盐酸反应速率的影响。图像显示氢气体积随时间的变化，分别是1.0 mol/dm³和2.0 mol/dm³ HCl。解释为什么高浓度曲线起始斜率更大，但两条曲线最终达到相同的气体体积。
With a higher concentration of HCl, there are more H⁺ ions per unit volume. This increases the frequency of effective collisions between H⁺ ions and magnesium atoms, so the initial rate of reaction is greater – the steeper gradient. However, the same mass of magnesium is used in both experiments, and it is the limiting reactant. Therefore, the total amount of hydrogen produced depends only on the amount of magnesium. Once all the magnesium has reacted, the gas volume stops increasing and is identical in both cases.
较高浓度的盐酸在单位体积中含有更多的H⁺离子，这增加了H⁺离子与镁原子之间有效碰撞的频率，因此初始反应速率更大——曲线更陡。但是，两个实验使用的镁质量相同，镁是限制反应物。因此，生成的氢气总量只取决于镁的量。一旦镁全部反应完，气体体积不再增加，二者最终体积相同。
6. Equilibrium and Le Chatelier’s Principle | 平衡与勒夏特列原理
Question: The Haber process is represented by the thermal equation: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol. Predict the direction of shift in equilibrium position when the temperature is raised and explain your reasoning.
题目：哈伯法表示为：N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = –92 kJ/mol。预测温度升高时平衡位置移动的方向并解释。
The forward reaction is exothermic (ΔH is negative), meaning it releases heat. The reverse reaction is therefore endothermic. According to Le Chatelier’s principle, if the temperature is increased, the system will attempt to oppose the change by absorbing the extra heat. It does so by favouring the endothermic reverse reaction. Thus, the equilibrium shifts to the left, producing more N₂ and H₂ and less NH₃. This is one reason why the Haber process uses a compromise temperature (about 450 °C) rather than a very high temperature.
正向反应是放热的（ΔH为负值），即释放热量；逆向反应则为吸热。根据勒夏特列原理，升高温度时，体系会通过吸收额外热量来对抗这一改变，因此会促进吸热的逆向反应。平衡向左移动，生成更多N₂和H₂，NH₃的产量减少。这正是哈伯法采用折中温度（约450 °C）而非极高温度的原因之一。
7. Acid–Base Titration Calculation | 酸碱滴定计算
Question: 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide solution is neutralised by 20.0 cm³ of sulfuric acid. Calculate the concentration of the sulfuric acid.
题目：25.0 cm³ 的0.100 mol/dm³氢氧化钠溶液被20.0 cm³硫酸完全中和。计算硫酸的浓度。
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Moles of NaOH = (0.100 mol/dm³) × (25.0 / 1000 dm³) = 0.00250 mol. From the equation, 2 mol of NaOH react with 1 mol of H₂SO₄, so moles of H₂SO₄ = 0.00250 / 2 = 0.00125 mol. Volume of H₂SO₄ = 20.0 / 1000 = 0.0200 dm³. Concentration = moles / volume = 0.00125 mol / 0.0200 dm³ = 0.0625 mol/dm³.
NaOH的物质的量 = 0.100 × (25.0/1000) = 0.00250 mol。根据方程式，2 mol NaOH 与 1 mol H₂SO₄ 反应，所以 H₂SO₄ 的物质的量 = 0.00250 / 2 = 0.00125 mol。硫酸体积 = 20.0/1000 = 0.0200 dm³。浓度 = 0.00125 / 0.0200 = 0.0625 mol/dm³。
8. Organic Chemistry: Alkanes and Alkenes | 有机化学：烷烃与烯烃
Question: Describe a chemical test that can be used to distinguish between ethane (C₂H₆) and ethene (C₂H₄). Include the expected observations for each compound.
题目：描述一种可用于区分乙烷（C₂H₆）和乙烯（C₂H₄）的化学测试。包括每种化合物的预期现象。
Add a few drops of orange‑brown bromine water to separate samples of the two gases. Ethene is an alkene and contains a C=C double bond. It will react with bromine in an addition reaction, causing the bromine water to decolourise immediately from orange‑brown to colourless. Ethane is an alkane with only C–C single bonds. It does not react with bromine water under normal laboratory conditions, so the orange‑brown colour remains unchanged. This test is specific to unsaturation.
向两种气体样品中分别滴加少量橙棕色的溴水。乙烯是烯烃，含有C=C双键，它将与溴发生加成反应，使溴水立即从橙棕色变为无色。乙烷是烷烃，只有C–C单键，在常规实验室条件下不与溴水反应，因此橙棕色保持不变。该测试专门用于检测不饱和键。
9. Energetics: Bond Energy Calculations | 能量学：键能计算
Question: Use the bond energies (in kJ/mol) to calculate the enthalpy change ΔH for the reaction: H₂(g) + Cl₂(g) → 2HCl(g). Bond energies: H–H = 436, Cl–Cl = 243, H–Cl = 432.
题目：利用键能（kJ/mol）计算反应 H₂(g) + Cl₂(g) → 2HCl(g) 的焓变ΔH。键能：H–H = 436，Cl–Cl = 243，H–Cl = 432。
Energy required to break bonds (endothermic): 1 × H–H (436) + 1 × Cl–Cl (243) = 679 kJ. Energy released when new bonds form (exothermic): 2 × H–Cl (2 × 432) = 864 kJ. Overall ΔH = energy absorbed – energy released = 679 – 864 = –185 kJ/mol. The negative sign indicates that the reaction is exothermic; 185 kJ of energy is released for every mole of H₂ that reacts according to the equation.
断裂化学键需吸收能量（吸热）：1 × H–H (436) + 1 × Cl–Cl (243) = 679 kJ。形成新键释放能量（放热）：2 × H–Cl (2 × 432) = 864 kJ。总ΔH = 吸收能量 – 释放能量 = 679 – 864 = –185 kJ/mol。负号表明反应放热；根据方程式，每摩尔H₂反应释放185 kJ能量。
10. Periodic Table Trends | 周期表规律
Question: Explain why the reactivity of Group 1 (alkali) metals increases as you move down the group from lithium to caesium. You should refer to atomic structure and the ease of electron loss.
题目：解释为什么第1族（碱金属）从上到下（从锂到铯）的反应性增强。需联系原子结构和失去电子的难易程度。
As you go down Group 1, each element has an extra electron shell compared with the one above. This means the outermost electron is further from the nucleus and is shielded by more inner electron shells. The electrostatic attraction between the positive nucleus and the outermost electron becomes weaker. Consequently, less energy is required to remove the outermost electron, so the metal loses its outer electron more easily and reacts more vigorously. Therefore, reactivity increases from Li to Cs.
沿着第1族向下，每个元素都比上一个多一个电子层。这意味着最外层电子离原子核更远，并受到更多内层电子的屏蔽。正电性的原子核与最外层电子之间的静电引力减弱。因此，移走最外层电子所需的能量减少，金属更容易失去其外层电子，反应更加剧烈。所以从锂到铯，反应性逐渐增强。
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