Year 10 CIE Chemistry: Unit Test Mock Paper Analysis | Year 10 CIE 化学：单元测试模拟卷解析

📚 Year 10 CIE Chemistry: Unit Test Mock Paper Analysis | Year 10 CIE 化学：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for Year 10 CIE Chemistry. We break down each question, explaining the core concepts, common mistakes, and step‑by‑step reasoning. Whether you are preparing for your end‑of‑topic assessment or building a solid foundation for IGCSE, this analysis will help you master the material.

本文对 Year 10 CIE 化学单元测试模拟卷进行详细解析。我们将逐题拆解，讲解核心概念、常见错误和逐步推理过程。无论你是在为单元测验备考，还是为 IGCSE 打基础，这份解析都能帮助你掌握相关内容。

1. Question 1: Atomic Structure | 第1题：原子结构

Question: An atom of element X has 12 protons and 12 neutrons. State the number of electrons in a neutral atom of X, its mass number, and its electronic configuration.

题目：元素 X 的原子有 12 个质子和 12 个中子。写出 X 中性原子的电子数、质量数和电子排布。

e electronic configuration for 12 electrons is 2,8,2 (filling shells: first shell holds up to 2, second up to 8, third holds the remaining 2).

中性原子的质子数等于电子数，因此电子数为 12。质量数是质子数加中子数：12 + 12 = 24。12 个电子的电子排布为 2,8,2（第一层最多 2 个，第二层最多 8 个，第三层放剩余的 2 个）。

Many students confuse the mass number with the atomic number. Remember, the atomic number is the number of protons only; the mass number includes both protons and neutrons.

许多学生会混淆质量数和原子序数。请记住，原子序数仅是质子数；质量数包括质子和中子。

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2. Question 2: Isotopes and Relative Atomic Mass | 第2题：同位素与相对原子质量

Question: Chlorine has two main isotopes: ³⁵Cl (abundance 75%) and ³⁷Cl (abundance 25%). Calculate the relative atomic mass (Aᵣ) of chlorine.

题目：氯有两种主要同位素：³⁵Cl（丰度 75%）和 ³⁷Cl（丰度 25%）。计算氯的相对原子质量（Aᵣ）。

dance₂) / 100. So Aᵣ = (35 × 75 + 37 × 25) / 100 = (2625 + 925) / 100 = 3550 / 100 = 35.5.

相对原子质量 = (质量数₁ × 丰度₁ + 质量数₂ × 丰度₂) / 100。因此 Aᵣ = (35 × 75 + 37 × 25) / 100 = (2625 + 925) / 100 = 3550 / 100 = 35.5。

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have identical chemical properties but slightly different physical properties.

同位素是质子数相同、中子数不同的同种原子。它们的化学性质相同，但物理性质略有差异。

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3. Question 3: Ionic and Covalent Bonding | 第3题：离子键与共价键

Question: Describe the formation of sodium chloride (NaCl) and explain why it conducts electricity when molten but not when solid. Then describe the bonding in a water molecule.

题目：描述氯化钠（NaCl）的形成过程，并解释为什么它在熔融态能导电而固态不能。然后描述水分子的成键。

Sodium (2,8,1) loses its one outer electron to become a Na⁺ ion with a full outer shell (2,8). Chlorine (2,8,7) gains that electron to become a Cl⁻ ion (2,8,8). The oppositely charged ions are held together by strong electrostatic forces in a giant ionic lattice. In the solid state, the ions are fixed in place and cannot move, so no conductivity. When molten, the ions are free to move and carry charge, allowing conduction.

钠（2,8,1）失去最外层一个电子，成为具有满外壳的 Na⁺ 离子（2,8）。氯（2,8,7）得到该电子，成为 Cl⁻ 离子（2,8,8）。相反电荷的离子通过强静电引力形成巨型离子晶格。固态时离子固定在位置上无法移动，因此不导电。熔融时离子可以自由移动并携带电荷，从而导电。

In a water molecule (H₂O), each hydrogen atom shares one electron with the oxygen atom, forming two single covalent bonds. Oxygen also has two lone pairs of electrons, giving the molecule a bent shape.

在水分子（H₂O）中，每个氢原子与氧原子共用一对电子，形成两个单共价键。氧原子还有两对孤对电子，使分子呈 V 形。

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4. Question 4: Balancing Equations | 第4题：配平化学方程式

Question: e : Balance the following equation: C₃H₈ + O₂ → CO₂ + H₂O. Then state what type of reaction this is.

题目：配平下列方程式：C₃H₈ + O₂ → CO₂ + H₂O。然后说明这是什么类型的反应。

: C₃H₈ + O₂ → 3CO₂ + 4H₂O. Step 3: Oxygen atoms. On the right: 3×2 = 6 from CO₂ plus 4×1 = 4 from H₂O, total 10 O atoms. O₂ is diatomic, so we need 5 O₂: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. This is a combustion reaction (burning a fuel in oxygen).

步骤1：碳原子。左边有3个C，在CO₂前加系数3：C₃H₈ + O₂ → 3CO₂ + H₂O。步骤2：氢原子。左边有8个H，在H₂O前加系数4：C₃H₈ + O₂ → 3CO₂ + 4H₂O。步骤3：氧原子。右边3×2=6来自CO₂，4×1=4来自H₂O，总共10个O原子。O₂是双原子分子，因此需要5个O₂：C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。这是燃烧反应（燃料在氧气中燃烧）。

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5. Question 5: Moles and Mass Calculations | 第5题：摩尔与质量计算

Question: : Calculate the mass of carbon dioxide produced when 8.8 g of propane (C₃H₈) undergoes complete combustion. (Aᵣ: C=12, H=1, O=16)

题目：计算 8.8 g 丙烷（C₃H₈）完全燃烧时产生的二氧化碳质量。（Aᵣ: C=12, H=1, O=16）

₂O, 1 mol C₃H₈ produces 3 mol CO₂. So moles of CO₂ = 0.20 × 3 = 0.60 mol. Molar mass of CO₂ = 12 + (2×16) = 44 g/mol. Mass of CO₂ = 0.60 mol × 44 g/mol = 26.4 g.

丙烷的摩尔质量 = (3×12) + (8×1) = 44 g/mol。丙烷的摩尔数 = 质量 / 摩尔质量 = 8.8 g / 44 g/mol = 0.20 mol。由配平方程式 C₃H₈ + 5O₂ → 3CO₂ + 4H₂O 可知，1 mol C₃H₈ 生成 3 mol CO₂。因此 CO₂ 的摩尔数 = 0.20 × 3 = 0.60 mol。CO₂ 的摩尔质量 = 12 + (2×16) = 44 g/mol。CO₂ 质量 = 0.60 mol × 44 g/mol = 26.4 g。

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6. Question 6: Acids, Bases and pH | 第6题：酸、碱和 pH 值

Question: A solution has a pH of 3. State whether it is acidic, alkaline or neutral. How would its pH change if water is added? Write the ionic equation for the neutralisation of hydrochloric acid with sodium hydroxide.

题目：某溶液的 pH 为 3。判断它是酸性、碱性还是中性。如果加水稀释，pH 会如何变化？写出盐酸与氢氧化钠中和反应的离子方程式。

sation reaction: HCl + NaOH → NaCl + H₂O. The ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l).

pH 3 是酸性的（pH < 7）。加水会稀释酸，降低 H⁺ 离子浓度，因此 pH 升高（趋近于 7）。中和反应：HCl + NaOH → NaCl + H₂O。离子方程式为 H⁺(aq) + OH⁻(aq) → H₂O(l)。

In neutralisation, the H⁺ from the acid and the OH⁻ from the alkali combine to form water. The other ions (Na⁺ and Cl⁻) remain in solution and do not participate in the reaction – they are spectator ions.

在中和反应中，酸中的 H⁺ 与碱中的 OH⁻ 结合生成水。其他离子（Na⁺ 和 Cl⁻）留在溶液中，不参与反应——它们是旁观离子。

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7. Question 7: Metals and Reactivity | 第7题：金属与反应性

Question: A student adds magnesium ribbon to dilute hydrochloric acid. State two observations and write a word equation. Then explain why copper does not react with dilute acids.

题目：学生将镁条加入到稀盐酸中。描述两个观察到的现象，并写出文字方程式。然后解释为什么铜不与稀酸反应。

dissolves, while the solution may become warm. The gas produced is hydrogen, which gives a 'pop' nesium chloride + hydrogen. Copper does not react with dilute acids because it is below hydrogen in the reactivity series; copper cannot displace hydrogen from the acid.

现象：冒泡（产生气体）和镁条溶解，同时溶液可能变热。生成的气体是氢气，点燃时会发出“噗”的爆鸣声。文字方程式：镁 + 盐酸 → 氯化镁 + 氢气。铜不与稀酸反应，因为它在金属活动性顺序中排在氢之后；铜不能从酸中置换出氢。

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8. Question 8: Electrolysis of Molten Sodium Chloride | 第8题：熔融氯化钠的电解

Question: Describe what happens during the electrolysis of molten sodium chloride, including the products at each electrode and the half‑equations.

题目：描述熔融氯化钠电解的过程，包括每个电极的产物和半反应方程式。

positive electrode), Cl⁻ ions are oxidised to chlorine gas: 2Cl⁻ → Cl₂ + 2e⁻. So overall: 2NaCl(l) → 2Na(l) + Cl₂(g). The sodium metal is silvery and floats on the melt; the chlorine gas is a pale green, pungent gas.

在熔融 NaCl 中，存在的离子是 Na⁺ 和 Cl⁻。在阴极（负极），Na⁺ 离子被还原为金属钠：Na⁺ + e⁻ → Na。在阳极（正极），Cl⁻ 离子被氧化为氯气：2Cl⁻ → Cl₂ + 2e⁻。总反应：2NaCl(l) → 2Na(l) + Cl₂(g)。金属钠呈银白色并浮在熔体上；氯气是黄绿色、有刺激性气味的气体。

Electrolysis breaks down a compound using electricity. The molten state is required so that ions are free to move. If aqueous NaCl were used, different products (hydrogen and chlorine) would form due to the presence of water.

电解是利用电来分解化合物。需要熔融状态以便离子自由移动。如果使用 NaCl 水溶液，由于水的存在，会生成不同的产物（氢气和氯气）。

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9. Question 9: Exothermic and Endothermic Reactions | 第9题：放热反应与吸热反应

Question: Classify the following as exothermic or endothermic: (a) combustion of methane, (b) photosynthesis, (c) dissolving ammonium nitrate in water, (d) neutralisation between HCl and NaOH. Explain one of these in terms of bond breaking and forming.

题目：将下列反应分类为放热或吸热：(a) 甲烷燃烧，(b) 光合作用，(c) 硝酸铵溶于水，(d) HCl 与 NaOH 的中和反应。选择其中一个，用断键和成键解释其能量变化。

(a) Exothermic – burning fuels releases heat. (b) Endothermic – photosynthesis absorbs energy from sunlight. (c) Endothermic – dissolving ammonium nitrate absorbs heat, making the beaker feel cold. (d) Exothermic – lease heat, the temperature rises. For combustion of methane: in reactants, bonds break (C–H and O=O), which absorbs energy. In products, new bonds form (C=O in CO₂ and O– ases energy. More energy is released than absorbed, so the reaction is exothermic overall.

(a) 放热——燃料燃烧释放热量。(b) 吸热——光合作用吸收太阳光能。(c) 吸热——硝酸铵溶于水吸收热量，使烧杯感觉变冷。(d) 放热——中和反应释放热量，温度升高。以甲烷燃烧为例：反应物中，旧键断裂（C–H 和 O=O）需要吸收能量。产物中，新键生成（CO₂ 中的 C=O 和 H₂O 中的 O–H）会释放能量。释放的能量大于吸收的能量，因此总反应为放热。

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10. Question 10: Practical Skills – Making Copper(II) Sulfate Crystals | 第10题：实验技能——制备硫酸铜晶体

Question: A student wants to prepare pure, dry copper(II) sulfate crystals from copper(II) oxide and dilute sulfuric acid. Outline the method, explaining why each step is necessary.

题目：学生希望用氧化铜和稀硫酸制备纯净干燥的硫酸铜晶体。简述操作步骤，并解释每一步的必要性。

Method: (1) Add excess black copper(II) oxide to warm dilute sulfuric acid and stir. The mixture turns blue as copper(II) sulfate forms. (2) Filter the mixture to remove unreacted copper(II) oxide. The filtrate is a blue solution of copper(II) sulfate. (3) Heat the filtrate gently to evaporate some water until crystallisation point is reached (when crystals just start to form on cooling). (4) Allow the solution to cool slowly. Blue crystals will form. (5) Filter the crystals, wash with a little cold distilled water, and dry between filter paper. Excess reagent ensures complete neutralisation; filtration removes solid impurities; slow cooling gives larger, pure crystals; washing removes any soluble impurities.

步骤：(1) 向温热的稀硫酸中加入过量黑色氧化铜并搅拌。混合物变蓝，生成硫酸铜。(2) 过滤混合物，除去未反应的氧化铜。滤液为蓝色的硫酸铜溶液。(3) 将滤液缓慢加热蒸发部分水分，直至达到结晶点（冷却时刚有晶体析出）。(4) 让溶液慢慢冷却。蓝色晶体会析出。(5) 过滤晶体，用少量冷蒸馏水洗涤，在滤纸间干燥。使用过量试剂确保完全中和；过滤除去固体杂质；缓慢冷却可获得较大、纯净的晶体；洗涤可去除可溶性杂质。

This practical is a typical example of preparing a soluble salt from an insoluble base. It tests understanding of filtration, evaporation, crystallisation, and safe handling of chemicals.

本实验是用不溶性碱制备可溶性盐的典型范例，考查对过滤、蒸发、结晶以及化学品安全操作的理解。

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Published by TutorHao | Chemistry Revision Series | aleveler.com

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