Cambridge IGCSE Exam Questions from Paper 3 | 剑桥IGCSE化学Paper 3考试题目详解

📚 Cambridge IGCSE Exam Questions from Paper 3 | 剑桥IGCSE化学Paper 3考试题目详解

Paper 3 is the extended theory paper for Cambridge IGCSE Chemistry. It tests your ability to apply knowledge, perform calculations, interpret data and explain chemical phenomena in depth. Questions are often structured and require clear, logical answers supported by chemical principles. In this article, we will explore typical Paper 3 question styles and show you how to approach them confidently, with model answers and step‑by‑step explanations.

Paper 3 是剑桥 IGCSE 化学的拓展理论试卷。它重点考查考生应用知识、完成计算、解释数据以及深入阐述化学现象的能力。题目多为结构化形式，要求作答清晰、逻辑严密并以化学原理为依据。本文将剖析 Paper 3 常见的题型，通过模型答案和逐步讲解，帮助你自信应对。

1. Atomic Structure and Isotopes | 原子结构与同位素

Questions often ask for the relative atomic mass (Aᵣ) from isotopic abundances or require you to draw electron configurations. A common calculation: Aᵣ = Σ (isotopic mass × % abundance) ÷ 100. Remember to express electron arrangements as 2.8.1 style for the first 20 elements.

这类题目常要求根据同位素丰度计算相对原子质量 (Aᵣ)，或让考生画出电子排布。常见的计算公式：Aᵣ = Σ (同位素质量 × 丰度百分比) ÷ 100。记得前 20 号元素用 2.8.1 的格式表示电子排布。

Example question: A sample of chlorine contains 75% of chlorine‑35 atoms and 25% of chlorine‑37 atoms. Calculate the relative atomic mass of chlorine in this sample.

题目示例：某氯元素样品含有 75% 的氯‑35 原子和 25% 的氯‑37 原子。计算该样品中氯的相对原子质量。

Aᵣ = (35 × 75 + 37 × 25) ÷ 100 = (2625 + 925) ÷ 100 = 35.5

The answer is 35.5. This matches the value in the Periodic Table because chlorine in nature is indeed a mixture of Cl‑35 and Cl‑37.

答案为 35.5。这与元素周期表中的数值相符，因为自然界中的氯正是 Cl‑35 和 Cl‑37 的混合物。

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2. Ionic and Covalent Bonding | 离子键与共价键

Paper 3 frequently asks you to draw ‘dot‑and‑cross’ diagrams for ionic compounds such as sodium oxide or magnesium chloride, and for simple covalent molecules like water and carbon dioxide. Always show the charges on ions and use different symbols for electrons from different atoms.

Paper 3 经常要求画出离子化合物（如氧化钠、氯化镁）以及简单共价分子（如水、二氧化碳）的点叉图。务必标明离子的电荷，并用不同符号表示来自不同原子的电子。

For MgCl₂, show Mg losing two electrons, each chlorine atom gaining one. The final diagram should display Mg²⁺ and two Cl⁻ ions with full outer shells. For covalent molecules, show shared pairs and lone pairs correctly; in H₂O, oxygen shares one electron with each hydrogen and has two lone pairs.

画 MgCl₂ 时，应体现 Mg 失去两个电子，每个氯原子得到一个电子。最终图中要展示 Mg²⁺ 和两个 Cl⁻ 离子，且都具有满壳层结构。对于共价分子，要正确画出共用电子对和孤对电子；在 H₂O 中，氧原子与每个氢原子共用一对电子，并有两对孤对电子。

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3. Moles, Mass and Concentration | 摩尔、质量与浓度

These numerical questions reward careful unit management. Key formulas: n = m/Mᵣ and n = c × V (dm³). Remember that 1 dm³ = 1000 cm³. Gas volume at r.t.p.: 1 mol of any gas occupies 24 dm³.

这类计算题要求仔细处理单位。核心公式：n = m/Mᵣ 及 n = c × V （单位 dm³）。记住 1 dm³ = 1000 cm³。常温常压下气体体积：任何气体 1 mol 的体积为 24 dm³。

Example: 2.8 g of iron reacts completely with dilute hydrochloric acid. Fe + 2HCl → FeCl₂ + H₂. Calculate the volume of hydrogen gas produced at r.t.p. (Aᵣ: Fe = 56)

示例：2.8 g 铁与稀盐酸完全反应。Fe + 2HCl → FeCl₂ + H₂。计算常温常压下生成氢气体积。（Aᵣ: Fe = 56）

n(Fe) = 2.8/56 = 0.050 mol. From equation, 1 mol Fe produces 1 mol H₂, so n(H₂) = 0.050 mol. Volume = 0.050 × 24 = 1.2 dm³.

n(Fe) = 2.8/56 = 0.050 mol。由方程式可知 1 mol Fe 生成 1 mol H₂，所以 n(H₂) = 0.050 mol。体积 = 0.050 × 24 = 1.2 dm³。

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4. Electrolysis and Half‑Equations | 电解与半反应式

You may be given a diagram of an electrolytic cell and asked to predict products at the electrodes. For molten ionic compounds, the metal forms at the cathode and the non‑metal at the anode. In aqueous solutions, you must compare the reactivity of the ions and the presence of water.

题目可能给出电解池示意图，要求预测两极产物。对于熔融离子化合物，金属在阴极生成，非金属在阳极生成。在水溶液中，则需要比较离子的反应性以及水分子的存在。

Typical Paper 3 question: Electrolysis of concentrated aqueous sodium chloride. Name the products at the cathode and anode, and write the half‑equations.

典型 Paper 3 题：电解浓氯化钠水溶液。说出阴、阳极产物并写出半反应式。

At cathode: hydrogen gas because H⁺ from water is easier to discharge than Na⁺. Half‑equation: 2H⁺ + 2e⁻ → H₂. At anode: chlorine gas because Cl⁻ concentration is high. Half‑equation: 2Cl⁻ → Cl₂ + 2e⁻. Sodium hydroxide remains in solution.

阴极产物为氢气，因为水中的 H⁺ 比 Na⁺ 更易放电。半反应式：2H⁺ + 2e⁻ → H₂。阳极产物为氯气，因为 Cl⁻ 浓度高。半反应式：2Cl⁻ → Cl₂ + 2e⁻。溶液中留下氢氧化钠。

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5. Energy Changes and Bond Energies | 能量变化与键能

Paper 3 can ask you to calculate the enthalpy change (ΔH) using bond energies. ΔH = total energy absorbed to break bonds – total energy released when forming bonds. A negative ΔH indicates an exothermic reaction.

Paper 3 可能要求使用键能计算焓变 (ΔH)。ΔH = 断裂化学键吸收的总能量 – 形成化学键释放的总能量。ΔH 为负值表示放热反应。

For the combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O. Given bond energies (kJ/mol): C–H 413, O=O 498, C=O 799, O–H 464. Calculate ΔH.

甲烷燃烧：CH₄ + 2O₂ → CO₂ + 2H₂O。已知键能 (kJ/mol)：C–H 413, O=O 498, C=O 799, O–H 464。计算 ΔH。

Energy in = (4×413) + (2×498) = 1652 + 996 = 2648 kJ

Energy out = (2×799) + (4×464) = 1598 + 1856 = 3454 kJ

ΔH = 2648 – 3454 = –806 kJ/mol

The negative sign confirms the reaction is exothermic, releasing 806 kJ per mole of methane burned.

负号确认该反应为放热反应，每摩尔甲烷燃烧释放 806 kJ 热量。

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6. Rates of Reaction and Collision Theory | 反应速率与碰撞理论

You must explain how temperature, concentration, surface area and catalysts affect rate in terms of particle collisions. A higher temperature increases both collision frequency and the proportion of particles with energy greater than activation energy.

你需要从粒子碰撞的角度解释温度、浓度、表面积和催化剂如何影响速率。升高温度不仅增加碰撞频率，还提高了能量超过活化能的粒子比例。

A common graph question: sketch a Maxwell‑Boltzmann distribution curve and show the effect of a catalyst. The catalyst lowers the activation energy, making a larger area under the curve to the right of the new Eₐ mark.

常见的图表题：画出麦克斯韦‑玻尔兹曼分布曲线并展示催化剂的作用。催化剂降低活化能，使得曲线下方位于新 Eₐ 标记右侧的面积增大。

Mark scheme tip: always include the phrase ‘more successful collisions per unit time’ in your explanation to score full marks.

评分窍门：解释时加上“单位时间内有效碰撞次数增多”往往能获得满分。

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7. Acids, Bases and Salt Preparation | 酸、碱与盐的制备

Paper 3 expects you to describe a method for preparing a pure, dry sample of a soluble salt, such as copper(II) sulfate, starting from an insoluble base. The method: add excess copper oxide to warm dilute sulfuric acid, filter off the excess, evaporate the filtrate to saturation, then leave to crystallise.

Paper 3 要求描述制备纯净干燥可溶性盐（例如硫酸铜）的方法，通常从不溶性碱开始。步骤：将过量氧化铜加入温热稀硫酸，滤除过量固体，将滤液蒸发至饱和，然后静置结晶。

You may also be asked to write ionic equations for neutralisation: H⁺ + OH⁻ → H₂O. For reactions with carbonates: 2H⁺ + CO₃²⁻ → H₂O + CO₂.

还可能要求书写中和反应的离子方程式：H⁺ + OH⁻ → H₂O。与碳酸盐反应：2H⁺ + CO₃²⁻ → H₂O + CO₂。

When naming salts, remember hydrochloric acid gives chlorides, sulfuric acid gives sulfates, and nitric acid gives nitrates.

给盐命名时记住：盐酸产生氯化物，硫酸产生硫酸盐，硝酸产生硝酸盐。

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8. Periodic Table Trends | 元素周期表趋势

Group trends are frequently tested. For Group 1, reactivity increases down the group because the outer electron becomes further from the nucleus and is more easily lost. For Group 7, reactivity decreases down the group as the ability to gain an electron diminishes.

族的变化趋势是常考点。第 1 族从上到下反应性增强，因为最外层电子离核越来越远，更易失去。第 7 族从上到下反应性减弱，因为获得电子的能力下降。

A typical extended question: describe the reaction of sodium, potassium and caesium with water, including observations and the resulting alkaline solution. State the trend and explain it in terms of atomic structure.

典型的拓展题：描述钠、钾和铯与水的反应，包括观察到的现象和生成的碱性溶液。陈述趋势并从原子结构角度解释。

Element	Observation	Product
Na	Floats, fizzes, melts into a ball	NaOH + H₂
K	Floats, violent fizzing, lilac flame	KOH + H₂
Cs	Explosive reaction, shatters container	CsOH + H₂

Why does reactivity increase? As you go down Group 1, the atomic radius increases, shielding by inner shells increases, the outer electron is held less strongly, so it is lost more readily.

为什么反应性增强？沿第 1 族向下，原子半径增大，内层电子屏蔽效应增强，最外层电子被原子核吸引的力减小，因此更容易失去。

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9. Organic Chemistry and Polymers | 有机化学与聚合物

The exam will ask you to recognise homologous series (alkanes, alkenes, alcohols, carboxylic acids) and their functional groups. You must be able to name and draw structural formulae for molecules up to four carbon atoms.

考试会要求识别同系列（烷烃、烯烃、醇、羧酸）及其官能团。你必须能命名并画出含最多四个碳原子的分子结构式。

Addition polymerisation is a key topic. For example, poly(ethene) is formed by the polymerisation of ethene monomers under high pressure with a catalyst. You should be able to draw the repeating unit and explain why polymers are non‑biodegradable.

加成聚合是重点。例如，聚乙烯由乙烯单体在高压和催化剂条件下聚合而成。你需要能画出重复单元，并解释为何聚合物难以生物降解。

n CH₂=CH₂ → –(–CH₂–CH₂–)–ₙ

Another favourite question: esterification between ethanol and ethanoic acid to form ethyl ethanoate and water, requiring a sulfuric acid catalyst.

另一常见考题：乙醇与乙酸在浓硫酸催化下发生酯化反应，生成乙酸乙酯和水。

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10. Experimental Design and Data Analysis | 实验设计与数据分析

Paper 3 often includes a question about a practical investigation. You might be asked to identify variables, plot a graph, evaluate the method and suggest improvements. Always distinguish between accuracy and reliability.

Paper 3 常包含实验探究题。可能要求识别变量、绘制图表、评价方法并提出改进。务必区分准确度与可靠性。

Example: In an experiment measuring the temperature change when different masses of NaOH dissolve in water, a student forgets to stir. What is the expected effect on the recorded temperature? Answer: Heat distributes unevenly, so the maximum temperature recorded is lower than the true value, reducing accuracy.

示例：某实验测定不同质量 NaOH 溶于水时的温度变化，学生忘记搅拌。这对记录温度有何影响？答：热量分布不均，记录到的最高温度低于真实值，降低了准确度。

To improve: use a polystyrene cup with a lid, stir continuously, and repeat three times to calculate a mean. This increases reliability.

改进方法：使用带盖的聚苯乙烯杯，持续搅拌，并重复三次取平均值，以提高可靠性。

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