📚 IGCSE Edexcel Chemistry: Unit Test Paper | IGCSE Edexcel 化学：单元测试卷
This unit test paper is designed for IGCSE Edexcel Chemistry students to self-assess their grasp of key topics, from atomic structure to organic chemistry. Each question is followed by a detailed answer and explanation to reinforce understanding of the core principles examined by Edexcel.
这份单元测试卷专为 IGCSE Edexcel 化学学生设计，用于自评从原子结构到有机化学等核心课题的掌握情况。每道题目后附有详尽的答案与解析，以巩固 Edexcel 考试所考查的核心原理。
1. States of Matter and Diffusion | 物态与扩散
Question: A student opens a bottle of perfume at one end of a room. After a few minutes, a person at the other end can smell it. Explain this observation in terms of particle movement and state the name of the process. (3 marks)
问题：一名学生在房间一端打开一瓶香水，几分钟后房间另一端的人就能闻到香味。请用粒子运动解释这一现象，并写出该过程的名称。（3分）
Answer: The perfume particles are in constant, random motion in the gas state. They spread out from the area of high concentration (near the bottle) to areas of low concentration throughout the room. This process is called diffusion.
答案：香水粒子在气态下处于持续、无规则的运动中。它们从高浓度区域（瓶子附近）向房间各处低浓度区域扩散。该过程称为扩散。
Explanation: In gases, particles move rapidly and collide with air particles, gradually dispersing. Diffusion does not require stirring or wind; it is driven by the kinetic energy of the particles. Heavier particles diffuse more slowly, which is why lighter perfume molecules travel faster.
解析：在气体中，粒子快速运动并与空气粒子碰撞，逐渐分散开。扩散不需要搅拌或风；它是由粒子的动能驱动的。较重的粒子扩散较慢，因此较轻的香水分子传播得更快。
2. Atomic Structure and Isotopes | 原子结构与同位素
Question: An atom of element X has a mass number of 35 and contains 18 neutrons. (a) Calculate its atomic (proton) number. (b) Write the electronic configuration of element X. (c) A different atom of element X has a mass number of 37. Explain why these two atoms are isotopes. (4 marks)
问题：某元素 X 的一个原子质量数为 35，含有 18 个中子。(a) 计算其原子序数（质子数）。(b) 写出元素 X 的电子排布。(c) 元素 X 的另一个原子质量数为 37。解释为什么这两种原子互为同位素。（4分）
Answer: (a) Proton number = mass number – number of neutrons = 35 – me number of protons (17) but different numbers of neutrons (18 vs. 20). They are isotopes – atoms of the same element with the same atomic number but different mass numbers.
答案：(a) 质子数 = 质量数 – 中子数 = 35 – 18 = 17。(b) 元素 X（氯）的电子排布为 2,8,7。(c) 两个原子质子数相同（17），但中子数不同（分别为 18 和 20）。它们是同位素——即同种元素、质子数相同但质量数不同的原子。
Explanation: The atomic number defines the element; here it is 17, which is chlorine. Isotopes share identical chemical properties because they have the same electron configuration. The two chlorine isotopes are Cl-35 and Cl-37, with relative abundances of about 75% and 25%, respectively.
解析：原子序数定义了元素种类；这里为 17，即氯元素。同位素因具有相同的电子排布而化学性质相同。两种氯同位素分别是 Cl-35 和 Cl-37，其相对丰度分别约为 75% 和 25%。
3. Periodic Table and Group 1 | 周期表与第1族
Question: Lithium, sodium and potassium are Group 1 metals. (a) Describe the trend in reactivity as you go down the group. (b) Write a balanced equation for the reaction of sodium with water, including state symbols. (c) Predict one observation you would make when a small piece of potassium is added to water and explain why the reaction is more vigorous than that of lithium. (5 marks)
问题：锂、钠、钾是第1族金属。(a) 描述该族自上而下反应活性的变化趋势。(b) 写出钠与水反应的配平方程式，并注明状态符号。(c) 预测将一小块钾加入水中会观察到什么现象，并解释为什么该反应比锂更剧烈。（5分）
Answer: + H₂(g). (c) Potassium would melt into a silvery ball, move rapidly on the surface, ignite with a lilac flame, and possibly explode. The outer electron in potassium is further from the nucleus and more shielded, so it is lost more easily than in lithium, making potassium more reactive.
答案：(a) 自上而下反应活性增强。(b) 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)。(c) 钾会熔化成银白色小球，在水面快速游动，产生淡紫色火焰，并可能发生爆炸。钾的最外层电子离核更远且受到更多屏蔽，比锂更容易失去，因此钾更活泼。
Explanation: Group 1 metals all have one electron in their outer shell. As atomic radius increases down the group, the attraction between the nucleus and outer electron weakens, lowering ionisation energy. This makes electron loss (oxidation) easier, increasing reactivity. Potassium's reaction is so exothermic that it ignites the hydrogen gas produced.
解析：第1族金属最外层均有一个电子。自上而下原子半径增大，原子核对外层电子的吸引力减弱，电离能降低，使得失电子（氧化）更容易，反应活性增强。钾的反应放热剧烈，足以引燃生成的氢气。
4. Ionic and Covalent Bonding | 离子键与共价键
Question: Magnesium oxide (MgO) has a high melting point and conducts electricity when molten but not when solid. Carbon dioxide (CO₂) is a gas at room temperature and does not conduct electricity in any state. Explain these properties in terms of structure and bonding. (6 marks)
问题：氧化镁（MgO）熔融时可以导电但固态时不导电，且熔点很高。二氧化碳（CO₂）在室温下为气体，任何状态下都不导电。请从结构和成键角度解释这些性质。（6分）
Answer: MgO is an ionic compound with a giant ionic lattice. Strong electrostatic forces between Mg²⁺ and O²⁻ ions require large amounts of energy to overcome, hence the high melting point. In solid state, ions are fixed in place and cannot move, so it does not conduct electricity. When molten, ions are free to move and carry charge, allowing conduction. CO₂ is a simple molecular covalent compound with weak intermolecular forces between molecules. Little energy is needed to overcome these forces, so it is a gas at room temperature. There are no mobile charged particles (ions or electrons), so it cannot conduct electricity in any state.
答案：MgO 是离子化合物，具有巨型离子晶格。Mg²⁺ 与 O²⁻ 之间的强静电引力需要大量能量才能克服，因此熔点高。固态时离子固定在原位无法移动，故不导电；熔融时离子可自由移动并携带电荷，因此能够导电。CO₂ 是简单分子共价化合物，分子间只存在微弱的分子间作用力，克服这些力所需能量很少，因而室温下为气体。由于任何状态下都没有可自由移动的带电粒子（离子或电子），故不能导电。
Explanation: The key distinction is between giant structures (ionic, metallic, giant covalent) and simple molecular structures. MgO's high melting point and conductivity when molten are classic ionic properties. For CO₂, even when dissolved in water it forms a weakly acidic solution but the conduction is due to the formation of H⁺ and HCO₃⁻ ions, not from CO₂ molecules themselves – the pure substance remains non-conductive.
解析：关键在于区分巨型结构（离子、金属、巨型共价）与简单分子结构。MgO 的高熔点与熔融导电性是典型的离子特性。对 CO₂ 而言，即使溶于水也仅形成弱酸性溶液，其导电性源于生成的 H⁺ 与 HCO₃⁻ 离子，而非 CO₂ 分子本身——纯物质仍不导电。
5. Mole Calculations | 摩尔计算
Question: es, of CaCO₃ used. (b) Calculate the mass of calcium oxide produced. (c) Calculate the volume of carbon dioxide produced at room temperature and pressure (rtp), where 1 mole of gas occupies 24.0 dm³. (Relative atomic masses: Ca = 40, C = 12, O = 16) (6 marks)
问题：5.00 g 碳酸钙（CaCO₃）被强热至完全分解：CaCO₃(s) → CaO(s) + CO₂(g)。(a) 计算所用 CaCO₃ 的物质的量（摩尔）。(b) 计算生成的氧化钙的质量。(c) 计算在室温常压（rtp）下生成的二氧化碳体积，1 摩尔气体占 24.0 dm³。（相对原子质量：Ca = 40，C = 12，O = 16）（6分）
Answer: ol CaO. Molar mass of CaO = 40 + 16 = 56 g/mol. Mass of CaO = moles × molar mass = 0.0500 × 56 = 2.80 g. (c) Moles of CO₂ = 0.0500 mol. Volume of CO₂ = moles × 24.0 = 0.0500 × 24.0 = 1.20 dm³.
答案：CaCO₃ 的摩尔质量 = 40 + 12 + (3 × 16) = 100 g/mol。(a) CaCO₃ 物质的量 = 质量 / 摩尔质量 = 5.00 / 100 = 0.0500 mol。(b) 由方程式知，1 mol CaCO₃ 生成 1 mol CaO。CaO 的摩尔质量 = 40 + 16 = 56 g/mol。CaO 质量 = 物质的量 × 摩尔质量 = 0.0500 × 56 = 2.80 g。(c) CO₂ 物质的量 = 0.0500 mol。CO₂ 体积 = 物质的量 × 24.0 = 0.0500 × 24.0 = 1.20 dm³。
Explanation: This question tests the three-step molar relationship: mass to moles, mole ratio from a balanced equation, and moles to mass or volume. Always check that the equation is balanced before using the mole ratio. At rtp, the molar gas volume is 24 dm³ for all gases.
解析：本题考察三步摩尔关系：质量与物质的量的换算、根据配平方程式的摩尔比、以及物质的量与质量或体积的换算。在使用摩尔比前务必确认方程式已配平。在室温常压下，任何气体的摩尔体积均为 24 dm³。
6. Electrolysis | 电解
Question: A student electrolyses molten lead(II) bromide (PbBr₂) using inert electrodes. (a) Write the half-equation for the reaction at the cathode and state the product. (b) Write the half-equation for the reaction at the anode and state the product. (c) Explain why electrolysis of molten lead(II) bromide produces lead and bromine, whereas electrolysis of aqueous lead(II) bromide could produce different substances. (5 marks)
问题：一名学生用惰性电极电解熔融溴化铅（PbBr₂）。(a) 写出阴极的半反应方程式并说明产物。(b) 写出阳极的半反应方程式并说明产物。(c) 解释为什么电解熔融溴化铅生成铅和溴，而电解溴化铅水溶液可能会生成不同物质。（5分）
Answer: roduct is bromine gas. (c) In molten PbBr₂, the only ions present are Pb²⁺ and Br⁻, so they are discharged. In aqueous solution, water is also present, meaning there are H⁺ and OH⁻ ions from the water. At the cathode, hydrogen can be produced instead of lead if H⁺ ions are easier to discharge, and at the anode, oxygen can be produced from OH⁻ instead of bromine, depending on concentration and electrode potentials.
答案：(a) 阴极（还原）：Pb²⁺ + 2e⁻ → Pb(l)；产物为金属铅。(b) 阳极（氧化）：2Br⁻ → Br₂(g) + 2e⁻；产物为溴气。(c) 在熔融 PbBr₂ 中，仅存在 Pb²⁺ 和 Br⁻ 离子，因此它们被放电。在溴化铅水溶液中，水分子也提供了离子，即溶液中含有 H⁺ 和 OH⁻。若 H⁺ 更容易在阴极放电，则生氢而非铅；若 OH⁻ 离子在阳极优先放电，则生成氧气而非溴，具体取决于浓度和电极电势。
Explanation: The products of electrolysis are determined by the ions present and their relative ease of discharge. In molten ionic compounds, it is straightforward. In aqueous solutions, the reactivity series at the cathode and the halide/OH⁻ competition at the anode must be considered. For lead(II) bromide solution, lead is still produced at the cathode (since lead is less reactive than hydrogen, actually lead can be discharged), but at the anode bromine is produced if the solution is concentrated; if dilute, oxygen may be produced.
解析：电解产物由存在的离子及其相对放电容易程度决定。在熔融离子化合物中，情况简单直接。在水溶液中，需考虑阴极的金属活动顺序和阳极的卤素/OH⁻ 竞争。对于溴化铅溶液，阴极仍生成铅（铅不如氢活泼，实际可放电）；阳极若溶液较浓则生成溴，若稀溶液则可能生成氧气。
7. Energy Changes and Rate of Reaction | 能量变化与反应速率
Question: ion can be made faster by adding a small amount of manganese(IV) oxide (MnO₂). (a) What is the role of MnO₂ in this reaction? rgies. (c) Explain, using collision theory, why increasing the temperature also increases the rate. (6 marks)
问题：过氧化氢（H₂O₂）在室温下缓慢分解：2H₂O₂(aq) → 2H₂O(l) + O₂(g)。加入少量二氧化锰（MnO₂）可加快分解。(a) MnO₂ 在该反应中起何作用？(b) 画出有催化剂和无催化剂时反应的能量分布图，并标注活化能。(c) 用碰撞理论解释升高温度为何也能加快反应速率。（6分）
Answer: (a) MnO₂ acts as a catalyst; it speeds up the reaction without being consumed. (b) [In a written response] The diagram shows the energy of reactants higher than products (exothermic). The uncatalysed pathway has a higher activation energy (Eₐ) hill; the catalysed pathway has a lower activation energy (Eₐ' < Eₐ). Both start and end at the same energy levels. (c) Increasing temperature gives particles more kinetic energy. They move faster, so collisions are more frequent, and a greater proportion of collisions have energy equal to or greater than the activation energy. This increases the frequency of successful collisions per unit time, raising the rate.
答案：(a) MnO₂ 作为催化剂，加速反应而自身不被消耗。(b) [在书面回答中] 图中反应物总能量高于生成物（放热）。无催化路径的活化能（Eₐ）较高；有催化路径的活化能（Eₐ' < Eₐ）较低。两条路径起点终点能量相同。(c) 升高温度使粒子动能增加。它们运动更快，碰撞更频繁，且更高比例的碰撞具有等于或超过活化能的能量，从而增加单位时间内有效碰撞的频率，提高反应速率。
Explanation: A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the proportion of collisions that are successful. This is different from temperature, which increases both collision frequency and the fraction of particles with sufficient energy. MnO₂ is a heterogeneous catalyst for H₂O₂ decomposition; it is unchanged at the end and can be filtered off and reused.
解析：催化剂提供了一条活化能较低的反应途径，提高了成功碰撞的比例。这与温度效应不同，温度同时增加了碰撞频率和能量充足粒子的比例。MnO₂ 是 H₂O₂ 分解的多相催化剂，反应前后不变，可过滤回收并重复使用。
8. Organic Chemistry: Alkanes and Alkenes | 有机化学：烷烃与烯烃
Question: Butane (C₄H₁₀) and but-1-ene (C₄H₈) are hydrocarbons. (a) State the general formula for alkanes and explain why butane is described as saturated. (b) Describe a chemical test to distinguish between butane and but-1-ene, including the result for each. (c) Write a balanced equation for the complete combustion of but-1-ene. (4 marks)
问题：丁烷（C₄H₁₀）与 1-丁烯（C₄H₈）都是碳氢化合物。(a) 写出烷烃的通式并解释为什么丁烷是饱和烃。(b) 描述一种鉴别丁烷和 1-丁烯的化学实验，包括各自的现象。(c) 写出 1-丁烯完全燃烧的配平方程式。（4分）
Answer: (a) Alkane general formula: CₙH₂ₙ₊₂. Butane is saturated because it contains only single C–C and C– he bromine water stays orange-brown (no reaction). With but-1-ene, the bromine water is decolourised (turns colourless) due to the addition reaction across the C=C double bond. (c) C₄H₈ + 6O₂ → 4CO₂ + 4H₂O.
答案：(a) 烷烃通式：CₙH₂ₙ₊₂。丁烷是饱和烃，因为它仅含 C–C 单键与 C–H 键，每个碳原子连接的氢原子数已达最大。(b) 在黑暗条件下向两种烃分别加入溴水（橙黄色）。丁烷中溴水保持橙黄色（不反应）。1-丁烯中溴水迅速褪色（变为无色），因为发生了 C=C 双键的加成反应。(c) C₄H₈ + 6O₂ → 4CO₂ + 4H₂O。
Explanation: The bromine water test is specific for unsaturated compounds containing double or triple bonds. Alkanes undergo substitution reactions only in the presence of UV light, not in the dark with bromine water. Combustion of any hydrocarbon in plenty of oxygen produces carbon dioxide and water.
解析：溴水实验专一性地检验含双键或三键的不饱和化合物。烷烃仅在有紫外光照射时才发生取代反应，黑暗中不与溴水反应。任何烃在充足氧气中完全燃烧均生成二氧化碳和水。
9. Acids, Bases and Salts | 酸、碱和盐
Question: Dilute sulfuric acid (H₂SO₄) is added to aqueous sodium hydroxide (NaOH) in a titration. (a) Write the balanced chemical equation for this neutralisation reaction. (b) 25.0 cm³ of NaOH solution required 18.60 cm³ of 0.100 mol/dm³ H₂SO₄ for complete neutralisation. Calculate the concentration of the NaOH solution in mol/dm³. (c) Name the salt formed and describe how you could obtain pure, dry crystals of this salt from the reaction mixture. (5 marks)
问题：在滴定实验中，将稀硫酸（H₂SO₄）加入氢氧化钠（NaOH）水溶液中。(a) 写出该中和反应的配平化学方程式。(b) 中和 25.0 cm³ NaOH 溶液需要 18.60 cm³ 0.100 mol/dm³ H₂SO₄。计算该 NaOH 溶液的浓度（mol/dm³）。(c) 写出生成的盐的名称，并说明如何从反应混合物中获得该盐的纯净干燥晶体。（5分）
Answer: NaOH, so moles of NaOH = 2 × 0.00186 = 0.00372 mol. Concentration of NaOH = moles / volume (dm³) = 0.00372 / (25.0 / 1000) = 0.1488 mol/dm³ ≈ 0.149 mol/dm³. (c) The salt is sodium sulfate (Na₂SO₄). To obtain pure dry crystals: evaporate the solution (from a titration using indicator, first remove indicator with charcoal) to about half volume, then leave to cool and crystallise. Filter the crystals, wash with a little cold distilled water, and dry between filter papers.
答案：(a) 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)。(b) H₂SO₄ 物质的量 = 0.100 × (18.60 / 1000) = 0.00186 mol。由方程式可知，1 mol H₂SO₄ 与 2 mol NaOH 反应，故 NaOH 物质的量 = 2 × 0.00186 = 0.00372 mol。NaOH 浓度 = 物质的量 / 体积（dm³）= 0.00372 / (25.0 / 1000) = 0.1488 mol/dm³ ≈ 0.149 mol/dm³。(c) 生成的盐是硫酸钠（Na₂SO₄）。获取纯净干燥晶体的方法：将溶液蒸发浓缩至原体积一半，然后冷却结晶。过滤出晶体，用少量冷蒸馏水洗涤，在滤纸之间压干。
Explanation: This titration calculation uses the stoichiometric ratio from the equation. Note that sulfuric acid is diprotic, providing two H⁺ ions per molecule, hence the 1:2 mole ratio. To obtain pure crystals, one must avoid heating to dryness, which would cause dehydration or impurity entrapment. Washing with cold water removes adhering solution without dissolving the crystals.
解析：滴定计算使用了方程式中的化学计量比。注意硫酸是二元酸，每分子提供两个 H⁺，因此物质的量比为 1:2。为获得纯晶体，不可蒸干，否则会导致脱水或包裹杂质。用冷水洗涤可去除晶体表面沾附的溶液而不溶解晶体。
10. Chemical Tests | 化学检验
Question: Describe chemical tests you would carry out to identify the following substances, giving the reagents and expected observations: (a) chloride ions in a solution; (b) hydrogen gas; (c) carbon dioxide gas; (d) sulfate ions in a solution. (4 marks)
问题：描述鉴别以下物质的化学实验，写出所用试剂和预期现象：(a) 溶液中的氯离子；(b) 氢气；(c) 二氧化碳气体；(d) 溶液中的硫酸根离子。（4分）
Answer: (a) Add a few drops of dilute nitric acid and then silver nitrate solution. A white precipitate of silver chloride forms. (b) Hold a lit splint near the mouth of the test tube. Hydrogen burns with a squeaky pop sound. (c) Bubble the gas through limewater (calcium hydroxide solution). The limewater turns milky (cloudy white precipitate). (d) Add a few drops of dilute hydrochloric acid and then barium chloride solution. A white precipitate of barium sulfate forms.
答案：(a) 加入几滴稀硝酸，然后加入硝酸银溶液。生成氯化银的白色沉淀。(b) 将点燃的木条靠近试管口。氢气燃烧并发出轻微的爆鸣声。(c) 将气体通入石灰水（氢氧化钙溶液）。石灰水变浑浊（出现白色沉淀）。(d) 加入几滴稀盐酸，然后加入氯化钡溶液。生成硫酸钡的白色沉淀。
Explanation: Acidification before adding the precipitating reagent is crucial for both chloride and sulfate tests to remove carbonate or sulfite ions that might give false positives. For chloride, nitric acid is used; for sulfate, hydrochloric acid is used because barium ions themselves could precipitate with carbonates. The limewater test is standard for CO₂. The squeaky pop test for hydrogen requires a small volume of gas.
解析：在加入沉淀剂前进行酸化对于氯离子和硫酸根离子检验至关重要，这可以排除碳酸根或亚硫酸根离子的干扰。检验氯离子用硝酸酸化；检验硫酸根离子用盐酸酸化，因为钡离子本身也会与碳酸根生成沉淀。石灰水是检验 CO₂ 的标准方法。氢气的爆鸣实验需用少量气体。
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