📚 IGCSE AQA Chemistry: Unit Test Paper | IGCSE AQA 化学：单元测试卷
This unit test paper covers key topics from the IGCSE AQA Chemistry syllabus. Each section presents a typical exam-style question followed by a detailed answer, helping you to revise atomic structure, bonding, quantitative chemistry, chemical changes, electrolysis, energy, rates, and organic chemistry. Use this paper to assess your understanding and identify areas for improvement.
本单元测试卷涵盖 IGCSE AQA 化学大纲中的核心主题。每部分都提供一道典型考题并附有详细解答，帮助你复习原子结构、化学键、定量化学、化学变化、电解、能量、反应速率和有机化学。通过这份试卷检验自己的掌握程度，找出薄弱环节。
1. Atomic Structure and the Periodic Table | 原子结构与元素周期表
Question: An atom of element Z has 17 protons, 18 neutrons, and 17 electrons. (a) State the mass number of this atom. (b) Write the electronic configuration of element Z. (c) In which group and period of the periodic table would you find element Z? Give a reason for your answer.
题目：元素 Z 的一个原子有 17 个质子、18 个中子和 17 个电子。(a) 写出该原子的质量数。(b) 写出元素 Z 的电子排布。(c) 元素 Z 位于周期表的哪个族和哪个周期？请说明理由。
Answer: (a) Mass number = protons + neutrons = 17 + 18 = 35. (b) Electrons are arranged as 2 in the first shell, 8 in the second shell, and 7 in the third shell, so the electronic configuration is 2,8,7. (c) Element Z is in Group 7 because it has 7 electrons in its outer shell. It is in Period 3 because it has three occupied electron shells. This element is chlorine.
答案：(a) 质量数 = 质子数 + 中子数 = 17 + 18 = 35。(b) 电子排布为第一层 2 个，第二层 8 个，第三层 7 个，因此电子排布写为 2,8,7。(c) 元素 Z 位于第 7 族，因为其最外层有 7 个电子；位于第 3 周期，因为它有三个电子层。该元素是氯。
2. Ionic Bonding | 离子键
Question: Sodium reacts with chlorine to form sodium chloride. Describe, in terms of electron transfer, how sodium and chlorine atoms form ions. Draw a dot-and-cross diagram to show the electronic structures of the ions formed. Include the charges on the ions.
题目：钠与氯反应生成氯化钠。请从电子转移的角度描述钠原子和氯原子如何形成离子。画出点叉图来表示所形成离子的电子结构，并标出离子的电荷。
Answer: A sodium atom has 1 electron in its outer shell. It loses this electron to achieve a full outer shell, forming a Na⁺ ion. A chlorine atom has 7 electrons in its outer shell. It gains 1 electron from sodium to fill its outer shell, forming a Cl⁻ ion. The ions are held together by strong electrostatic forces of attraction in a giant ionic lattice. In a dot-and-cross diagram, the sodium ion is shown with no outer electrons but an empty outer shell (still drawn as a circle) and a + charge. The chloride ion is shown with 8 outer electrons (dots and crosses representing the original electrons and the transferred electron) and a – charge. Both ions achieve the electronic structure of a noble gas.
答案：钠原子最外层有 1 个电子，它失去这个电子后形成满壳层结构，成为 Na⁺ 离子。氯原子最外层有 7 个电子，它从钠原子得到 1 个电子填满最外层，成为 Cl⁻ 离子。离子之间以强大的静电吸引力结合在一起，形成巨大的离子晶格。在点叉图中，钠离子外层没有电子，但仍画出空的外层圆圈，并标注 + 电荷；氯离子外层画有 8 个电子（用点和叉表示原有的电子和转移来的电子），并标注 – 电荷。两种离子都达到了稀有气体的电子结构。
3. Covalent Bonding | 共价键
Question: Methane (CH₄) is a simple covalent molecule. Draw a dot-and-cross diagram for methane, showing only the outer shell electrons. Explain why methane has a low boiling point.
题目：甲烷 (CH₄) 是一种简单的共价分子。画出甲烷的点叉图，只展示最外层电子。解释为什么甲烷的沸点很低。
Answer: Carbon has 4 outer electrons and hydrogen has 1 outer electron. In the dot-and-cross diagram, the carbon atom shares one electron with each of the four hydrogen atoms, forming four C–H covalent bonds. This gives carbon a full outer shell of 8 electrons and each hydrogen a full shell of 2 electrons. Methane has a low boiling point because it consists of small, simple molecules with weak intermolecular forces. Only a small amount of energy is needed to overcome these forces, so the substance boils at a low temperature.
答案：碳原子最外层有 4 个电子，氢原子最外层有 1 个电子。在点叉图中，碳原子与每个氢原子共用一对电子，形成四个 C–H 共价键。这使得碳的最外层达到 8 电子稳定结构，每个氢达到 2 电子稳定结构。甲烷沸点低，是因为它由简单的小分子组成，分子间的分子间作用力很弱。只需要很少的能量就能克服这些力，因此该物质在低温下即沸腾。
4. Metallic Bonding | 金属键
Question: Copper is a metal that conducts electricity. Describe the structure and bonding in copper and explain why it can conduct electricity in both the solid and liquid states.
题目：铜是一种能导电的金属。描述铜的结构和键合方式，并解释为什么它在固态和液态时都能导电。
Answer: Copper has a giant metallic lattice structure. The atoms are arranged in regular layers, and the outer electrons become delocalised, forming a ‘sea’ of free electrons. Positive metal ions are held together by strong electrostatic attraction to these delocalised electrons. Copper conducts electricity because the delocalised electrons are free to move throughout the lattice and carry charge. In the solid state, the layers are fixed but electrons can still flow. In the liquid state, both ions and electrons become mobile, so conduction is also possible.
答案：铜具有巨大的金属晶格结构。原子按规则层排列，外层电子离域，形成自由电子的“海洋”。带正电的金属离子与这些离域电子之间强烈的静电吸引力将整个结构维系在一起。铜能导电，是因为离域电子可在整个晶格中自由移动并携带电荷。在固态时，虽然原子层位置固定，但电子仍能流动；在液态时，离子和电子都变得可移动，因此也能导电。
5. Quantitative Chemistry | 定量化学
Question: Magnesium reacts with oxygen to form magnesium oxide: 2Mg + O₂ → 2MgO. Calculate the mass of magnesium oxide produced when 4.8 g of magnesium is completely burnt in excess oxygen. (Relative atomic masses: Mg = 24, O = 16)
题目：镁与氧气反应生成氧化镁：2Mg + O₂ → 2MgO。计算将 4.8 g 镁在过量的氧气中完全燃烧后生成的氧化镁的质量。（相对原子质量：Mg = 24，O = 16）
Answer: Step 1: Moles of Mg = mass / Mᵣ = 4.8 / 24 = 0.20 mol. Step 2: From the equation, 2 mol Mg produce 2 mol MgO, so the mole ratio is 1:1. Therefore, moles of MgO = 0.20 mol. Step 3: Mᵣ of MgO = 24 + 16 = 40. Step 4: Mass of MgO = moles × Mᵣ = 0.20 × 40 = 8.0 g.
答案：步骤 1：镁的物质的量 = 质量 / 相对原子质量 = 4.8 / 24 = 0.20 mol。步骤 2：根据化学方程式，2 mol Mg 生成 2 mol MgO，物质的量之比为 1:1，因此 MgO 的物质的量也是 0.20 mol。步骤 3：MgO 的相对分子质量 = 24 + 16 = 40。步骤 4：MgO 的质量 = 物质的量 × 相对分子质量 = 0.20 × 40 = 8.0 g。
6. Chemical Changes – Making Salts | 化学变化 – 制取盐
Question: Describe a safe method to prepare a pure, dry sample of copper(II) sulfate crystals from copper(II) oxide and dilute sulfuric acid. Include the names of the apparatus you would use.
题目：描述用氧化铜和稀硫酸制备纯净干燥的硫酸铜晶体的一种安全方法，并写出所用仪器的名称。
Answer: Warm a measured volume of dilute sulfuric acid in a beaker. Add copper(II) oxide powder a little at a time, stirring continuously, until no more dissolves and some black powder remains unreacted. This ensures the acid is fully neutralised. Filter the mixture using filter paper and a funnel to remove the excess copper(II) oxide. Collect the filtrate (copper(II) sulfate solution) in an evaporating basin. Heat gently over a water bath to evaporate some of the water until crystals just begin to form. Leave the solution to cool and crystallise. Dry the crystals by pressing them between pieces of filter paper.
答案：在烧杯中预热一定体积的稀硫酸。分批加入氧化铜粉末，不断搅拌，直到不再溶解且有少量黑色粉末剩余为止，这样可确保酸完全被中和。用滤纸和漏斗过滤混合物，除去过量的氧化铜。将滤液（硫酸铜溶液）收集到蒸发皿中。在水浴上温和加热，蒸发掉部分水，直至刚好开始有晶体析出。让溶液冷却结晶。最后用滤纸压干晶体，获得纯净干燥的硫酸铜晶体。
7. Electrolysis | 电解
Question: Describe what happens during the electrolysis of molten lead(II) bromide. Include the products at each electrode and the half-equations. Why must the lead(II) bromide be molten?
题目：描述电解熔融溴化铅时发生的现象。写出两极产物及半反应方程式。为什么溴化铅必须处于熔融状态？
Answer: Lead(II) bromide (PbBr₂) must be molten so that the ions are free to move and carry charge. During electrolysis: At the cathode (negative electrode), Pb²⁺ ions gain electrons and are reduced to lead metal: Pb²⁺ + 2e⁻ → Pb. Silvery lead droplets form. At the anode (positive electrode), Br⁻ ions lose electrons and are oxidised to bromine gas: 2Br⁻ → Br₂ + 2e⁻. Red-brown bromine gas bubbles off. Overall, lead is produced at the cathode and bromine at the anode.
答案：溴化铅 (PbBr₂) 必须是熔融状态，这样离子才能自由移动并传导电荷。电解过程中：在阴极（负极），Pb²⁺ 离子得到电子，被还原成金属铅：Pb²⁺ + 2e⁻ → Pb，产生银白色铅滴。在阳极（正极），Br⁻ 离子失去电子，被氧化成溴气：2Br⁻ → Br₂ + 2e⁻，冒出红棕色的溴气。总的来说，阴极生成铅，阳极生成溴。
8. Energy Changes in Reactions | 反应中的能量变化
Question: The reaction between citric acid and sodium hydrogencarbonate is endothermic. Draw and label a simple energy level diagram for this reaction. Mark the activation energy and the overall energy change (ΔH). Explain what is happening to the energy of the chemicals and the surroundings.
题目：柠檬酸与碳酸氢钠的反应是吸热反应。画出并标注该反应的能量示意图，标明活化能和反应热 (ΔH)。解释反应中物质和环境的能量变化。
Answer: In an endothermic reaction, the products have more energy than the reactants. The energy level diagram shows reactants at a lower energy level and products at a higher energy level. An upward arrow is drawn from the energy level of the reactants to the peak (activation complex), labelled as activation energy. The overall energy change ΔH is the vertical distance between reactants and products; it is positive, so energy is taken in from the surroundings. This causes the temperature of the surroundings to decrease.
答案：在吸热反应中，生成物的能量高于反应物。能量变化图中，反应物处于较低能级，生成物处于较高能级。从反应物能级到顶峰（活化络合物）画向上的箭头，标注为活化能。总能量变化 ΔH 是反应物与生成物之间的垂直距离，为正值，表示从环境中吸收能量，因此环境温度会下降。
9. Rates of Reaction – Collision Theory | 反应速率 – 碰撞理论
Question: Explain, using collision theory, why increasing the temperature increases the rate of a chemical reaction between magnesium ribbon and dilute hydrochloric acid.
题目：用碰撞理论解释为什么升高温度会增加镁条与稀盐酸反应的速率。
Answer: Increasing the temperature gives the reactant particles more kinetic energy. This has two effects: (1) Particles move faster, so they collide more frequently. (2) A greater proportion of the collisions have energy equal to or greater than the activation energy, meaning more collisions are successful and lead to a reaction. Both factors together increase the rate of reaction significantly.
答案：升高温度使反应物粒子获得更多的动能。这产生两个影响：(1) 粒子运动更快，因此碰撞频率增加。(2) 更多碰撞的能量达到或超过活化能，意味着有更多的碰撞是有效的并能引起反应。这两个因素共同作用，显著提高了反应速率。
10. Organic Chemistry – Alkanes and Carbon Dioxide Test | 有机化学 – 烷烃与二氧化碳检验
Question: Propane (C₃H₈) is an alkane used as a fuel. (a) Write a balanced chemical equation for the complete combustion of propane. (b) Describe a chemical test for the carbon dioxide gas produced, giving the observation and the result.
题目：丙烷 (C₃H₈) 是一种用作燃料的烷烃。(a) 写出丙烷完全燃烧的配平化学方程式。(b) 描述检验生成的二氧化碳气体的化学方法，写出操作现象和结论。
Answer: (a) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. (b) Bubble the gas through limewater (calcium hydroxide solution). If the limewater turns milky or cloudy, carbon dioxide is present. This is because a white precipitate of calcium carbonate forms: CO₂ + Ca(OH)₂ → CaCO₃ + H₂O.
答案：(a) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。(b) 将气体通入石灰水（氢氧化钙溶液）中。若石灰水变浑浊或呈乳白色，则说明有二氧化碳存在。这是因为生成了碳酸钙白色沉淀：CO₂ + Ca(OH)₂ → CaCO₃ + H₂O。
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