📚 GCSE WJEC Chemistry Unit 1 Test Paper: Practice & Revision | GCSE WJEC 化学单元一测试卷：练习与复习
This resource is designed as a complete unit test paper for WJEC GCSE Chemistry Unit 1: Chemical Substances, Reactions and Essential Resources. It blends targeted revision with authentic exam-style questions, enabling you to test your understanding and sharpen your exam technique. Work through each section carefully, apply your knowledge, and use the detailed answers to identify areas for improvement.
本资源是一套完整的 WJEC GCSE 化学单元一（化学物质、反应与重要资源）测试卷，将针对性复习与真实的考试风格题目相结合，帮助您检验理解并提升应试技巧。请仔细完成每个部分，运用所学知识，并参考详细答案找出需要改进的地方。
1. Overview of the WJEC Unit 1 Exam | WJEC 单元一考试概述
Unit 1 accounts for 45% of the total GCSE Chemistry qualification. The written paper lasts 1 hour 45 minutes and carries 80 marks. Questions range from multiple-choice items to short structured responses and one extended six-mark question that assesses the quality of written communication. A solid grasp of practical skills and mathematical requirements is essential, as they form an integral part of the assessment.
单元一占总成绩的 45%。笔试时间为 1 小时 45 分钟，满分 80 分。题型包括选择题、简答题以及一道考查书面表达质量的六分扩展题。扎实掌握实验技能与数学要求至关重要，因为它们构成了考试不可或缺的一部分。
2. Exam Format and Question Types | 考试格式与题型
The paper is divided into two sections. Section A contains approximately six to eight structured questions, each mixing short recall and data analysis tasks. Section B presents one or two longer-response questions that demand linking ideas across topics. All questions are compulsory, and marks are clearly indicated for each part. Common command words include ‘state’, ‘describe’, ‘explain’, ‘calculate’ and ‘evaluate’.
试卷分为两个部分。A 部分包含约六至八道结构化问题，每道题都融合了简单的知识回忆与数据分析任务。B 部分通常有一至两道需要联系不同主题的长答题。所有题目均为必答，每部分的分值都清晰标注。常见的指令词有“写出”、“描述”、“解释”、“计算”和“评价”。
Tip: Read each question twice, highlight key terms and pay close attention to the number of marks. If a question is worth three marks, your answer should contain three distinct scientific points or steps in a calculation.
提示：每道题读两遍，标出关键词，并特别留意分值。如果一道题是 3 分，你的答案应包含三个独立的科学要点或计算步骤。
3. Key Topic 1: Atomic Structure and the Periodic Table | 关键主题一：原子结构与元素周期表
Question 1: An atom of element X has 15 protons and 16 neutrons. State its mass number and draw its electronic configuration in shells.
题目 1：元素 X 的原子含有 15 个质子和 16 个中子。请写出其质量数，并画出它的核外电子排布。
Answer: Mass number = protons + neutrons = 15 + 16 = 31. The electronic configuration is 2,8,5. This means two electrons in the first shell, eight in the second, and five in the outermost shell.
答案：质量数 = 质子数 + 中子数 = 15 + 16 = 31。电子排布为 2,8,5，即第一层有 2 个电子，第二层有 8 个，最外层有 5 个。
Question 2: In the periodic table, why are elements in Group 7 (the halogens) described as having similar chemical properties?
题目 2：在元素周期表中，为什么第 7 族元素（卤素）被描述为具有相似的化学性质？
Answer: All halogens have seven electrons in their outermost shell. Chemical properties are largely determined by the number of outer electrons, so they all gain one electron in reactions and form ions with a 1⁻ charge.
答案：所有卤素原子最外层都有 7 个电子。化学性质主要由最外层电子数决定，因此它们在反应中都倾向于得到一个电子，形成带一个单位负电荷的离子。
4. Key Topic 2: Bonding, Structure and Properties | 关键主题二：化学键、结构与性质
Question 3: Sodium chloride (NaCl) has a high melting point while carbon tetrachloride (CCl₄) is a liquid at room temperature. Explain this difference in terms of structure and bonding.
题目 3：氯化钠 (NaCl) 的熔点很高，而四氯化碳 (CCl₄) 在室温下为液体。请从结构与化学键的角度解释这一差异。
Answer: NaCl is a giant ionic lattice held together by strong electrostatic forces between Na⁺ and Cl⁻ ions. A large amount of energy is needed to break these bonds, so the melting point is high. CCl₄ consists of simple covalent molecules with weak intermolecular forces between them. Only a small amount of energy is required to overcome these weak forces, giving it a low boiling point.
答案：NaCl 是由 Na⁺ 和 Cl⁻ 之间强大的静电作用力构成的巨型离子晶格，破坏这些化学键需要大量能量，因此熔点很高。CCl₄ 由简单的共价分子组成，分子间只有微弱的分子间作用力，只需少量能量就能克服，因此沸点低。
Key comparison table:
| Property | Ionic compound | Simple covalent compound |
|---|---|---|
| Melting/boiling point | High | Low |
| Conductivity | When molten or in solution | Non-conductor |
| State at room temp. | Solid | Often gas/liquid |
比对表格：
| 性质 | 离子化合物 | 简单共价化合物 |
|---|---|---|
| 熔点/沸点 | 高 | 低 |
| 导电性 | 熔融或水溶液可导电 | 不导电 |
| 室温状态 | 固体 | 常为气体/液体 |
5. Key Topic 3: Chemical Formulae and Equations | 关键主题三：化学式和化学方程式
Question 4: Write the balanced chemical equation for the reaction between magnesium and oxygen to form magnesium oxide, including state symbols.
题目 4：写出镁与氧气反应生成氧化镁的配平化学方程式，并标注状态符号。
Answer: 2Mg(s) + O₂(g) → 2MgO(s)
答案：2Mg(s) + O₂(g) → 2MgO(s)
Question 5: A student prepares zinc sulfate by reacting zinc with dilute sulfuric acid. Construct the balanced ionic equation for this reaction.
题目 5：一名学生通过锌与稀硫酸反应制备硫酸锌。写出该反应的离子方程式。
Answer: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g). Note that sulfate ions are spectator ions and are omitted from the ionic equation.
答案：Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)。注意硫酸根离子是旁观离子，在离子方程式中省略。
6. Key Topic 4: Reactivity Series and Extraction of Metals | 关键主题四：金属活动性顺序与金属提炼
Question 6: Place the metals copper, magnesium, zinc and iron in the correct reactivity order, starting with the most reactive. Explain why gold is found native in the earth’s crust but iron is extracted from its ore by reduction with carbon.
题目 6：将铜、镁、锌和铁按照活动性由强到弱的顺序排列。解释为什么金在自然界中以单质形式存在，而铁需要用碳从其矿石中还原提炼。
Answer: Reactivity order: magnesium > zinc > iron > copper. Gold is very unreactive and does not combine with other elements, so it occurs uncombined. Iron is more reactive and exists as iron oxide; carbon is more reactive than iron and can displace it from its oxide: 2Fe₂O₃ + 3C → 4Fe + 3CO₂.
答案：活动性顺序为：镁 > 锌 > 铁 > 铜。金非常不活泼，不易与其他元素化合，因此以单质形式存在。铁相对较活泼，以氧化铁形式存在；碳比铁更活泼，能从铁的氧化物中将其置换出来：2Fe₂O₃ + 3C → 4Fe + 3CO₂。
7. Key Topic 5: Acids, Bases and Salts | 关键主题五：酸、碱和盐
Question 7: Describe how to prepare a pure, dry sample of copper(II) sulfate crystals from copper(II) oxide and dilute sulfuric acid. Name the type of reaction taking place.
题目 7：描述如何用氧化铜和稀硫酸制备一份纯净干燥的硫酸铜晶体。并指出所发生的反应类型。
Answer: Add black copper(II) oxide powder to warm dilute sulfuric acid while stirring until no more dissolves. Filter the mixture to remove unreacted solid. Heat the filtrate gently to evaporate some water until crystallisation point is reached, then leave to cool. Collect the blue crystals, wash with a little cold water and dry between filter papers. This is a neutralisation reaction: CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l).
答案：向热的稀硫酸中加入黑色氧化铜粉末，搅拌至不再溶解。过滤除去未反应固体。将滤液缓慢加热蒸发一部分水，直到接近结晶点，然后静置冷却。收集蓝色晶体，用少量冷水洗涤，再用滤纸吸干。这是一个中和反应：CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)。
Question 8: Write the balanced equation for the reaction of calcium carbonate with nitric acid and identify the gas released.
题目 8：写出碳酸钙与硝酸反应的配平方程式，并指出生成的气体。
Answer: CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g). Carbon dioxide is released, which turns limewater milky.
答案：CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g)。释放出的气体是二氧化碳，可使石灰水变浑浊。
8. Key Topic 6: Electrolysis | 关键主题六：电解
Question 9: Predict the products at the cathode and anode during the electrolysis of molten lead(II) bromide, giving electrode half-equations.
题目 9：预测电解熔融溴化铅时，阴极和阳极的产物，并写出电极半反应式。
Answer: Cathode product: lead (a grey metal). Anode product: bromine (a reddish-brown gas). Half-equations: at cathode, Pb²⁺ + 2e⁻ → Pb(l); at anode, 2Br⁻ → Br₂(g) + 2e⁻. The overall reaction is PbBr₂(l) → Pb(l) + Br₂(g).
答案：阴极产物是铅（灰色金属），阳极产物是溴（红棕色气体）。半反应式为：阴极 Pb²⁺ + 2e⁻ → Pb(l)；阳极 2Br⁻ → Br₂(g) + 2e⁻。总反应为 PbBr₂(l) → Pb(l) + Br₂(g)。
Question 10: Explain why in the electrolysis of aqueous sodium chloride, hydrogen gas is formed at the cathode rather than sodium metal.
题目 10：解释为什么在电解氯化钠水溶液时，阴极生成的是氢气而不是金属钠。
Answer: In aqueous solution, water molecules are present alongside sodium ions. At the cathode, H⁺ ions from water are more easily reduced than Na⁺ ions. The half-equation is 2H⁺(aq) + 2e⁻ → H₂(g). Sodium ions remain in solution.
答案：在水溶液中，水分子与钠离子同时存在。在阴极，水产生的 H⁺ 离子比 Na⁺ 离子更容易被还原。半反应式为 2H⁺(aq) + 2e⁻ → H₂(g)，钠离子留在溶液中。
9. Key Topic 7: Energy Changes in Reactions | 关键主题七：反应中的能量变化
Question 11: When zinc powder is added to copper(II) sulfate solution, the temperature rises. State whether the reaction is exothermic or endothermic and sketch a simple energy level diagram labelling the activation energy and ΔH.
题目 11：将锌粉加入硫酸铜溶液中，温度升高。指出该反应是放热还是吸热，并画出一个简单的能级图，标出活化能和 ΔH。
Answer: The reaction is exothermic because heat is released to the surroundings, causing a temperature rise. In the energy diagram, the products have lower energy than the reactants. ΔH is negative. Activation energy is the minimum energy required for a successful collision.
答案：该反应为放热反应，因为热量释放到周围环境，导致温度升高。在能级图中，生成物的能量低于反应物，ΔH 为负值。活化能是发生有效碰撞所需的最低能量。
ΔH for an exothermic reaction is typically expressed with a negative sign, e.g., ΔH = −217 kJ/mol.
放热反应的 ΔH 通常用负值表示，例如 ΔH = −217 kJ/mol。
10. Sample Structured Question Walkthrough: Limestone Cycle and Atmospheric Chemistry | 样题解答示范：石灰石循环与大气化学
The following six-mark question tests your ability to bring together concepts from different topics. Read the context carefully and then work through the model answer.
以下是一个六分题，考查您综合不同主题概念的能力。请仔细阅读背景信息，然后研读参考答案。
Context: Limestone (calcium carbonate) is used in the manufacture of quicklime and slaked lime. Carbon dioxide released during this process contributes to the greenhouse effect, but it is also the same gas that turns limewater milky.
背景：石灰石（碳酸钙）用于制造生石灰和熟石灰。在此过程中释放的二氧化碳会导致温室效应，但它同样是能使石灰水变浑浊的气体。
Q: Outline the three-step thermal decomposition and cycle of limestone. Explain why carbon dioxide can be both useful as a test reagent and harmful as a greenhouse gas. Include balanced equations for all reactions described.
问：概述石灰石热分解及其三步循环过程。解释为什么二氧化碳既可以作为有用的检测试剂，又是一种有害的温室气体。请写出所有反应的配平方程式。
Model answer:
Step 1: CaCO₃(s) → CaO(s) + CO₂(g) – thermal decomposition of limestone to quicklime. Step 2: CaO(s) + H₂O(l) → Ca(OH)₂(s) – slaking to form slaked lime. Step 3: Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l) – carbonation, which is also the limewater test. CO₂ turns limewater milky due to the formation of insoluble calcium carbonate, making it an excellent test for the gas. However, CO₂ absorbs infrared radiation and traps heat in the atmosphere, enhancing the natural greenhouse effect and causing climate change. Therefore, the same molecule has both beneficial and detrimental environmental impacts.
参考答案：第一步：CaCO₃(s) → CaO(s) + CO₂(g) —— 石灰石热分解生成生石灰。第二步：CaO(s) + H₂O(l) → Ca(OH)₂(s) —— 生石灰与水消化生成熟石灰。第三步：Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l) —— 碳酸化反应，也就是石灰水检测反应。CO₂ 因生成不溶于水的碳酸钙使石灰水变浑浊，是检测该气体的绝佳方法。然而，CO₂ 会吸收红外辐射并将热量禁锢在大气层中，增强自然温室效应，导致气候变化。因此，同一分子对环境既有益又有害。
11. Top Tips for Unit 1 Test Success | 单元一测试高分技巧
Start by allocating time based on marks — roughly one minute per mark. If you are stuck on a question, move on and return later. For calculations, always show full working; even if the final answer is wrong, you can earn marks for correct method and units. In six-mark questions, structure your answer into logical steps using connectives such as ‘firstly’, ‘consequently’ and ‘therefore’.
首先要根据分值分配时间——大约一分钟一题。如果某道题卡住，先跳过去，稍后再回做。计算题务必展示完整步骤；即使最终答案有误，正确的方法和单位也能得分。对于六分题，用“首先”、“因此”、“于是”等连接词将答案组织成逻辑清晰的步骤。
Pay special attention to practical-based questions. Revise key experiments such as preparation of salts, reactivity series investigations and electrolysis setups. Memorise the tests for common gases: hydrogen (pop with a lit splint), oxygen (relights a glowing splint), carbon dioxide (turns limewater milky) and chlorine (bleaches damp litmus paper).
特别留意实验类题目。复习关键实验，如盐的制备、金属活动性探究和电解装置。熟记常见气体的检验方法：氢气（点燃的木条有爆鸣声），氧气（使带火星的木条复燃），二氧化碳（使石灰水变浑浊），氯气（使湿润的石蕊试纸褪色）。
12. Final Practice Checklist | 最终练习清单
Before the exam, ensure you can confidently: write electronic configurations for the first 20 elements; explain the properties of ionic and covalent substances; construct and balance symbol equations, including ionic equations; use the reactivity series to predict displacement reactions; describe the extraction of iron in the blast furnace and aluminium by electrolysis; prepare a pure dry salt; label an electrolytic cell and predict products for both molten and aqueous electrolytes; interpret simple energy level diagrams; and link the limestone cycle to carbon dioxide testing and climate change.
考试前，请确保您能够自信地完成以下任务：书写前 20 号元素的电子排布；解释离子化合物和共价化合物的性质；书写并配平化学方程式，包括离子方程式；运用金属活动性顺序预测置换反应；描述高炉炼铁和电解法炼铝的过程；制备纯净干燥的盐；绘制并标注电解池，预测熔融态和水溶液电解产物；解读简单能级图；以及将石灰石循环与二氧化碳检测及气候变化联系起来。
Self-test: Close the book and sketch an A4 revision map covering all seven Unit 1 topics. Add equations, diagrams and key terms without referring to notes. Pinpoint gaps and review those sections again.
自测：合上书本，在一张 A4 纸上画出涵盖单元一全部七个主题的复习导图。不参考笔记，补充方程式、示意图和关键术语。找出知识缺口，然后再次复习相应章节。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导