📚 Year 10 WJEC Chemistry: Unit Test Mock Paper Walkthrough | Year 10 WJEC 化学：单元测试模拟卷解析
Preparing for your Year 10 WJEC Chemistry unit test requires more than just rereading notes – you need to apply your knowledge to exam-style questions. This walkthrough takes you through a mock paper covering atomic structure, bonding, mole calculations, acids and bases, electrolysis, and reactivity. Each question is carefully explained so you can understand the model answers, pick up marks for working, and learn how to avoid common pitfalls.
准备 Year 10 WJEC 化学单元测试，仅仅重读笔记是不够的——你需要将知识应用到考试题型中。本次解析将带你完成一份涵盖原子结构、化学键、摩尔计算、酸碱、电解和反应活性的模拟卷。每道题都配有详细的解释，帮助你理解标准答案、学会争取步骤分，并掌握如何避开常见陷阱。
1. Structure of the WJEC Unit Test | WJEC 单元测试卷结构
A typical WJEC Year 10 Chemistry unit test lasts 45–60 minutes and contains a mixture of short-answer, structured, and calculation questions. Marks are awarded for correct chemical terminology, balanced equations, logical working in calculations, and precise diagrams. Questions are often set in familiar practical contexts, such as making salts or electrolysis. The mock paper discussed here samples those key skills.
典型的 WJEC Year 10 化学单元测试时长为 45–60 分钟，包含简答题、结构化问题和计算题。评分侧重于正确的化学术语、配平的方程式、清晰的计算步骤以及准确的示意图。题目常设置在熟悉的实验情境中，如制盐或电解。本模拟卷就涵盖了这些核心技能。
2. Question 1: Atomic Number and Mass Number | 题目1：原子序数与质量数
Question: A neutral atom has 15 protons, 15 electrons, and 16 neutrons. State its atomic number, its mass number, identify the element, and draw the electron configuration using numbers (e.g. 2,8,1).
题目： 一个中性原子含有 15 个质子、15 个电子和 16 个中子。写出它的原子序数、质量数，辨认该元素，并用数字表示法画出电子排布（如 2,8,1）。
Answer & Explanation: er 15 is phosphorus (P). With 15 electrons, the configuration is 2,8,5. Many students incorrectly give mass number as 15 or 16; remember mass number is the total of protons and neutrons.
答案与解析： 原子序数 = 质子数 = 15。质量数 = 质子数 + 中子数 = 31。原子序数为 15 的元素是磷 (P)。15 个电子的排布为 2,8,5。许多同学错误地将质量数写成 15 或 16；请记住，质量数是质子与中子之和。
3. Question 2: Formation of Ionic Bonds | 题目2：离子键的形成
Question: Magnesium (atomic number 12) reacts with chlorine (atomic number 17) to form magnesium chloride. Describe the bonding in this compound by stating what happens to the electrons. Include half-equations for the ion formation and a diagram showing the electron transfer.
题目： 镁（原子序数 12）与氯（原子序数 17）反应生成氯化镁。通过描述电子的转移来描述该化合物的键合。写出离子形成的半方程式，并画出电子转移示意图。
Answer: 8,7) gains one electron to become a Cl⁻ ion with an octet. The half-equations are: Mg → Mg²⁺ + 2e⁻ and Cl₂ + 2e⁻ → 2Cl⁻ (or Cl + e⁻ → Cl⁻). The oppositely charged ions attract strongly in a giant ionic lattice. The dot-and-cross diagram must show Mg with no outer electrons and two Cl⁻ ions each with eight outer electrons, brackets and charges.
答案： 镁的电子排布为 2,8,2。它失去两个最外层电子形成稳定八隅体，变成 Mg²⁺ 离子。每个氯原子（2,8,7）获得一个电子成为 Cl⁻ 离子。半方程式为：Mg → Mg²⁺ + 2e⁻ 和 Cl₂ + 2e⁻ → 2Cl⁻（或 Cl + e⁻ → Cl⁻）。相反电荷的离子在巨型离子晶格中强烈吸引。点叉图必须显示 Mg 没有最外层电子，而两个 Cl⁻ 离子各有八个最外层电子，并带上方括号和电荷。
4. Question 3: Relative Formula Mass (Mᵣ) | 题目3：相对式量 (Mᵣ)
Question: ) of aluminium sulfate, Al₂(SO₄)₃. (Aᵣ values: Al = 27, S = 32, O = 16) Show your working.
题目： 计算硫酸铝 Al₂(SO₄)₃ 的相对式量 (Mᵣ)。（Aᵣ 值：Al = 27，S = 32，O = 16）请写出计算过程。
Answer: s no unit for relative formula mass. A common mistake is forgetting to multiply the oxygen inside the brackets; note that (SO₄)₃ means 3 S and 12 O.
答案： Al₂(SO₄)₃ 含有 2 个 Al、3 个 S 和 12 个 O 原子。计算：(2 × 27) + (3 × 32) + (12 × 16) = 54 + 96 + 192 = 342。相对式量为 342，没有单位。一个常见错误是忘记将括号内的氧原子数量乘以 3；(SO₄)₃ 表示 3 个 S 和 12 个 O。
5. Question 4: Moles, Mass and Molar Mass | 题目4：摩尔、质量与摩尔质量
Question: of the solid. (Aᵣ: Na = 23, O = 16, H = 1) Give your answer to three significant figures.
题目： 氢氧化钠 (NaOH) 常用于实验室酸的中和。计算需要多少质量的 NaOH 来制备 0.250 摩尔的固体。（Aᵣ：Na = 23，O = 16，H = 1）答案保留三位有效数字。
Answer: wer is 10.0 g (to three significant figures). Always check whether the question requires g or kg, and look out for significant figures instructions. Many students omit the unit or write 10 g, losing a mark for precision.
答案： NaOH 的 Mᵣ = 23 + 16 + 1 = 40。质量 = 摩尔数 × 摩尔质量 = 0.250 × 40 = 10.0 g。答案为 10.0 克（三位有效数字）。时刻看清题目要求的是克还是千克，并关注有效数字的指示。很多同学漏掉单位或写成 10 g，因精度不够而失分。
mass = moles × Mᵣ
6. Question 5: Acid–Base Neutralisation | 题目5：酸碱中和反应
Question: Hydrochloric acid (HCl) is neutralised by potassium hydroxide (KOH) solution. Write a word equation, a balanced symbol equation, and the ionic equation for this reaction. State the name of the salt produced.
题目： 盐酸 (HCl) 会被氢氧化钾 (KOH) 溶液中和。写出该反应的文字表达式、配平的符号方程式和离子方程式。说出生成的盐的名称。
Answer: n: H⁺ + OH⁻ → H₂O. The salt is potassium chloride. The key insight is that the reaction of a strong acid with a strong base has the same essential ionic equation: hydrogen ions react with hydroxide ions to produce water. Spectator ions (K⁺ and Cl⁻) are omitted.
答案： 文字表达式：盐酸 + 氢氧化钾 → 氯化钾 + 水。配平的符号方程式：HCl + KOH → KCl + H₂O。离子方程式：H⁺ + OH⁻ → H₂O。生成的盐是氯化钾。核心在于强酸与强碱的反应本质上离子方程式相同：氢离子与氢氧根离子结合生成水。旁观离子（K⁺ 和 Cl⁻）被省略。
7. Question 6: Electrolysis of Molten Lead(II) Bromide | 题目6：熔融溴化铅的电解
Question: Molten lead(II) bromide (PbBr₂) is electrolysed using inert graphite electrodes. Describe what you would observe at the anode and the cathode. Write half-equations for both electrode reactions and explain why the lead(II) bromide must be molten.
题目： 使用惰性石墨电极电解熔融溴化铅 (PbBr₂)。描述在阳极和阴极会观察到什么现象。写出两个电极反应的半方程式，并解释为什么溴化铅必须是熔融状态。
Answer: lectrode), Br⁻ ions are oxidised: 2Br⁻ → Br₂ + 2e⁻. Observation: reddish-brown fumes of bromine gas are seen. The compound must be molten so that the ions are free to move and carry charge. Solid lead(II) bromide does not conduct electricity because the ions are locked in the lattice.
答案： 在阴极（负电极），Pb²⁺ 离子被还原：Pb²⁺ + 2e⁻ → Pb。观察现象：生成银灰色金属（铅）。在阳极（正电极），Br⁻ 离子被氧化：2Br⁻ → Br₂ + 2e⁻。观察现象：可见红棕色的溴蒸气。化合物必须处于熔融状态，这样离子才能自由移动并传导电荷。固态溴化铅不导电，因为离子被束缚在晶格中。
8. Question 7: Reactivity Series and Displacement | 题目7：活动性顺序与置换反应
Question: A student adds a strip of zinc metal to a blue copper(II) sulfate solution. After a few minutes, the blue colour fades and a reddish-brown solid appears on the zinc strip. Explain these observations. Write a balanced symbol equation and an ionic equation for the reaction.
题目： 一名学生将锌条放入蓝色的硫酸铜溶液中。几分钟后，蓝色褪去，锌条表面出现红棕色固体。解释这些现象。写出该反应的配平符号方程式和离子方程式。
Answer: ns are reduced to copper metal. The reddish-brown solid is copper. Zinc atoms are oxidised to Zn²⁺ ions. Balanced equation: Zn + CuSO₄ → ZnSO₄ + Cu. Ionic equation: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). The sulfate ions are spectators.
答案： 锌比铜活泼，因此它能从硫酸铜溶液中置换出铜。蓝色（由 Cu²⁺ 离子导致）逐渐褪去，因为铜离子被还原成金属铜。红棕色固体即为铜。锌原子被氧化成 Zn²⁺ 离子。配平方程式：Zn + CuSO₄ → ZnSO₄ + Cu。离子方程式：Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。硫酸根离子是旁观离子。
9. Question 8: Balancing Chemical Equations | 题目8：配平化学方程式
Question: Balance these equations by writing the correct numbers in the blanks:
(a) __ Al + __ O₂ → __ Al₂O₃
(b) __ C₂H₆ + __ O₂ → __ CO₂ + __ H₂O
题目： 在横线上填入正确数字，配平下列方程式：
(a) __ Al + __ O₂ → __ Al₂O₃
(b) __ C₂H₆ + __ O₂ → __ CO₂ + __ H₂O
Answer & Explanation: C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O. Count: left 4 C, 12 H, 14 O; right 4 C, 12 H, 14 O. Always balance O₂ last as it is a single element. A common error is changing subscripts instead of putting coefficients in front.
答案与解析： (a) 先配平铝：4 Al + 3 O₂ → 2 Al₂O₃。检查：左右各 4 个 Al、6 个 O。(b) 对于烃类燃烧，先配平碳，再氢，最后氧。2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O。点数：左侧 4 C、12 H、14 O；右侧 4 C、12 H、14 O。氧气是单质，通常留到最后配平。一个常见的错误是去改化学式的小下标，而不是在前面配上适当的系数。
10. Common Mistakes to Avoid | 常见易错点
Error 1: c number with mass number. Atomic number = protons only. Mass number = protons + neutrons.
错误 1： 混淆原子序数与质量数。原子序数仅等于质子数，质量数 = 质子数 + 中子数。
Error 2: Forgetting state symbols when required, especially in ionic equations. Include (s), (l), (g), (aq) to gain full marks.
错误 2： 在需要时忘记标注状态符号，尤其是在离子方程式中。应注明 (s)、(l)、(g)、(aq) 以获得满分。
Error 3: relative formula mass calculations. Mᵣ of Ca(OH)₂ means Ca + 2×O + 2×H, not Ca + O₂ + H₂.
错误 3： 计算相对式量时处理括号不当。Ca(OH)₂ 的 Mᵣ 表示 Ca + 2×O + 2×H，而不是 Ca + O₂ + H₂。
Error 4: Missing out units or not following significant figure instructions in calculation answers. Always include g, kg, cm³ etc., and round as instructed.
错误 4： 计算题答案遗漏单位或未按要求保留有效数字。务必修带单位（g、kg、cm³ 等），并按要求进行四舍五入。
Error 5: Confusing oxidation and reduction at electrodes. Remember OIL RIG: Oxidation Is Loss of electrons; Reduction Is Gain. At anode: oxidation; at cathode: reduction.
错误 5： 混淆电极上的氧化和还原反应。记住“生氧还原”：氧化是失去电子，还原是得到电子。阳极发生氧化，阴极发生还原。
11. Final Exam Tips | 考试终极贴士
Always read the question stem twice – underline command words such as 'describe', 'explain', 'calculate', and 'state'. Show all your working in calculations; even if the final answer is wrong, you can still earn method marks. Write ionic equations with correct charges and inspect whether spectator ions need to be removed. Manage your time so you leave a few minutes at the end to check balancing, units, and significant figures. Finally, be confident – you’ve practised the core skills, and steady revision will pay off.
务必审题两遍——划出‘描述’、‘解释’、‘计算’和‘写出’等指令词。在计算题中展示所有步骤；即便最终答案有误，也能拿到过程分。写离子方程式时注意电荷正确，并判断是否需要删去旁观离子。合理分配时间，最后留出几分钟检查配平、单位和有效数字。最后，保持自信——你已经练习了核心技能，扎实的复习一定会有回报。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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