📚 WJEC Year 10 Chemistry Unit Test Mock Paper Analysis | WJEC 十年级化学单元测试模拟卷解析
This article provides a detailed walkthrough of a typical WJEC Year 10 Chemistry unit test mock paper. It covers answers, explanations and examiner tips for key question types from atomic structure to quantitative chemistry. The content aligns with the WJEC GCSE Chemistry specification and helps you practise the skills needed to succeed in class tests and mocks.
本文详细解析一份典型的 WJEC 十年级化学单元测试模拟卷，涵盖从原子结构到定量化学的主要题型的答案、讲解及考官提示。内容紧扣 WJEC GCSE 化学大纲，帮助你练习课堂测验和模拟考所需的关键技能。
1. Mock Paper Structure Overview | 模拟卷结构概览
The mock paper is designed to mimic a 60-minute WJEC Unit 1 test. It includes multiple-choice questions, short structured questions, calculations, and a data analysis task. Topics span atomic structure, the periodic table, ionic and covalent bonding, balancing equations, moles, acids and bases, electrolysis and the reactivity series.
该模拟卷模拟 60 分钟的 WJEC 单元 1 测试，题型包括选择题、简答题、计算题和数据分析题。内容涵盖原子结构、周期表、离子键与共价键、配平方程式、摩尔、酸碱、电解及金属活动性顺序。
| Section | Question Types | Marks | Main Topics |
|---|---|---|---|
| A | Multiple-choice (Q1-5) | 5 × 2 = 10 | Atomic structure, periodic table, bonding |
| B | Structured response (Q6-9) | 28 | Equations, moles, acids, electrolysis, reactivity |
| C | Data analysis (Q10) | 12 | Titration curves, indicator choice |
The total mark is 50, and grade boundaries typically follow: 45+ for A*, 38+ for A, 30+ for B. Use this breakdown to identify your weaker areas during revision.
试卷总分 50 分，常见的等级分数线为：45 分及以上为 A*，38 分及以上为 A，30 分及以上为 B。在复习时可根据上述划分找出自己的薄弱环节。
2. Questions 1–5: Multiple Choice on Atomic Structure and Periodic Table | 选择题 1–5：原子结构与周期表
Question 1: Which particle has a relative charge of -1 and a mass that is often considered negligible?
第 1 题：哪种粒子相对电荷为 -1 且质量通常可忽略不计？
Answer: Electron. The electron carries a charge of -1 relative to a proton, and its mass is about 1/1836 of a proton's mass. In most GCSE calculations, we treat its mass as negligible compared with protons and neutrons.
答案：电子。电子的相对电荷为 -1，其质量约为质子的 1/1836。在 GCSE 阶段的大多数计算中，与质子和中子相比，电子的质量可忽略不计。
Question 2: An atom of element X has 20 protons and 20 neutrons. What is its mass number, and in which group and period of the periodic table is it found?
第 2 题：某元素 X 的原子有 20 个质子和 20 个中子，它的质量数是多少？它位于周期表的哪一族和哪一周期？
e electron configuration is 2,8,8,2. This places it in Group 2 (2 outer electrons) and Period 4 (4 occupied shells). The element is calcium.
答案：质量数 = 质子数 + 中子数 = 40。原子序数为 20，电子排布为 2,8,8,2。它在第 2 族（最外层 2 个电子）和第 4 周期（4 个电子层）。该元素是钙。
Question 3: Which of the following substances has a giant covalent structure? A) Sodium chloride B) Water C) Diamond D) Copper
第 3 题：下列哪种物质具有巨型共价结构？A) 氯化钠 B) 水 C) 金刚石 D) 铜
Answer: C, diamond. In diamond, each carbon atom is covalently bonded to four others in a continuous 3D network. Sodium chloride is ionic, water is simple molecular, and copper is metallic.
答案：C，金刚石。金刚石中每个碳原子都与其他四个碳原子形成共价键，构成连续的三维网络。氯化钠是离子化合物，水是简单分子，铜是金属。
Question 4: Why do elements in the same group of the periodic table have similar chemical properties?
第 4 题：为什么同族元素具有相似的化学性质？
Answer: They have the same number of electrons in their outermost shell. For example, all Group 1 elements have one outer electron, so they all react vigorously with water to form an alkali and hydrogen.
答案：因为它们的最外层电子数相同。例如，所有第 1 族元素最外层都有一个电子，因此它们都能与水剧烈反应生成碱和氢气。
Question 5: Which type of bonding holds the atoms together in a molecule of oxygen (O₂)?
第 5 题：氧分子（O₂）中原子之间是哪一类化学键？
wo pairs of electrons with the other, giving a stable octet. The bond can be shown as O=O.
答案：双共价键。每个氧原子与另一个氧原子共享两对电子，从而获得稳定的八电子结构。化学键可表示为 O=O。
3. Question 6: Balancing Equations and Mole Calculations | 第 6 题：化学方程式配平与摩尔计算
The question asks: ' dium reacts completely with excess water. (Relative atomic masses: Na = 23, O = 16, H = 1)'
题目要求：“配平化学方程式：__Na + __H₂O → __NaOH + __H₂。然后计算 4.6 g 钠与过量水完全反应时生成的氢氧化钠的物质的量（相对原子质量：Na=23，O=16，H=1）。”
Step 1 - Balancing: Count atoms on each side. Place a 2 in front of Na on the left and a 2 in front of NaOH on the right to balance sodium. Then balance hydrogen by placing a 2 in front of H₂O. The balanced equation is:
步骤 1 - 配平：数清两边原子个数。在左侧 Na 前配 2，右侧 NaOH 前配 2 使钠原子守恒。然后在 H₂O 前配 2 使氢原子守恒。完整的配平方程式为：
2Na + 2H₂O → 2NaOH + H₂
Na to NaOH is 2:2, which simplifies to 1:1. Therefore, moles of NaOH produced = 0.20 mol.
步骤 2 - 摩尔计算：先求 Na 的物质的量。n = m / M = 4.6 g ÷ 23 g mol⁻¹ = 0.20 mol。由方程式可知，Na 与 NaOH 的系数比为 2:2，即 1:1。因此生成 NaOH 的物质的量为 0.20 mol。
Examiners often award marks for showing the working formula, substitution, correct answer and appropriate unit. Always write g mol⁻¹ for molar mass and mol for amount.
考官通常会给出公式、代入数据、正确答案和单位的分数。务必使用 g mol⁻¹ 表示摩尔质量，用 mol 表示物质的量。
4. Question 7: Acids, Bases and Neutralisation | 第 7 题：酸、碱与中和反应
The task: 'Hydrochloric acid (HCl) is neutralised by sodium hydroxide solution. Write a balanced symbol equation with state symbols, and then write the ionic equation for the reaction. Explain why the temperature of the mixture increases.'
题目：“盐酸 (HCl) 与氢氧化钠溶液发生中和反应。写出配平的带状态符号的化学方程式，再写出离子方程式。并解释混合物温度为何升高。”
The full balanced equation is:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
The ionic equation shows only the species that change. Sodium ions (Na⁺) and chloride ions (Cl⁻) remain as spectator ions. The reaction is:
H⁺(aq) + OH⁻(aq) → H₂O(l)
The temperature rises because neutralisation is exothermic. Energy is released when the strong H⁺—OH⁻ bonds form in water, and the total energy of the products is lower than that of the reactants. In the lab, you would observe the thermometer reading increase by several degrees Celsius.
温度升高是因为中和反应放热。当水中的强 O—H 共价键形成时释放能量，产物的总能量低于反应物。在实验中你会发现温度计的读数升高数摄氏度。
t (aq) next to ions. Always include state symbols unless instructed otherwise by the question.
常见错误：忘记写状态符号，或写为 H⁺ + OH⁻ → H₂O 而未给离子标注 (aq)。除非题目另有说明，否则必须始终标注状态符号。
5. Question 8: Electrolysis Predictions | 第 8 题：电解预测
The question: 'When a direct current is passed through molten lead(II) bromide using inert electrodes, lead forms at the cathode and bromine forms at the anode. Write half-equations for both electrode reactions and explain why the lead(II) ions are reduced.'
题目：“用惰性电极电解熔融溴化铅时，阴极生成铅，阳极生成溴。写出两个电极反应的半方程式，并解释为何铅离子被还原。”
In molten lead(II) bromide, the ions are free to move. At the cathode (negative electrode), Pb²⁺ ions gain electrons (reduction):
Pb²⁺ + 2e⁻ → Pb
At the anode (positive electrode), Br⁻ ions lose electrons (oxidation):
2Br⁻ → Br₂ + 2e⁻
Reduction is the gain of electrons. Each Pb²⁺ ion gains two electrons to become a neutral lead atom, so it is reduced. Remember the mnemonic OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons.
还原是指得到电子。每个 Pb²⁺ 离子得到两个电子变为中性铅原子，因此被还原。记住口诀 OIL RIG：氧化失电子，还原得电子。
If inert electrodes (like graphite or platinum) are not used, electrode reactivity matters, but that is not required here.
若不使用惰性电极（如石墨或铂），电极本身的活泼性会影响产物，但本题无需考虑。
6. Question 9: Reactivity Series and Displacement | 第 9 题：活动性顺序与置换反应
Here is the question: 'A piece of zinc metal is added to blue copper(II) sulfate solution. State the colour change expected and write a balanced equation, including state symbols, for the reaction. Name the type of reaction.'
题目：“将一块锌放入蓝色的硫酸铜溶液中。描述预期的颜色变化，写出配平并带有状态符号的化学方程式，并指出反应类型。”
Zinc is more reactive than copper, so it displaces copper from its compound. The balanced equation is:
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
The colour change: the blue colour of the copper(II) sulfate solution fades and eventually becomes colourless (zinc sulfate solution is colourless). A reddish-pink solid (copper) coats the surface of the zinc.
颜色变化：硫酸铜溶液的蓝色逐渐褪去，最终变为无色（硫酸锌溶液为无色）。锌的表面会覆盖一层红棕色的固体铜。
ion: zinc is oxidised (Zn → Zn²⁺ + 2e⁻) and copper(II) ions are reduced (Cu²⁺ + 2e⁻ → Cu).
该反应属于置换反应，同时也是氧化还原反应：锌被氧化 (Zn → Zn²⁺ + 2e⁻)，铜离子被还原 (Cu²⁺ + 2e⁻ → Cu)。
Writing the ionic equation can earn extra marks:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
仔细写出离子方程式可额外得分。
7. Question 10: Data Analysis – Titration Curves | 第 10 题：数据分析 – 滴定曲线
In this section, students are given a titration curve of pH against volume of sodium hydroxide added to hydrochloric acid. The pH rises slowly at first, then shoots up steeply, and finally levels off. The question: 'Mark the equivalence point and justify why phenolphthalein is a suitable indicator for this titration.'
本题提供了一张向盐酸中逐滴加入氢氧化钠溶液时 pH 随体积变化的滴定曲线。曲线起初缓慢上升，随后急剧升高，最后趋于平缓。题目要求：“标出滴定终点，并说明酚酞为何是此滴定的合适指示剂。”
The equivalence point lies at the midpoint of the steep vertical rise, where the number of moles of H⁺ equals the number of moles of OH⁻ added. For a strong acid–strong base titration, the pH at the equivalence point is 7.
滴定终点位于曲线垂直陡升段的中点，该点处加入的 OH⁻ 的物质的量与溶液中 H⁺ 的物质的量恰好相等。强酸–强碱滴定的终点 pH 值为 7。
Phenolphthalein changes colour over the pH range of about 8.2 to 10.0. Because this range falls entirely within the steep portion of the curve, the colour change from colourless to pink is sharp and coincides with a tiny addition of base. This makes it a perfect indicator.
酚酞的变色范围约为 pH 8.2～10.0。由于该范围完全处于曲线的陡升段内，从无色到粉红的颜色变化非常敏锐，只需极少量的碱即会变色，因此是理想的指示剂。
Be careful not to confuse the end point (where the indicator changes colour) with the equivalence point. The end point should be as close as possible to the equivalence point.
注意勿将滴定终点（指示剂变色点）与等当点混淆。所选的终点应尽可能接近等当点。
8. Common Mistakes and Examiner Tips | 常见错误与考官提示
Mistake 1: Writing ionic equations without cancelling spectator ions. Exam boards expect simplified ionic equations. For instance, Na⁺ and Cl⁻ should be removed from the neutralisation ionic equation.
错误 1：书写离子方程式时未删去旁观离子。考试要求写出最简离子方程式，例如中和反应的离子方程式中需删去 Na⁺ 和 Cl⁻。
Mistake 2: Using 'moles' incorrectly in unit handling. Students sometimes write 'number of moles = mass × Mr' instead of mass / Mr. Always use n = m / M with the correct units.
错误 2：摩尔计算中的单位处理不当。常有学生误写为“物质的量 = 质量 × 相对分子质量”，而正确公式应为 n = m / M。务必使用正确公式和单位。
Mistake 3: Confusing oxidation and reduction. Remember OIL RIG, and when identifying redox reactions, state which species is oxidised and which is reduced with electron half-equations if possible.
错误 3：混淆氧化与还原。牢记 OIL RIG 口诀。在分析氧化还原反应时，尽量指出哪种物质被氧化、哪种被还原，并写出电子半方程式。
Examiner tip: State symbols (s), (l), (g), (aq) carry marks. Unless told otherwise, always include them in symbol equations.
考官提示：状态符号 (s)、(l)、(g)、(aq) 都是有分的。除非题目明确说明可以不写，否则务必在符号方程式中标注。
Stay precise when describing colour changes. For example, 'the blue solution turns colourless' is better than 'the colour disappears'.
描述颜色变化时要准确。例如，“蓝色溶液变为无色”比“颜色消失”更符合得分要求。
9. Key Formulas and Constants
Published by TutorHao | Year 10 Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图