Year 11 SQA Chemistry: Unit Test Mock Paper Analysis | 十一年级 SQA 化学：单元测试模拟卷解析

📚 Year 11 SQA Chemistry: Unit Test Mock Paper Analysis | 十一年级 SQA 化学：单元测试模拟卷解析

This article walks you through a typical SQA National 5 Chemistry unit test mock paper, breaking down the most common question types and the key knowledge required to answer them confidently. Each section focuses on a core topic from Chemical Changes & Structure, Nature's Chemistry, or Chemistry in Society. You will find paired English and Chinese explanations for every concept, helping you master both the subject content and the examination technique.

本文带你解析一份典型的 SQA National 5 化学单元测试模拟卷，逐类攻克高频题型，结合知识点精准突破。每个专题都提炼自《化学变化与结构》《自然中的化学》和《社会中的化学》三个单元，采用中英双语逐点讲解，帮助你同时把握学科内容与应试方法。

---

1. Atomic Structure and Isotope Questions | 原子结构与同位素题型

Mock Question: An atom of element X has 17 protons and 20 neutrons. What is the mass number of X, and which particle has the same number of electrons as X⁻?

模拟题：某 X 元素的原子含有 17 个质子和 20 个中子。X 的质量数是多少？与 X⁻ 具有相同电子数的粒子是哪一个？

Step 1: The mass number is the sum of protons and neutrons. Here, 17 + 20 = 37.

第一步：质量数等于质子数与中子数之和，本例中 17 + 20 = 37。

Step 2: A neutral atom of X has 17 electrons. The negative ion X⁻ has gained one extra electron, giving 18 electrons in total.

第二步：中性 X 原子含有 17 个电子。带一个负电荷的 X⁻ 离子得到一个电子，电子总数变为 18。

Step 3: Identify a particle with 18 electrons. For example, an argon atom (atomic number 18) or a sulfide ion S²⁻ (16 protons + 2 extra electrons). The mock paper often asks you to compare with noble gas configurations.

第三步：寻找含有 18 个电子的粒子，例如氩原子（原子序数 18）或硫离子 S²⁻（16 个质子加上 2 个额外电子）。模拟卷常要求考生与稀有气体电子构型对比。

Step 4: Remember that isotopes have the same number of protons but a different number of neutrons. The mass spectrometer data can identify isotopes.

第四步：牢记同位素质子数相同而中子数不同。质谱仪数据可用于区分同位素。

---

2. Ionic and Covalent Formulae | 离子与共价化合物化学式

Mock Question: Write the formula for aluminium oxide and draw a diagram showing the bonding in a molecule of water.

模拟题：写出氧化铝的化学式，并画出水分子中共价键的示意图。

Step 1: Aluminium is in Group 13 and forms Al³⁺ ions. Oxygen is in Group 16 and forms O²⁻ ions. To balance charges, two Al³⁺ (total +6) combine with three O²⁻ (total -6). The formula is Al₂O₃.

第一步：铝位于第 13 族，形成 Al³⁺ 离子；氧位于第 16 族，形成 O²⁻ 离子。为使电荷平衡，需要 2 个 Al³⁺（共 +6）与 3 个 O²⁻（共 −6）结合，化学式为 Al₂O₃。

Step 2: Water is a covalent molecule. Oxygen shares two pairs of electrons — one pair with each hydrogen atom. Draw the oxygen in the centre with two bonding pairs and two lone pairs, and each hydrogen with one bonding pair.

第二步：水是共价分子。氧原子与两个氢原子各共用一对电子。中心氧原子周围画出两对成键电子和两对孤对电子，每个氢原子周围只有一对成键电子。

Step 3: In the mock paper, you may be asked to name compounds from given formulae, such as naming CuSO₄ as copper(II) sulfate, showing your understanding of valency and transition metals.

第三步：模拟卷中可能要求根据化学式命名，例如将 CuSO₄ 命名为硫酸铜(II)，考察对化合价和过渡金属命名规则的掌握。

---

3. Balancing Chemical Equations | 配平化学方程式

Mock Question: Balance the equation: __C₃H₈ + __O₂ → __CO₂ + __H₂O

模拟题：配平下列方程式：__C₃H₈ + __O₂ → __CO₂ + __H₂O

n. There are 3 carbon atoms in C₃H₈, so place a 3 in front of CO₂: C₃H₈ + O₂ → 3CO₂ + H₂O.

第一步：从碳原子入手。C₃H₈ 中含 3 个碳原子，因此在 CO₂ 前配系数 3：C₃H₈ + O₂ → 3CO₂ + H₂O。

There are 8 hydrogen atoms in C₃H₈, so place a 4 in front of H₂O: C₃H₈ + O₂ → 3CO₂ + 4H₂O.

第二步：接着配平氢原子。C₃H₈ 中有 8 个氢原子，在 H₂O 前配 4：C₃H₈ + O₂ → 3CO₂ + 4H₂O。

de has 3×2 + 4×1 = 10 oxygen atoms. So place a 5 in front of O₂: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

第三步：最后配氧。右侧共有 3×2 + 4×1 = 10 个氧原子，因此在 O₂ 前配 5：C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。

Step 4: Double-check that all atoms balance. State symbols may be required: (g) for gases, (l) for liquids, (aq) for aqueous solutions.

第四步：重新核对所有原子是否守恒。模拟题有时会要求标注物质状态：(g) 气体，(l) 液体，(aq) 水溶液。

---

4. Mole and Mass Calculations | 摩尔与质量计算

Mock Question: ced when 8.8 g of propane (C₃H₈) is burned completely. (Formula mass: C₃H₈ = 44, CO₂ = 44)

模拟题：8.8 g 丙烷 (C₃H₈) 完全燃烧后生成二氧化碳的质量是多少？（式量：C₃H₈ = 44，CO₂ = 44）

n = m / M

Step 1: Calculate the number of moles of propane. n(C₃H₈) = 8.8 g / 44 g mol⁻¹ = 0.20 mol.

第一步：计算丙烷的物质的量。n(C₃H₈) = 8.8 g / 44 g mol⁻¹ = 0.20 mol。

1 mole of propane gives 3 moles of carbon dioxide. So moles of CO₂ = 0.20 × 3 = 0.60 mol.

第二步：由配平方程式 C₃H₈ + 5O₂ → 3CO₂ + 4H₂O 知，1 mol 丙烷生成 3 mol 二氧化碳。因此 n(CO₂) = 0.20 × 3 = 0.60 mol。

Step 3: Mass of CO₂ = n × M = 0.60 mol × 44 g mol⁻¹ = 26.4 g.

第三步：CO₂ 的质量 m = n × M = 0.60 mol × 44 g mol⁻¹ = 26.4 g。

/ M and the proportional reasoning clearly; mock markers look for well-structured working.

第四步：务必清晰展示公式 n = m / M 及比例推导过程，阅卷人看重结构化的计算步骤。

---

5. Acids, Bases and pH | 酸碱与 pH 计算

Mock Question: A student adds hydrochloric acid to sodium hydroxide solution. Write the neutralisation equation and explain why the pH of the resulting solution might be 7, less than 7 or greater than 7.

模拟题：一名学生向氢氧化钠溶液中加入盐酸。写出中和反应方程式，并解释最终溶液的 pH 为何可能等于 7、小于 7 或大于 7。

Step 1: The neutralisation reaction is: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l).

第一步：中和反应方程式为：HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)。

Step 2: pH 7 occurs when the exact stoichiometric amount of acid and alkali have been mixed — no excess H⁺ or OH⁻ remains.

第二步：当酸与碱恰好完全反应、无多余的 H⁺ 或 OH⁻ 剩余时，pH 等于 7。

Step 3: If more acid is added, there is an excess of H⁺ ions, lowering the pH below 7. If more alkali is added, excess OH⁻ ions raise the pH above 7. The concept of limiting reactant is crucial here.

第三步：若酸过量，溶液中 H⁺ 浓度增加，pH 小于 7；若碱过量，OH⁻ 浓度增加，pH 大于 7。这里的核心考点是限制反应物的判断。

Step 4: The pH scale is logarithmic. A change of 1 pH unit represents a tenfold change in H⁺ ion concentration. The mock paper may test this idea coupled with indicators like phenolphthalein.

第四步：pH 为对数标度，每变化 1 个单位，H⁺ 浓度改变 10 倍。模拟题可能结合酚酞等指示剂进行考查。

---

广告 / Sponsored

6. Naming Hydrocarbons and Homologous Series | 烃的命名与同系物

Mock Question: Name the compound with the molecular formula C₅H₁₂ and draw the full structural formula of its branched isomer.

模拟题：写出分子式为 C₅H₁₂ 的化合物名称，并画出其支链异构体的完整结构式。

Step 1: C₅H₁₂ follows the general formula CₙH₂ₙ₊₂, so it is an alkane. The straight-chain isomer is pentane.

第一步：C₅H₁₂ 符合通式 CₙH₂ₙ₊₂，属于烷烃。直链异构体称为戊烷（pentane）。

Step 2: The branched isomer is 2-methylbutane. The longest continuous chain has four carbons (butane), and a methyl group is attached to the second carbon atom.

第二步：支链异构体为 2-甲基丁烷（2-methylbutane）。最长碳链含四个碳原子（丁烷），在第二个碳上连接一个甲基。

Step 3: When drawing the full structural formula, show all atoms and bonds: CH₃–CH(CH₃)–CH₂–CH₃. Ensure each carbon forms four bonds.

第三步：绘制完整结构式时需标出所有原子和键：CH₃–CH(CH₃)–CH₂–CH₃，并确保每个碳原子形成四个共价键。

Step 4: The homologous series share the same functional group and similar chemical properties. The mock may ask you to predict the formula of the next member, e.g., hexane (C₆H₁₄).

第四步：同系物具有相同的官能团和相似的化学性质。模拟题可能要求预测下一个同系物的分子式，如己烷 (C₆H₁₄)。

---

7. Alcohols, Carboxylic Acids and Their Reactions | 醇、羧酸及其反应

Mock Question: Ethanol can be oxidised to a carboxylic acid. Name the carboxylic acid and state a reagent that could be used for this oxidation.

模拟题：乙醇可被氧化为一种羧酸。命名该羧酸，并写出一种可用于此氧化反应的试剂。

Step 1: Ethanol (CH₃CH₂OH) belongs to the alcohol homologous series. Oxidation converts the –OH group into a –COOH group, producing ethanoic acid (CH₃COOH).

第一步：乙醇 (CH₃CH₂OH) 属于醇类同系物。氧化反应将 –OH 转变为 –COOH 基团，生成乙酸 (CH₃COOH)。

Step 2: Common oxidising agents include acidified potassium dichromate (K₂Cr₂O₇ / H⁺) or heating with copper(II) oxide. The colour change from orange to green indicates oxidation has occurred.

第二步：常用的氧化剂有酸化重铬酸钾 (K₂Cr₂O₇ / H⁺) 或与氧化铜共热。反应中橙色变为绿色表明氧化作用发生。

Step 3: Carboxylic acids react with alcohols to form esters in the presence of a concentrated sulfuric acid catalyst. For example, ethanoic acid + ethanol ⇌ ethyl ethanoate + water.

第三步：羧酸与醇在浓硫酸催化下发生酯化反应，例如乙酸 + 乙醇 ⇌ 乙酸乙酯 + 水。

Step 4: The mock may include questions on naming esters or identifying the alcohol and acid used to make a specific ester. Remember that the alcohol part gives the first part of the ester name.

第四步：模拟题可能要求命名酯，或由特定酯反推所用的醇和酸。需注意酯的名称中醇的部分在前。

---

8. Rates of Reaction and Collision Theory | 反应速率与碰撞理论

Mock Question: Explain why increasing the concentration of hydrochloric acid speeds up the reaction with magnesium ribbon, using collision theory.

模拟题：用碰撞理论解释，为什么提高盐酸的浓度会加快其与镁带的反应速率。

Step 1: For a reaction to occur, particles must collide with sufficient energy (activation energy) and with the correct orientation.

第一步：发生反应的前提是粒子必须碰撞，且碰撞的能量不低于活化能，同时取向正确。

Step 2: Increasing the concentration of HCl means there are more H⁺ ions per unit volume. This increases the frequency of collisions between H⁺ ions and magnesium atoms.

第二步：提高盐酸浓度意味着单位体积内含有更多的 H⁺ 离子，增大了 H⁺ 离子与镁原子之间的碰撞频率。

Step 3: More frequent successful collisions per second lead to a higher rate of reaction. The total number of successful collisions increases, even though the activation energy remains unchanged.

第三步：单位时间内发生更多有效碰撞，反应速率随之提高。虽然活化能不变，但有效碰撞总数上升。

Step 4: Other factors that affect rate include temperature (increases particle energy and collision frequency), particle size (increases surface area) and the use of a catalyst (lowers activation energy). Mock papers often combine a graph interpretation with these factors.

第四步：影响速率的其他因素有温度（增加粒子能量和碰撞频率）、颗粒大小（增大表面积）以及催化剂（降低活化能）。模拟卷常将图表分析与这些因素结合考查。

---

9. Electrolysis of Molten Compounds and Solutions | 熔融物与溶液的电解

Mock Question: Describe what happens at the cathode during the electrolysis of molten lead(II) bromide, and write the relevant ion-electron equation.

模拟题：描述电解熔融溴化铅(II)时阴极发生的变化，并写出相应的离子-电子方程式。

Step 1: In molten lead(II) bromide, the ions Pb²⁺ and Br⁻ are free to move. During electrolysis, positive Pb²⁺ ions migrate to the negative cathode.

第一步：熔融溴化铅中，Pb²⁺ 和 Br⁻ 离子可自由移动。电解时，阳离子 Pb²⁺ 向负极（阴极）移动。

(reduction) to form neutral lead atoms: Pb²⁺ + 2e⁻ → Pb. Silvery liquid lead is observed.

第二步：在阴极，Pb²⁺ 离子得到电子（被还原），生成中性铅原子：Pb²⁺ + 2e⁻ → Pb。观察到银白色液态铅。

e anode, bromide ions lose electrons: 2Br⁻ → Br₂ + 2e⁻. Red-brown bromine gas is produced.

第三步：在阳极，溴离子失去电子：2Br⁻ → Br₂ + 2e⁻，产生红棕色溴气。

Step 4: For aqueous solutions, you must compare the reactivity of the metal ion with hydrogen and the halide ion with hydroxide. Use the electrochemical series. Simpler ions like Na⁺ are more difficult to discharge than H⁺, so hydrogen gas is released instead.

第四步：电解水溶液时，必须根据电化序比较金属离子与氢离子、卤离子与氢氧根离子的放电顺序。如 Na⁺ 比 H⁺ 更难放电，因此阴极释放氢气。

---

10. Titration Calculation and Analysis | 滴定计算与分析

Mock Question: alculate the concentration of the sodium hydroxide solution. 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

模拟题：一名学生用 0.100 mol l⁻¹ 硫酸滴定 25.0 cm³ 氢氧化钠溶液，平均滴定体积为 20.5 cm³。计算氢氧化钠溶液的浓度。2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

es of H₂SO₄ used: n = c × V (in litres) = 0.100 mol l⁻¹ × (20.5 / 1000) dm³ = 0.00205 mol.

第一步：计算所用硫酸的物质的量：n = c × V (单位为升) = 0.100 mol l⁻¹ × (20.5/1000) dm³ = 0.00205 mol。

quation, 1 mol H₂SO₄ reacts with 2 mol NaOH. So moles of NaOH = 2 × 0.00205 = 0.00410 mol.

第二步：由化学计量数可知，1 mol H₂SO₄ 与 2 mol NaOH 反应，故 n(NaOH) = 2 × 0.00205 = 0.00410 mol。

Step 3: Concentration of NaOH = n / V = 0.00410 mol / (25.0/1000) dm³ = 0.164 mol l⁻¹.

第三步：NaOH 的浓度 c = n / V = 0.00410 mol / 0.0250 dm³ = 0.164 mol l⁻¹。

Step 4: In the unit test, you may also be asked to explain the use of a suitable indicator, such as phenolphthalein (pink to colourless in acid-base titrations), and to identify concordant titres. Always show your working clearly and round the final answer appropriately.

第四步：单元测试还可能要求解释指示剂的选择，例如酚酞（酸性滴定中由粉红色变为无色），以及识别平行滴定结果。务必清晰展示计算过程并合理修约最终答案。

---

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）