📚 Cambridge Year 9 Chemistry: Case Study Practice | 剑桥九年级化学：案例分析实战演练
This article guides you through a series of real-world case studies designed to sharpen your ability to apply core Year 9 Cambridge Chemistry ideas. Each scenario is followed by a step-by-step analysis in both English and Chinese, helping you link theory to practical problem-solving, just as you would in an examination or laboratory setting.
本文通过一系列真实案例分析，帮助你提高应用剑桥九年级化学核心知识的能力。每个情境都配有中英双语的分步解析，引导你将理论与实际解题联系起来，像在考试或实验室中那样思考。
1. Investigating Reaction Rate with Magnesium and Acid | 镁与酸反应速率的探究
A student adds a 5 cm strip of clean magnesium ribbon to 50 cm³ of dilute hydrochloric acid (1.0 mol/dm³) in a conical flask. The flask is immediately connected to a gas syringe, and the volume of hydrogen gas collected is recorded every 10 seconds. The results are shown in the table below.
一位学生将一条 5 cm 长的洁净镁带加入 50 cm³ 稀盐酸（1.0 mol/dm³）中，迅速将锥形瓶与气体注射器连接，每隔 10 秒记录收集到的氢气体积。结果如下表。
| Time / s | Volume of H₂ / cm³ |
|---|---|
| 0 | 0 |
| 10 | 24 |
| 20 | 42 |
| 30 | 52 |
| 40 | 58 |
| 50 | 60 |
a) Write a balanced word equation for the reaction.
a) 写出该反应的配平文字方程式。
b) Use the data to explain why the rate of reaction is fastest in the first 10 seconds.
b) 利用数据解释为什么前 10 秒的反应速率最快。
c) Suggest how and why the shape of the graph would change if the experiment were repeated at a higher temperature.
c) 如果在更高温度下重复实验，预测曲线的形状会如何变化并解释原因。
Analysis
oric acid → magnesium chloride + hydrogen. To write a symbol equation: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
a) 镁与盐酸反应生成氯化镁和氢气。文字方程式为：镁 + 盐酸 → 氯化镁 + 氢气。符号方程式：Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)。
b) The rate is highest initially because the concentration of hydrochloric acid is at its maximum, so there are more H⁺ ions per unit volume to collide with the magnesium surface. In the first 10 seconds, 24 cm³ of hydrogen is produced, giving an average rate of 2.4 cm³/s, compared with only 0.2 cm³/s in the final 10 seconds. As the acid is used up, the concentration decreases and the frequency of effective collisions becomes lower, slowing down the reaction.
b) 反应初期速率最快是因为盐酸的浓度处于最高点，单位体积内有更多的 H⁺ 离子与镁表面发生碰撞。前 10 秒产生 24 cm³ 氢气，平均速率约 2.4 cm³/s，而最后 10 秒只有 0.2 cm³/s。随着酸的消耗，浓度下降，有效碰撞频率降低，反应变慢。
c) At a higher temperature, the particles have greater kinetic energy, move faster, and collide more frequently and with more energy. This means a larger proportion of collisions exceed the activation energy. The graph would be steeper initially and reach the final volume more quickly, but the total volume of gas collected would remain the same because the same mass of magnesium is used.
c) 在更高温度下，粒子动能增大，运动加快，碰撞更频繁且能量更高，超过活化能阈值的碰撞比例增加。因此反应曲线初始斜率更陡，更快达到最终体积，但由于镁的质量相同，最终收集的气体总体积不变。
2. Identifying White Solids Using Laboratory Tests | 利用实验室检验鉴别白色固体
A student is given four unlabelled white powders: table salt (sodium chloride), sugar, baking soda (sodium hydrogen carbonate) and chalk (calcium carbonate). By using only water, dilute hydrochloric acid and a Bunsen burner, design a sequence of tests to distinguish each substance. Record expected observations for each step.
一位学生拿到四种没有标签的白色粉末：食盐（氯化钠）、白糖、小苏打（碳酸氢钠）和粉笔屑（碳酸钙）。只允许使用水、稀盐酸和本生灯，设计一系列检验方案区分它们，并记录每一步的预期现象。
Analysis
Step 1 – Solubility test: Add a small amount of each powder to separate test tubes containing about 5 cm³ of distilled water and shake. Sugar and salt will dissolve completely, forming colourless solutions. Chalk and baking soda may appear to dissolve partially or give a cloudy mixture, but chalk is actually insoluble in water, while baking soda is sparingly soluble. However, the key observation is that sugar and salt give clear solutions, while the other two leave some solid residue.
步骤一 – 溶解性试验：分别取少量粉末加入装有约 5 cm³ 蒸馏水的试管中振荡。糖和盐会完全溶解，形成无色溶液。粉笔屑和小苏打可能看似部分溶解或变浑浊，其实粉笔屑不溶于水，小苏打微溶。关键现象是糖和盐得到澄清溶液，另两种有明显固体残留。
Step 2 – Acid test on the two insoluble/semi-soluble samples: Add dilute hydrochloric acid to the baking soda and chalk residues. Baking soda effervesces vigorously, producing carbon dioxide gas, which turns limewater milky. Chalk (calcium carbonate) also effervesces with acid, giving CO₂. Therefore, this step alone does not separate them; however, the rate of fizzing may differ – baking soda reacts more vigorously with cold acid. To distinguish, we can use the flame test. But the problem limits us to a Bunsen burner: make a paste of the solids, dip a nichrome wire, and hold in a blue flame. Chalk gives a brick-red flame (calcium), while baking soda gives a yellow flame (sodium). So the solid that gives a yellow flame is baking soda, and the one with a brick-red flame is chalk.
步骤二 – 对两种不易溶的样品进行酸检：分别向小苏打和粉笔屑残渣中加入稀盐酸。小苏打剧烈冒泡，产生使石灰水变浑浊的二氧化碳气体。粉笔屑（碳酸钙）遇酸也冒泡生成 CO₂。因此单靠此步无法完全区分；但冒泡速率可能不同，小苏打在冷酸中反应更剧烈。为确定，可利用焰色反应：将固体调成糊状，用镍铬丝蘸取在蓝色火焰中灼烧。粉笔屑呈现砖红色火焰（钙），小苏打呈现黄色火焰（钠）。因此，焰色为黄色的是小苏打，砖红色的是粉笔屑。
Step 3 – gar will caramelise, turning brown and giving a distinct sweet smell, while salt will simply leave white crystals with no odour. Alternatively, salt can be identified by a yellow flame test, but the smell and colour change of heated sugar are sufficient. In summary: substance that dissolves, gives no colour change on heating, yellow flame test = salt; dissolves, caramelises = sugar; insoluble, acid effervescence, brick-red flame = chalk; sparingly soluble, acid effervescence, yellow flame = baking soda.
步骤三 – 区分盐和糖：将两种澄清溶液小心加热蒸发。糖会焦化，变成棕色并散发特有的甜味，而盐仅析出白色晶体且无气味。也可用焰色反应，食盐呈黄色火焰，但加热时的颜色和气味变化足以区分。总结：溶于水，加热不变色，黄色火焰 = 食盐；溶于水，加热焦化 = 白糖；不溶，遇酸冒泡，砖红色火焰 = 粉笔屑；微溶，遇酸冒泡，黄色火焰 = 小苏打。
3. pH and Household Solutions | pH 与家用溶液的测定
A group of students use universal indicator solution to test five common household liquids: vinegar, baking soda solution, distilled water, lemon juice and soap solution. They record the colours and approximate pH values as shown:
一组学生用通用指示剂测试五种常见家用液体：醋、小苏打溶液、蒸馏水、柠檬汁和肥皂水，记录颜色和近似 pH 值如下：
| Liquid | Colour with Universal Indicator | Estimated pH |
|---|---|---|
| Vinegar | Red | 2 |
| Lemon juice | Orange | 3 |
| Distilled water | Green | 7 |
| Baking soda solution | Blue | 9 |
| Soap solution | Violet | 11 |
a) Arrange the liquids in order of increasing acidity (most acidic first).
a) 按酸性由强到弱排列这些液体。
b) Explain why distilled water gives a green colour and how its pH explains its neutral nature.
b) 解释为什么蒸馏水呈绿色，以及其 pH 如何体现中性。
c) If vinegar is diluted ten times with water, predict the colour change of universal indicator and the new estimated pH range.
c) 若将醋用水稀释十倍，预测通用指示剂的颜色变化以及新的近似 pH 范围。
Analysis
ing soda solution (pH 9) → soap solution (pH 11). Vinegar is the most acidic because it contains ethanoic acid; lemon juice contains citric acid.
a) 酸性随 pH 降低而增强。因此由强到弱的顺序为：醋 (pH 2) → 柠檬汁 (pH 3) → 蒸馏水 (pH 7) → 小苏打溶液 (pH 9) → 肥皂水 (pH 11)。醋含乙酸，柠檬汁含柠檬酸，故酸度较高。
tral solution has pH = 7 at 25°C. Distilled water undergoes slight self-ionisation: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq), giving [H⁺] = 1 × 10⁻⁷ mol/dm³.
b) 通用指示剂在 pH=7 时呈绿色，因为纯水中 H⁺ 和 OH⁻ 离子浓度相等。中性溶液在 25°C 下 pH 值为 7。蒸馏水存在微弱自电离：H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)，得出 [H⁺] = 1 × 10⁻⁷ mol/dm³。
c) Vinegar is an acidic solution. When diluted ten times, the H⁺ ion concentration decreases, causing the pH to increase by approximately 1 unit (since pH is logarithmic). The colour would change from red to orange-red or orange, indicating a pH of around 3. The actual pH depends on whether the acid is strong or weak; for a weak acid like ethanoic acid, the shift is slightly less than 1 unit, but the colour would visibly lighten.
c) 醋是酸性溶液。稀释十倍后，H⁺ 离子浓度下降，pH 值大约增加 1 个单位（因 pH 为对数标度）。颜色将从红色变为橙红色或橙色，pH 约升至 3。实际变化因乙酸是弱酸会略有偏差，但颜色明显变淡。
4. Separating Mixtures Using a Sequence of Techniques | 分离混合物方法的组合使用
A mixture contains sand, common salt (sodium chloride), iodine crystals and ethanol. You are provided with standard laboratory equipment including filter paper, evaporating dish, Bunsen burner, distillation apparatus and a fume cupboard. Design a step-by-step method to obtain each component in a relatively pure form. Justify each separation technique chosen.
一种混合物含有沙子、食盐（氯化钠）、碘晶体和乙醇。实验室提供滤纸、蒸发皿、本生灯、蒸馏装置和通风橱。请设计一个分步方法以获得每种成分的相对纯净形态，并说明每一步选择分离技术的理由。
Analysis
Step 1 – Filtration: Add the mixture to a beaker of water and stir. Ethanol will dissolve, salt will dissolve, iodine is slightly soluble but likely remains as crystals with sand. Sand is insoluble. Reassess: the mixture contains four components – ethanol (liquid), iodine (solid), salt (soluble solid), sand (insoluble solid). If we add water, ethanol will mix completely, salt dissolves, iodine may partially dissolve (but is more soluble in ethanol). Instead, we can first add the mixture directly to a beaker and use gentle warming to separate iodine by sublimation. Iodine turns from solid to purple vapour when heated and can be collected on a cold surface. But this must be done in a fume cupboard. So first: heat the mixture gently in an evaporating dish covered with a filter funnel with a cold water flask on top to collect iodine crystals by sublimation. Then the remaining mixture (salt + sand + ethanol) can be separated further. Another approach: add ethanol to the original mixture, iodine and ethanol dissolve, filter to remove sand; then distil the filtrate to recover ethanol, leaving iodine and salt mixture, then sublimation to get iodine, leaving salt. Many routes exist. A clear plan: use the fact that ethanol is volatile, iodine sublimes, salt dissolves in water, sand is insoluble.
步骤一 – 升华分离碘：将混合物放入蒸发皿，在通风橱内微热，碘升华产生紫色蒸气，遇冷表面凝华成碘晶体，收集。剩余为沙子、盐和乙醇。
步骤二 – 溶解与过滤：向剩余物中加入水，乙醇和盐溶解，沙子不溶。过滤，沙子留在滤纸上，冲洗干燥得纯净沙子。滤液含盐、乙醇和水。
步骤三 – 蒸馏分离乙醇：用蒸馏装置加热滤液。乙醇沸点 78°C，先蒸出，经冷凝回收液态乙醇。控制温度不超过 100°C，残余液体为盐水溶液。
步骤四 – 蒸发结晶获得盐：将残余盐水溶液转移至蒸发皿，加热蒸干水分，得到白色食盐晶体。若需要更纯，可用少量水重结晶。
Justifications: Sublimation exploits iodine's ability to change from solid to gas directly, avoiding dissolving other components. Filtration is used because sand is insoluble. Distillation works because ethanol has a lower boiling point than water, and evaporation crystallisation works because salt is soluble and thermally stable.
理由：升华利用碘可直接从固态变气态的特性，避免引入溶剂；过滤基于沙不溶于水；蒸馏因乙醇沸点低于水；蒸发结晶则因盐可溶且热稳定。
5. Displacement Reactions and the Reactivity Series | 置换反应与活动性顺序
A student sets up three test tubes. In tube 1, a strip of zinc is placed in copper(II) sulfate solution. In tube 2, a strip of copper is placed in zinc sulfate solution. In tube 3, a strip of magnesium is placed in zinc sulfate solution. Predict observations and write balanced symbol equations where a reaction occurs. Also explain any reaction in terms of electron transfer.
一位学生准备了三支试管。试管 1 中将锌条放入硫酸铜溶液中；试管 2 中将铜条放入硫酸锌溶液中；试管 3 中将镁条放入硫酸锌溶液中。预测观察到的现象，并写出发生反应的配平符号方程式，同时从电子转移角度解释反应。
Analysis
of copper metal deposits on the zinc strip. Equation: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). Ionic equation: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Zinc atoms lose two electrons each (oxidation) while Cu²⁺ ions gain two electrons (reduction).
试管 1：锌比铜活泼，锌置换铜。硫酸铜溶液的蓝色逐渐褪去，锌条表面覆盖一层红棕色金属铜。方程式：Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)。离子方程式：Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)。锌原子失去两个电子（氧化），Cu²⁺ 离子得到两个电子（还原）。
Tube 2: Copper is less reactive than zinc, so no reaction occurs. The blue colour of copper sulfate would not change significantly, though a very slight mixing of colours may occur due to diffusion. No solid deposit forms.
试管 2：铜不如锌活泼，不发生反应。溶液颜色基本不变，没有固体沉积。
ur change is less dramatic but the solution may become slightly warmer as it is an exothermic displacement. Equation: Mg(s) + ZnSO₄(aq) → MgSO₄(aq) + Zn(s). Magnesium atoms oxidise to Mg²⁺, zinc ions reduce to Zn atoms. This confirms the reactivity series: Mg > Zn > Cu.
试管 3：镁比锌活泼，镁置换锌。灰色镁条可能被一层暗色的锌覆盖，溶液颜色变化不明显，但反应可能微热。方程式：Mg(s) + ZnSO₄(aq) → MgSO₄(aq) + Zn(s)。镁原子氧化成 Mg²⁺，锌离子还原成锌原子。这验证了活动性顺序：Mg > Zn > Cu。
6. Combustion Reactions – Comparing Magnesium and a Candle | 燃烧反应 – 比较镁与蜡烛
A student burns a piece of magnesium ribbon in air and observes a brilliant white flame, leaving a white ash. Separately, a candle is burned under a cold metal plate, collecting a black soot and a colourless liquid. Analyse both processes with balanced equations, identify the products, and state whether each combustion is complete or incomplete.
一位学生在空气中燃烧一段镁带，看到耀眼白光并留下白色灰烬。另将蜡烛在冷金属板下燃烧，收集到黑色烟灰和无色液体。分析两个过程，写出配平方程式，确认产物并判断燃烧是否充分。
Analysis
s is a complete combustion, as magnesium oxide is the only product. Magnesium is oxidised, and the high temperature produces bright light. The product is basic in nature, turning moist red litmus blue.
镁燃烧发出耀眼白光。白色灰烬是氧化镁。镁与空气中氧气反应：2Mg(s) + O₂(g) → 2MgO(s)。这是充分燃烧，唯一的产物为氧化镁。镁被氧化，高温产生强光。产物呈碱性，可使湿红色石蕊试纸变蓝。
ack soot (carbon) and liquid water observed on the cold plate indicate incomplete combustion. The equation for complete combustion would be: 2C₂₀H₄₂(s) + 61O₂(g) → 40CO₂(g) + 42H₂O(l). In reality, limited oxygen leads to some carbon monoxide and carbon particulates. The presence of soot is evidence of incomplete combustion. The liquid condensed is water, demonstrating that the wax contains hydrogen.
蜡烛主要成分为碳氢化合物蜡，如 C₂₀H₄₂。充分燃烧时仅产生二氧化碳和水。但冷板上收集到的黑色烟灰（碳粒）和液态水表明发生了不充分燃烧。充分燃烧方程式为：2C₂₀H₄₂(s) + 61O₂(g) → 40CO₂(g) + 42H₂O(l)。实际供氧不足会生成一氧化碳和碳微粒。烟灰即为不充分燃烧的证据。冷凝的无色液体是水，说明蜡含氢。
7. Concentration Calculations | 浓度计算
A student dissolves 11.7 g of sodium chloride (NaCl) in enough distilled water to make 250 cm³ of solution. The molar mass of NaCl is 58.5 g/mol. (a) Calculate the concentration of the solution in g/dm³. (b) Calculate the amount of NaCl in moles. (c) Hence determine the concentration in mol/dm³.
一位学生将 11.7 g 氯化钠 (NaCl) 溶于足量蒸馏水中，制成 250 cm³ 溶液。NaCl 的摩尔质量为 58.5 g/mol。(a) 计算该溶液的质量浓度 (g/dm³)。(b) 计算 NaCl 的物质的量 (mol)。(c) 由此求算摩尔浓度 (mol/dm³)。
Analysis
f solution (dm³). 250 cm³ = 0.250 dm³. So concentration = 11.7 g / 0.250 dm³ = 46.8 g/dm³.
(a) 质量浓度 = 溶质质量 (g) / 溶液体积 (dm³)。250 cm³ = 0.250 dm³。因此质量浓度 = 11.7 g / 0.250 dm³ = 46.8 g/dm³。
(b) Number of moles = mass / molar mass = 11.7 g / 58.5 g/mol = 0.200 mol.
(b) 物质的量 = 质量 / 摩尔质量 = 11.7 g / 58.5 g/mol = 0.200 mol。
/dm³. Many examination questions require the formula: c = n / V, so keep units consistent.
(c) 摩尔浓度 = 物质的量 / 体积 (dm³) = 0.200 mol / 0.250 dm³ = 0.800 mol/dm³。很多试题要求使用公式 c = n / V，注意单位统一。
c = m ÷ V and n = m ÷ M
8. Balancing Equations and Conservation of Mass | 配平方程式与质量守恒
this equation and show how the total mass of reactants equals the total mass of products, using relative atomic masses: C=12, H=1, O=16.
考虑甲烷 (CH₄)
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