📚 Year 9 CIE Chemistry: Case Study Practice | Year 9 CIE 化学：案例分析实战演练
This article presents a series of case studies designed to build analytical and problem-solving skills for Year 9 CIE Chemistry. Each scenario is rooted in core syllabus topics such as atomic structure, bonding, chemical reactions, stoichiometry, and practical techniques. Working through these real-world style problems will help you deepen conceptual understanding and boost exam confidence.
本文提供一系列精心设计的案例分析，旨在培养 Year 9 CIE 化学的分析与解题能力。每个案例紧扣原子结构、化学键、化学反应、化学计量和实验技术等核心大纲内容。通过演练这些贴近实际的问题，你将加深概念理解，并提升应试信心。
1. Identifying an Unknown White Solid | 识别未知白色固体
A student finds an unlabelled white powder. She tests it:
一名学生发现一瓶未贴标签的白色粉末。她进行了如下测试：
It dissolves in water to give a colourless solution.
粉末溶于水，得到无色溶液。
Adding dilute hydrochloric acid produces vigorous effervescence; the gas turns limewater milky.
加入稀盐酸后剧烈冒泡；产生的气体使石灰水变浑浊。
A flame test on the solid gives a bright yellow flame.
对固体进行焰色试验，火焰呈亮黄色。
Identify the white solid and explain your reasoning.
请鉴定该白色固体，并解释推理过程。
The milky limewater proves the gas is carbon dioxide, so the solid is a carbonate. The yellow flame test reveals the presence of sodium ions. Hence the unknown solid is sodium carbonate, Na₂CO₃.
石灰水变浑浊证明气体是二氧化碳，因此该固体是碳酸盐。黄色焰色表明存在钠离子。所以未知固体是碳酸钠 Na₂CO₃。
2. Electrolysis of Molten Lead(II) Bromide | 熔融溴化铅的电解
During the electrolysis of molten lead(II) bromide, PbBr₂, using inert electrodes, a silvery liquid forms at the cathode and a red-brown gas is released at the anode. Write half-equations for both electrodes and explain which ions are oxidised or reduced.
用惰性电极电解熔融溴化铅 PbBr₂ 时，阴极生成银白色液体，阳极释放出红棕色气体。写出两极的半反应方程式，并解释哪种离子被氧化、哪种被还原。
Br₂ + 2e⁻. Lead(II) ions gain electrons (reduced), bromide ions lose electrons (oxidised).
阴极（还原）：Pb²⁺ + 2e⁻ → Pb（熔融铅）。阳极（氧化）：2Br⁻ → Br₂ + 2e⁻。铅离子得到电子被还原，溴离子失去电子被氧化。
3. Calculating Gas Volumes in Reactions | 计算反应产生的气体体积
6.5 g of zinc granules are added to excess dilute hydrochloric acid at room temperature and pressure (r.t.p.), where the molar gas volume is 24 dm³ mol⁻¹. The relative atomic mass of Zn is 65. Write the balanced equation and calculate the volume of hydrogen gas produced.
室温常压下，将 6.5 g 锌粒加入过量稀盐酸中，气体摩尔体积为 24 dm³ mol⁻¹。锌的相对原子质量为 65。写出配平的化学方程式，并计算产生的氢气体积。
H₂, so moles of H₂ = 0.10 mol. Volume of H₂ = moles × 24 dm³ mol⁻¹ = 0.10 × 24 = 2.4 dm³.
方程式：Zn + 2HCl → ZnCl₂ + H₂。锌的物质的量 = 质量 / 相对原子质量 = 6.5 / 65 = 0.10 mol。根据方程式，1 mol 锌产生 1 mol 氢气，故氢气物质的量 = 0.10 mol。氢气体积 = 物质的量 × 24 dm³ mol⁻¹ = 0.10 × 24 = 2.4 dm³。
4. Reaction Rate Comparison: Metals with Acid | 金属与酸反应的速率比较
Equal-sized strips of magnesium, zinc, and iron are placed in separate test tubes of the same concentration of hydrochloric acid. Hydrogen gas is collected, and the time for each strip to dissolve completely is recorded. State the order of reaction rate from fastest to slowest and explain using the reactivity series.
将大小相同的镁条、锌条和铁条分别放入盛有相同浓度盐酸的试管中，收集氢气并记录完全溶解所需时间。按反应速率从快到慢排序，并用金属活动性顺序加以解释。
Fastest to slowest: Mg > Zn > Fe. Magnesium is the most reactive among the three because it loses electrons more easily, placing it higher in the reactivity series. Zinc reacts moderately, and iron the slowest. This trend matches their tendency to form positive ions.
从快到慢：Mg > Zn > Fe。三者中镁最活泼，失电子更容易，在活动性顺序中位置最高；锌反应适中；铁最慢。这一趋势与它们形成阳离子的倾向一致。
5. Choosing Separation Techniques for a Mixture | 为混合物选择分离技术
A mixture contains sand, salt, and water. A second mixture is a solution of two different food colourings. For each mixture, identify the most suitable separation technique and describe the principle behind it.
一种混合物含有沙子、食盐和水。另一种混合物是两种不同食用色素的溶液。针对每种混合物，确定最合适的分离技术，并描述其背后的原理。
Mixture 1 (sand, salt, water): Filtration first to separate sand (insoluble) from salt solution, followed by evaporation or crystallisation to obtain solid salt. Mixture 2 (food colourings): Chromatography. The dyes separate because they have different solubilities in the solvent and different affinities for the paper, moving at different rates.
混合物1（沙子、盐、水）：先过滤分离不溶的沙子和盐溶液，然后通过蒸发或结晶得到固态食盐。混合物2（食用色素）：色谱法。色素因在溶剂中的溶解度不同、对滤纸的吸附力不同，从而以不同速度迁移而分离。
6. Flame Tests and Precipitation Reactions for Ion Identification | 焰色试验和沉淀反应鉴别离子
An unknown aqueous solution might contain Ba²⁺, Cu²⁺, Cl⁻ or SO₄²⁻ ions. A flame test shows no characteristic colour. Adding silver nitrate and nitric acid gives a white precipitate. Adding barium chloride and dilute HCl also gives a white precipitate. Deduce which ions are present and explain each observation.
某未知水溶液可能含有 Ba²⁺、Cu²⁺、Cl⁻ 或 SO₄²⁻。焰色试验无特征焰色。加入硝酸银和稀硝酸产生白色沉淀。加入氯化钡和稀盐酸也产生白色沉淀。推断存在哪些离子，并解释每个观察结果。
No flame colour rules out Cu²⁺ (blue-green) and Ba²⁺ (apple-green). White precipitate with acidified AgNO₃ indicates Cl⁻ (AgCl). White precipitate with acidified BaCl₂ indicates SO₄²⁻ (BaSO₄). Both Cl⁻ and SO₄²⁻ are present; no barium or copper ions are detected.
无焰色排除了 Cu²⁺（蓝绿色）和 Ba²⁺（苹果绿）。加酸性硝酸银产生白色沉淀表明有 Cl⁻（生成 AgCl）。加酸性氯化钡产生白色沉淀表明有 SO₄²⁻（生成 BaSO₄）。因此溶液同时存在 Cl⁻ 和 SO₄²⁻，未检出钡或铜离子。
7. Determining the Composition of Water by Electrolysis | 通过电解确定水的组成
When acidified water is electrolysed using platinum electrodes, two gases are collected. The volume ratio is 2:1 by volume. The gas with the larger volume burns with a squeaky pop, while the other relights a glowing splint. Name the gases and explain how this experiment confirms that water is H₂O.
用铂电极电解酸化水，收集到两种气体，体积比为 2:1。体积较大的气体燃烧发出爆鸣声，另一种气体能复燃带火星的木条。命名这两种气体，并说明该实验如何证实水的化学式为 H₂O。
The larger-volume gas is hydrogen (2 volumes); the smaller is oxygen (1 ws that water molecules consist of two hydrogen atoms for every one oxygen atom, supporting the formula H₂O.
体积较大的气体是氢气（2 体积），体积较小的是氧气（1 体积）。水分解的方程式为 2H₂O(l) → 2H₂(g) + O₂(g)。实验表明水分子中氢、氧原子个数比为 2:1，从而证实水的化学式为 H₂O。
8. Explaining Properties of Ionic vs. Covalent Compounds | 解释离子化合物与共价化合物的性质差异
Study the properties of sodium chloride and carbon dioxide. One has a high melting point and conducts electricity when molten; the other is a gas at room temperature and does not conduct. Explain these differences in terms of structure and bonding.
比较氯化钠和二氧化碳的性质。一种熔点高且熔融态能导电；另一种在室温下为气体且不导电。从结构和化学键角度解释这些差异。
| Property | 性质 | NaCl (ionic | 离子) | CO₂ (covalent | 共价) |
|---|---|---|
| Melting point | 熔点 | High (strong ionic bonds in giant lattice) | Low (weak intermolecular forces) |
| Electrical conductivity when molten | 熔融态导电性 | Yes (ions free to move) | No (no mobile ions or electrons) |
In NaCl, oppositely charged ions are held by strong electrostatic forces throughout the giant lattice; melting requires much energy. Molten NaCl has mobile ions that carry charge. CO₂ consists of small molecules with weak intermolecular forces, so it sublimes easily and cannot conduct electricity as there are no free charge carriers.
氯化钠中，阴、阳离子通过强烈的静电作用形成巨型离子晶格，熔化需大量能量；熔融态离子可自由移动而导电。二氧化碳由小分子组成，分子间作用力微弱，易升华，且没有自由移动的离子或电子，故不导电。
9. Using the Reactivity Series to Predict Displacement Reactions | 利用活动性顺序预测置换反应
A student dips a zinc rod into copper(II) sulfate solution. A brown coating appears on the zinc, and the blue solution fades. When a copper rod is dipped into zinc sulfate solution, no change occurs. Explain both observations using the reactivity series.
一名学生将锌棒浸入硫酸铜溶液中，锌棒表面出现棕色覆盖层，蓝色溶液变浅。而将铜棒浸入硫酸锌溶液中则无变化。利用金属活动性顺序解释这两个现象。
he brown coating is copper metal, and the blue colour fades as Cu²⁺ ions are removed. Copper is less reactive than zinc, so it cannot displace zinc; hence no reaction occurs.
锌比铜活泼，能将铜从其盐溶液中置换出来：Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)。棕色覆盖层是铜，溶液蓝色因 Cu²⁺ 浓度降低而变浅。铜的活泼性弱于锌，无法置换锌，因此不发生反应。
10. Analysis of Combustion Products of Hydrocarbons | 烃类燃烧产物的分析
Methane, CH₄, is burned in a plentiful supply of air. The combustion products are passed through a U-tube containing anhydrous copper(II) sulfate and then through limewater. Describe the expected observations and write the balanced equation for the reaction.
甲烷 CH₄ 在充足的空气中燃烧。将燃烧产物依次通过装有无水硫酸铜的 U 形管和石灰水。描述预期现象，并写出反应的化学方程式。
on: CH₄ + 2O₂ → CO₂ + 2H₂O. Complete combustion of a hydrocarbon always yields carbon dioxide and water.
无水硫酸铜由白色变为蓝色，表明有水蒸气生成。石灰水变浑浊，证明生成二氧化碳。方程式：CH₄ + 2O₂ → CO₂ + 2H₂O。烃的完全燃烧总是生成二氧化碳和水。
11. Determining the Formula of Magnesium Oxide by Experiment | 实验测定氧化镁的化学式
(Ar: Mg = 24, O = 16) Calculate the simplest whole-number ratio of magnesium to oxygen and hence the empirical formula.
将 0.48 g 镁带在坩埚中强热至与氧气完全反应，所得氧化镁质量为 0.80 g。（相对原子质量：Mg = 24，O = 16）计算镁与氧的最简整数比，并给出实验式。
Mass of oxygen = 0.80 – .020 mol. Moles O = 0.32 / 16 = 0.020 mol. Ratio Mg:O = 1:1. The empirical formula is MgO.
氧的质量 = 0.80 – 0.48 = 0.32 g。镁的物质的量 = 0.48 / 24 = 0.020 mol。氧的物质的量 = 0.32 / 16 = 0.020 mol。镁与氧的物质的量之比为 1:1，实验式为 MgO。
12. Interpreting a Chromatogram | 解读色谱图
A chromatogram shows three spots from an unknown mixture along with spots of pure substances A, B, and C run under the same conditions. The solvent front moved 8.0 cm. Spot X from the mixture moved 3.2 cm, and spot Y moved 6.4 cm. Calculate Rf values for X and Y. If A has Rf 0.40 and B has Rf 0.80, identify which spots correspond to A and B, and state whether substance C is present (C has Rf 0.60).
某色谱图显示未知混合物的三个斑点，以及相同条件下标准物质 A、B、C 的斑点。溶剂前沿移动了 8.0 cm。混合物中的斑点 X 移动了 3.2 cm，斑点 Y 移动了 6.4 cm。计算 X 和 Y 的 Rf 值。若 A 的 Rf 为 0.40，B 的 Rf 为 0.80，请指出 X 和 Y 分别对应哪种物质，并判断 C（Rf = 0.60）是否存在。
0.80, matching substance B. No spot at Rf 0.60, so substance C is absent from the mixture.
Rf = 物质移动距离 / 溶剂前沿距离。Rf (X) = 3.2 / 8.0 = 0.40，与物质 A 匹配。Rf (Y) = 6.4 / 8.0 = 0.80，与物质 B 匹配。没有 Rf 为 0.60 的斑点，因此混合物中不含物质 C。
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