📚 KS3 CCEA Chemistry: Case Study Practical Exercises | KS3 CCEA 化学：案例分析实战演练
This article presents a series of practical case studies designed to reinforce key concepts in KS3 CCEA Chemistry. Each study focuses on real-world applications of chemical reactions, separation techniques, and data analysis, helping students develop essential investigative skills. By working through these examples, you will learn how to design experiments, interpret results, and draw evidence-based conclusions – all critical for success in chemistry.
本文呈现了一系列旨在巩固 KS3 CCEA 化学核心概念的实践案例分析。每个案例聚焦于化学反应、分离技术和数据分析的实际应用，帮助学生培养基本的探究能力。通过完成这些示例，你将学会如何设计实验、解读结果并得出基于证据的结论——这些都是化学学习成功的关键。
1. Acid Rain and Building Materials | 酸雨与建筑材料
Acid rain forms when sulfur dioxide (SO₂) and nitrogen oxides (NOₓ) from fossil fuel combustion dissolve in rainwater, producing weak sulfuric and nitric acids. Marble, a metamorphic rock composed primarily of calcium carbonate (CaCO₃), is especially vulnerable to acid rain. In this case study, we simulate acid rain impact by monitoring mass loss of a marble chip placed in dilute sulfuric acid over several days.
酸雨是由于化石燃料燃烧产生的二氧化硫 (SO₂) 和氮氧化物 (NOₓ) 溶解在雨水中，生成稀硫酸和稀硝酸而形成的。大理石是一种主要由碳酸钙 (CaCO₃) 组成的变质岩，特别容易受到酸雨的侵蚀。在本案例分析中，我们通过将一块大理石碎片放入稀硫酸中，连续数天监测其质量损失，来模拟酸雨的影响。
The reaction is:CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂. A student recorded the following data:
反应方程式为：CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂。一名学生记录的数据如下：
| Day | Mass of marble (g) |
|---|---|
| 0 | 10.00 |
| 3 | 9.85 |
| 6 | 9.70 |
| 9 | 9.55 |
The consistent mass loss indicates that calcium carbonate is reacting with the acid, producing soluble calcium sulfate and releasing carbon dioxide gas. This explains why historic marble statues and buildings erode faster in polluted environments.
质量持续下降表明碳酸钙在与酸反应，生成可溶性硫酸钙并释放二氧化碳气体。这解释了为什么受污染环境中的历史大理石雕像和建筑物会更快被侵蚀。
To control the investigation, an identical marble chip was placed in pure water. No mass change was observed, confirming that the acidity is the cause of erosion. The rate of mass loss can be calculated as: average mass loss per day = (initial mass – final mass) / total days. Here, (10.00 – 9.55) / 9 = 0.05 g/day.
为控制变量，将一块相同的大理石碎片放入纯水中，未观察到质量变化，证实酸性是侵蚀的原因。质量损失速率可计算为：平均每日质量损失 = (初始质量 – 最终质量) / 总天数。此处为 (10.00 – 9.55) / 9 = 0.05 克/天。
2. Comparing Corrosion of Metals | 比较金属的腐蚀
Corrosion is a natural process that deteriorates metals, often through reaction with oxygen and water. Rusting, the corrosion of iron, is a common and costly problem. This case study examines how different metals – iron, copper, and zinc – behave under humid conditions, linking findings to their reactivity.
腐蚀是金属因与环境物质（如氧气和水）反应而发生劣化的自然过程。铁的生锈是常见且代价高昂的腐蚀问题。本案例分析研究不同金属——铁、铜和锌——在潮湿条件下的行为表现，并将其与金属活泼性联系起来。
Three clean metal nails (one iron, one copper, one zinc) were each wrapped in moist cotton wool and left in open test tubes for two weeks. Observations were recorded:
- Iron nail: reddish-brown flaky rust appeared; mass increased slightly.
- Copper nail: surface turned from shiny red-brown to dark brown/black, minimal mass change.
- Zinc nail: dull grey coating formed; no significant rust-like flaking.
三枚洁净的金属钉（铁、铜、锌各一枚）分别包裹在潮湿的棉花中，置于敞口试管中并放置两周。观察记录如下：
- 铁钉：出现红棕色片状铁锈，质量略有增加。
- 铜钉：表面从光亮红棕色变为暗棕色/黑色，质量变化极小。
- 锌钉：形成暗灰色涂层，未出现明显的片状锈蚀。
Rusting of iron requires both oxygen and water. The increase in mass for iron is because rust is hydrated iron(III) oxide (Fe₂O₃·xH₂O), which incorporates water molecules. Copper reacts slowly with oxygen and moisture to form a protective layer of copper oxide and eventually green patina, but this process is much slower than rusting. Zinc corrodes slightly, forming a layer of zinc oxide that adheres tightly and protects the underlying metal – a process called sacrificial protection if used to coat iron.
铁生锈需要氧气和水同时存在。铁的质量增加是因为铁锈是水合氧化铁 (Fe₂O₃·xH₂O)，其中结合了水分子。铜与氧气和水分缓慢反应，形成具有保护作用的氧化铜层，并最终变成绿色铜绿，但这一过程远慢于生锈。锌轻微腐蚀，形成紧密附着的氧化锌层，保护内部金属——若用于镀铁，便称为牺牲保护。
Students can extend the investigation by placing zinc wrapped around an iron nail (galvanising) to observe that the iron does not rust. This demonstrates the reactivity series: zinc > iron > copper. The more reactive zinc corrodes preferentially.
学生可以通过将锌片缠绕在铁钉上（模拟镀锌）来扩展研究，观察到铁不会生锈。这体现了金属活泼性顺序：锌 > 铁 > 铜。更活泼的锌优先被腐蚀。
3. Temperature and Reaction Rate – Magnesium with Acid | 温度与反应速率——镁与酸
The rate of a chemical reaction can be affected by several factors, including temperature. When magnesium ribbon reacts with dilute hydrochloric acid, hydrogen gas is produced. This case study uses the volume of gas collected over time at three different temperatures to explore how temperature changes reaction speed.
化学反应速率可受多种因素影响，包括温度。当镁条与稀盐酸反应时，产生氢气。本案例分析通过记录在三个不同温度下随时间收集的气体体积，探究温度如何改变反应速率。
Reaction: Mg + 2HCl → MgCl₂ + H₂. A student measured the volume of hydrogen gas produced every 20 seconds using a gas syringe, keeping the magnesium length, acid volume, and acid concentration constant. The water bath was set at 20 °C, 30 °C, and 40 °C.
反应方程式：Mg + 2HCl → MgCl₂ + H₂。一名学生使用气体注射器，每 20 秒测量一次产生的氢气体积，并保持镁条长度、酸体积和酸浓度不变。水浴温度分别设置为 20 °C、30 °C 和 40 °C。
Results table (excerpt at 60 seconds):
| Temperature (°C) | Volume of H₂ at 60 s (cm³) |
|---|---|
| 20 | 15 |
| 30 | 28 |
| 40 | 45 |
The data clearly show that as temperature increases, the volume of gas produced in the same time is larger, meaning the reaction is faster. This is because particles have more kinetic energy, move faster, and collide more frequently and with greater energy, exceeding the activation energy more often.
数据清晰地表明，随着温度升高，相同时间内产生的气体体积更大，意味着反应更快。这是因为粒子具有更多的动能，运动更快，碰撞更频繁且能量更高，从而更经常地超过活化能。
The initial rate can be estimated by the slope of the volume-time graph in the first 20 seconds. A student calculated the rate at 30 °C as approximately 1.4 cm³/s. At higher temperatures, the curve is steeper initially. This investigation demonstrates the particle collision theory.
初期速率可通过体积-时间图前 20 秒的斜率估算。一名学生计算出 30 °C 时的速率约为 1.4 cm³/s。温度越高，初始曲线越陡峭。本项研究展示了粒子碰撞理论。
4. Chromatography of Inks | 墨水的色谱分析
Paper chromatography is a technique used to separate mixtures of soluble substances, such as the dyes in inks. It works because different dyes have different solubilities in a solvent and adhere to the paper to varying extents. This case study analyses a black ink to determine if it is a pure substance or a mixture.
纸色谱法是一种用于分离混合物中可溶性物质（如墨水中的染料）的技术。其原理在于不同染料在溶剂中的溶解度不同，且对纸张的吸附程度也各不相同。本案例分析对一种黑色墨水进行分析，以判断它是纯净物还是混合物。
A pencil line was drawn 2 cm from the bottom of filter paper. A spot of black ink and spots of three known colour dyes (A, B, C: blue, yellow, red) were placed on the line. The paper was placed in a beaker with water as the solvent, ensuring the pencil line stayed above the water surface. After 15 minutes, the chromatogram was removed and dried.
在滤纸底部 2 厘米处用铅笔画一条线。将黑色墨水和三种已知颜色染料（A、B、C：蓝、黄、红）的点样点在此线上。将滤纸放入盛有作为溶剂的水的烧杯中，确保铅笔线位于水面上方。15 分钟后取出色谱图并干燥。
Observations: The black ink separated into two spots – one matching dye A (blue) and one matching dye C (red). Dye B (yellow) moved the highest, indicating the highest solubility. Rf values were calculated:
观察结果：黑色墨水分离成两个斑点——一个与染料 A（蓝色）相同，另一个与染料 C（红色）相同。染料 B（黄色）移动距离最大，表明其溶解度最高。计算 Rf 值如下：
Rf = distance moved by substance / distance moved by solvent front
Rf = 物质移动的距离 / 溶剂前沿移动的距离
- Blue (A): 3.2 cm / 8.0 cm = 0.40
- Red (C): 5.6 cm / 8.0 cm = 0.70
- Black components: 0.40 and 0.70, confirming it is a mixture of blue and red dyes.
蓝色 (A)：3.2 cm / 8.0 cm = 0.40；红色 (C)：5.6 cm / 8.0 cm = 0.70；黑色成分：0.40 和 0.70，证实它是蓝色与红色染料的混合物。Rf 值是无量纲的，可用于物质鉴定。
This case study shows that chromatography can reveal the composition of seemingly uniform colourings. Using a lid on the beaker prevents solvent evaporation and ensures a saturated atmosphere for consistent separation.
本案例分析表明，色谱法能揭示看似均匀的着色剂的组成成分。烧杯加盖可防止溶剂蒸发，确保用于分离的饱和气氛保持一致。
5. Testing Water Purity – Evaporation and Distillation | 检测水纯度——蒸发与蒸馏
Pure water is a compound containing only H₂O molecules, boiling at exactly 100 °C at standard pressure. Natural water samples, such as seawater and tap water, contain dissolved salts and other substances. This case study compares simple evaporation with distillation to assess purity and recover pure water.
纯水是仅含 H₂O 分子的化合物，在标准压力下沸点恰好为 100 °C。天然水样（如海水和自来水）含有溶解的盐和其他物质。本案例分析比较简单蒸发与蒸馏，以评估纯度并回收纯水。
Sample A (tap water) and Sample B (seawater) were each divided into two portions. One portion of each was evaporated to dryness in an evaporating dish on a water bath; the other portion was distilled using a Liebig condenser.
将样品 A（自来水）和样品 B（海水）各分为两份。每份中的一部分在水浴上的蒸发皿中蒸干；另一部分使用李比希冷凝器进行蒸馏。
Results: Evaporation of seawater left a white solid residue (mainly sodium chloride), while tap water left a very faint trace. Distillation of seawater produced a colourless liquid that boiled at 100 °C and left no residue on evaporation – pure water. The boiling point of the original seawater was higher than 100 °C due to dissolved salts.
结果：蒸干海水后留下白色固体残留物（主要为氯化钠），而自来水残留极微量的痕迹。蒸馏海水产生了一种无色液体，沸点为 100 °C，蒸发后无残留——即纯水。原始海水由于含有溶解盐，其沸点高于 100 °C。
This demonstrates that evaporation can reveal the presence of dissolved solids, but it cannot separate the water from them; distillation, however, effectively separates the solvent from the solute by boiling and condensation. The distillate is pure water. Seawater desalination plants use distillation (or reverse osmosis) to obtain fresh water.
这表明蒸干能揭示溶解固体的存在，但不能将水与其分离；而蒸馏则通过沸腾和冷凝，有效地将溶剂与溶质分离。馏出物是纯水。海水淡化工厂使用蒸馏（或反渗透）来获取淡水。
6. Carbon Dioxide Production and Properties | 二氧化碳的制备与性质
Carbon dioxide (CO₂) is a colourless, odourless gas that plays a crucial role in photosynthesis and respiration. In the laboratory, it is commonly prepared by reacting a carbonate with an acid. This case study assesses the collection method and classic limewater test.
二氧化碳 (CO₂) 是一种无色无味的气体，在光合作用和呼吸作用中扮演关键角色。在实验室中，通常通过碳酸盐与酸反应来制备。本案例分析评估了收集方法及经典的石灰水检验。
A student added dilute hydrochloric acid to marble chips (calcium carbonate) in a conical flask. The gas produced was collected via downward delivery (upward displacement of air) because CO₂ is denser than air. Observations of the reaction include bubbling and the disappearance of marble chips.
学生将稀盐酸加入盛有大理石碎片（碳酸钙）的锥形瓶中。由于 CO₂ 密度大于空气，生成的气体通过向下排空气法（向上排空气收集）进行收集。反应现象包括冒泡和大理石碎片逐渐消失。
Equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
The gas was tested by bubbling it through limewater (calcium hydroxide solution), which turned milky. A second test showed that placing a burning splint in the gas caused it to extinguish, confirming it does not support combustion. However, the definitive test is the limewater test, as many gases can extinguish a flame.
将气体通入石灰水（氢氧化钙溶液）中，石灰水变浑浊。第二个检验是将燃着的木条伸入气体中，火焰熄灭，证实该气体不支持燃烧。但确定性的检验仍是石灰水试验，因为许多气体都能使火焰熄灭。
Limewater reaction: CO₂ + Ca(OH)₂ → CaCO₃ (white precipitate) + H₂O. If excess CO₂ is bubbled, the precipitate may dissolve because soluble calcium hydrogencarbonate forms. This highlights the importance of controlled testing.
石灰水反应：CO₂ + Ca(OH)₂ → CaCO₃ (白色沉淀) + H₂O。若通入过量 CO₂，沉淀可能会溶解，因为生成了可溶的碳酸氢钙。这突显了控制性试验的重要性。
7. Burning Fuels – Products of Combustion | 燃料的燃烧——燃烧产物
Combustion is a rapid reaction between a substance and oxygen, releasing heat and light. Fossil fuels like methane and candle wax are hydrocarbons. Complete combustion produces carbon dioxide and water. This case study uses candle burning to detect the products experimentally.
燃烧是物质与氧气之间发生的快速反应，释放出热和光。化石燃料如甲烷和蜡烛蜡是碳氢化合物。完全燃烧生成二氧化碳和水。本案例分析利用蜡烛燃烧实验来检测这些产物。
A candle was placed under a funnel connected to a vacuum pump. A cold, dry test tube was held in the stream of gases to condense water vapour. Separately, a hand pump drew the products through limewater.
将蜡烛置于连接抽气泵的漏斗下方。在气流中放置一个冷而干的试管，使水蒸气冷凝。另外，使用手动泵将燃烧产物抽入石灰水中。
Observations: The cold test tube visibly misted, and when wiped, colourless liquid was collected that turned anhydrous cobalt chloride paper from blue to pink, confirming water. The limewater turned milky, confirming carbon dioxide.
观察结果：冷试管明显起雾，擦拭后收集到的无色液体使无水氯化钴试纸由蓝变粉，证实是水。石灰水变浑浊，证实了二氧化碳。
These results show that candle wax (a hydrocarbon mixture) contains carbon and hydrogen. Equation for methane combustion: CH₄ + 2O₂ → CO₂ + 2H₂O. Incomplete combustion would produce soot (carbon) and carbon monoxide, a toxic gas. The case study can be extended by partially covering the candle to observe black soot formation.
这些结果表明，蜡烛蜡（碳氢化合物混合物）含有碳和氢。甲烷燃烧的方程式：CH₄ + 2O₂ → CO₂ + 2H₂O。不完全燃烧会产生碳黑（烟尘）和有毒气体一氧化碳。可通过部分覆盖蜡烛观察到黑烟产生，从而扩展本案例分析。
8. Neutralisation in Everyday Life – Indigestion Tablets | 日常生活中的中和反应——消食片
Acid indigestion results from excess hydrochloric acid in the stomach. Antacid tablets contain bases such as magnesium hydroxide or calcium carbonate, which neutralise the acid. This case study evaluates the effectiveness of two different brands via pH change and titration.
胃酸过多引起胃灼热。抗酸药片中含有碱，如氢氧化镁或碳酸钙，可中和胃酸。本案例分析通过 pH 变化和滴定，评估两种不同品牌消食片的效果。
A student crushed one tablet of Brand X (containing calcium carbonate) and added it to 50 cm³ of 0.1 mol/dm³ hydrochloric acid (simulating stomach acid). After stirring, the pH increased from 1 to 6. Brand Y (containing magnesium hydroxide) was tested similarly and raised the pH to 7.5.
一名学生将一片 X 品牌药片（含碳酸钙）碾碎，加入 50 cm³ 0.1 摩尔/立方分米的盐酸（模拟胃酸）中。搅拌后，pH 从 1 升至 6。对 Y 品牌（含氢氧化镁）进行相同测试，pH 升至 7.5。
Neutralisation reaction for X: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. For Y: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O. Both reactions produce water and a salt, but carbonate-based tablets can cause burping due to CO₂ production.
X 的中和反应：CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂。Y 的反应：Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O。两种反应都生成水和盐，但碳酸盐类药片由于产生 CO₂ 可能会引起打嗝。
Using an indicator, the amount of acid neutralised per tablet was calculated. Brand X neutralised 0.045 mol HCl per tablet, while Brand Y neutralised 0.050 mol HCl. Results are aligned with the mass of active ingredient. This links to stoichiometry and reinforces the concept that bases neutralise acids to form salts and water.
通过使用指示剂，计算出每片药片所中和的酸的量。X 品牌每片中和 0.045 摩尔 HCl，而 Y 品牌为 0.050 摩尔 HCl。结果与活性成分的质量一致。这关联到化学计量学，并强化了碱中和酸生成盐和水这一概念。
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