📚 Year 10 CCEA Chemistry: Case Study Practice Drills | Year 10 CCEA 化学：案例分析实战演练
Case studies help you connect textbook chemistry to the real world. In this article, we walk through a series of applied problems that mirror the CCEA Year 10 specification. Each scenario strengthens your ability to analyse data, write equations, and carry out calculations – just as you will need to in the exam.
案例分析能帮你把课本上的化学和现实世界联系起来。本文通过一系列与 CCEA Year 10 考纲对应的应用性问题，训练你分析数据、书写方程式和进行计算的能力，这些正是考试中需要掌握的技能。
1. Atomic Structure & Carbon-14 Dating | 原子结构与碳-14 定年法
A team of archaeologists discovers a wooden tool in a bog. Carbon-14 dating shows it contains only 25% of the original ¹⁴C atoms. Carbon-14 has a half-life of 5730 years. The lab also identifies tiny salt crystals on the surface, composed of Na⁺ and Cl⁻ ions.
一支考古队在沼泽中发现了一件木制工具。碳-14 定年显示它只含有当初 ¹⁴C 原子数的 25%。碳-14 的半衰期为 5730 年。实验室还在表面发现了微小的盐晶体，由 Na⁺ 和 Cl⁻ 离子组成。
-life of a radioactive isotope is the time taken for half of its nuclei to decay. Since 25% remains, two half-lives have passed: after one half-life 50% remains, after two half-lives 25% remains. Therefore, the age = 2 × 5730 = 11460 years.
放射性同位素的半衰期是其一半原子核发生衰变所用的时间。剩余 25% 说明已经过了两个半衰期：一个半衰期后剩 50%，两个半衰期后剩 25%。因此年龄 = 2 × 5730 = 11460 年。
Solid sodium chloride forms when Na⁺ and Cl⁻ ions attract each other strongly in a giant ionic lattice. The electrostatic attraction between oppositely charged ions explains its high melting point and solubility in water.
氯化钠固体中 Na⁺ 和 Cl⁻ 离子在巨大的离子晶格中强烈吸引。异号离子间的静电引力解释了它具有高熔点以及可溶于水的性质。
The archaeologist also tests a white crust on the wood using a flame test. A bright yellow flame indicates the presence of sodium ions. Adding dilute nitric acid and silver nitrate solution gives a white precipitate, confirming chloride ions.
考古人员还用焰色反应检验木头上的白色结壳。亮黄色的火焰表明存在钠离子。加入稀硝酸和硝酸银溶液产生白色沉淀，证实有氯离子。
2. Ionic vs Covalent Substances at the Scene | 现场中的离子型与共价型物质
A crime scene investigator finds two white powders: one is glucose (C₆H₁₂O₆) and the other is potassium chloride (KCl). The investigator carries out simple lab tests to identify them. Glucose dissolves easily in water but does not conduct electricity. Potassium chloride dissolves and the solution conducts electricity well.
犯罪现场调查员发现两种白色粉末：一种是葡萄糖 (C₆H₁₂O₆)，另一种是氯化钾 (KCl)。调查员通过简单的实验室测试来鉴别它们。葡萄糖易溶于水但不导电，氯化钾溶于水后溶液导电性很好。
Potassium chloride is an ionic compound consisting of K⁺ and Cl⁻ ions. When dissolved, the ions are free to move and carry charge, so the solution conducts electricity. In the solid state, ions are locked in place and cannot conduct.
氯化钾是离子化合物，由 K⁺ 和 Cl⁻ 离子组成。溶解后离子可以自由移动并携带电荷，因此溶液能导电。固态时离子被固定在晶格中，无法导电。
Glucose consists of covalent molecules held together by weak intermolecular forces. It dissolves because water molecules form hydrogen bonds with –OH groups, but no ions are produced, so the solution does not conduct.
葡萄糖由共价分子组成，分子间存在微弱的分子间作用力。它能溶解是因为水分子与 –OH 基团形成氢键，但没有离子产生，因此溶液不导电。
Melting point tests also help identify the compounds. Potassium chloride has a high melting point (about 770 °C) due to strong ionic bonds throughout the lattice. Glucose melts around 146 °C because only weak intermolecular attractions must be overcome.
熔点测试也有助鉴别。氯化钾的熔点很高（约 770 °C），因为晶格中存在强大的离子键。葡萄糖在约 146 °C 熔化，只需克服微弱的分子间作用力。
3. Calculating Fertilisers: Mass & Moles | 肥料中的化学计算：质量与摩尔
A farmer applies ammonium nitrate (NH₄NO₃) to a field. The chosen fertiliser must provide a high percentage of nitrogen by mass. The farm technician calculates the percentage of nitrogen in NH₄NO₃ to compare it with urea, CO(NH₂)₂.
一位农民给田地施用硝酸铵 (NH₄NO₃)。所选肥料必须提供高的氮质量分数。农场技术员计算 NH₄NO₃ 中的氮含量并与尿素 CO(NH₂)₂ 进行比较。
+ 62 = 80. Two nitrogen atoms contribute 2 × 14 = 28. Percentage N = (28 ÷ 80) × 100 = 35%.
相对原子质量：N = 14，H = 1，O = 16。NH₄NO₃ 的 Mᵣ = (14 + 1×4) + (14 + 16×3) = 18 + 62 = 80。两个氮原子的总质量为 2 × 14 = 28。氮的质量分数 = (28 ÷ 80) × 100 = 35%。
)×2 = 12 + 16 + 32 = 60. Two nitrogen atoms give 28. Percentage N = (28 ÷ 60) × 100 = 46.7%. Urea is a more nitrogen-rich fertiliser.
对于尿素 CO(NH₂)₂：Mᵣ = 12 + 16 + (14+2)×2 = 12+16+32 = 60。两个氮原子共 28。氮的质量分数 = (28 ÷ 60) × 100 = 46.7%。尿素是氮含量更高的肥料。
s of N required = 12 000 g. Mass of NH₄NO₃ = 12 000 ÷ 0.35 = 3.43 × 10⁴ g (about 34.3 kg).
农民每公顷需要施用 1.2×10⁴ g 氮。需要硝酸铵的质量 = 12000 ÷ 0.35 = 3.43×10⁴ g（约 34.3 公斤）。
4. Neutralising Stomach Acid | 胃酸的中和
A patient takes an antacid tablet containing magnesium hydroxide, Mg(OH)₂, to relieve indigestion. The stomach produces hydrochloric acid, HCl(aq). Each tablet contains 0.580 g of Mg(OH)₂. The chemist checks whether one tablet can neutralise 1.0 dm³ of 0.020 mol/dm³ HCl.
一位病人服用含有氢氧化镁 Mg(OH)₂ 的抗酸药片来缓解消化不良。胃分泌盐酸 HCl(aq)。每片含 0.580 g Mg(OH)₂。化学家验证一片药能否中和 1.0 dm³ 0.020 mol/dm³ 的 HCl。
₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l). Mᵣ of Mg(OH)₂ = 24.3 + (16+1)×2 = 24.3 + 34 = 58.3.
中和反应：Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l)。Mg(OH)₂ 的 Mᵣ = 24.3 + (16+1)×2 = 58.3。
that can be neutralised = 2 × 9.95×10⁻³ = 0.0199 mol (since 1 Mg(OH)₂ reacts with 2 HCl).
一片药中 Mg(OH)₂ 的摩尔数 = 0.580 ÷ 58.3 = 9.95×10⁻³ mol。可中和的 HCl 摩尔数 = 2 × 9.95×10⁻³ = 0.0199 mol（因为 1 个 Mg(OH)₂ 与 2 个 HCl 反应）。
s 0.0199 mol of neutralisation capacity, so it almost exactly neutralises the acid. A slight excess of acid remains, but the tablet is effective enough.
1.0 dm³ 0.020 mol/dm³ 酸中 HCl 的摩尔数 = 0.020 mol。药片提供的中和能力为 0.0199 mol，几乎能完全中和，仅剩极少量酸，效果足够。
5. Electrolysis of Brine – Industry in Action | 盐水的电解——工业生产实例
A chemical plant produces chlorine, hydrogen and sodium hydroxide by electrolysing concentrated sodium chloride solution (brine). The products are used to make PVC, bleach and soap. The technician monitors the gas collected at each electrode.
化工厂通过电解浓氯化钠溶液（盐水）生产氯气、氢气和氢氧化钠。这些产品用于制造 PVC、漂白剂和肥皂。技术人员监测每个电极收集到的气体。
ecules are reduced: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq). Colourless hydrogen gas is produced.
在钛阳极上，氯离子被氧化：2Cl⁻(aq) → Cl₂(g) + 2e⁻。收集到浅绿色的氯气。在钢阴极上，水分子被还原：2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq)，产生无色的氢气。
dium hydroxide. The overall reaction is: 2NaCl(aq) + 2H₂O(l) → H₂(g) + Cl₂(g) + 2NaOH(aq).
钠离子留在溶液中，与氢氧根离子结合形成氢氧化钠。总反应为：2NaCl(aq) + 2H₂O(l) → H₂(g) + Cl₂(g) + 2NaOH(aq)。
After electrolysis, the remaining solution turns universal indicator purple, confirming the presence of a strong alkali (NaOH). The plant pipes chlorine to a bleach plant, where it reacts with NaOH to make sodium hypochlorite (NaClO).
电解后剩余溶液使通用指示剂变紫，证实存在强碱氢氧化钠。工厂将氯气输送至漂白剂车间，与 NaOH 反应生成次氯酸钠 (NaClO)。
6. Rate of Reaction – Catalytic Converters | 反应速率——催化转化器
Car exhaust gases contain toxic carbon monoxide and nitrogen oxides. A catalytic converter uses a platinum-rhodium honeycomb to speed up the conversion to harmless nitrogen and carbon dioxide. Without a catalyst, the activation energy is too high.
汽车尾气中含有有毒的一氧化碳和氮氧化物。催化转化器利用铂铑蜂窝状载体加速转化为无害的氮气和二氧化碳。没有催化剂时活化能太高。
g) + N₂(g). Platinum provides a surface where reactant molecules adsorb, weakening their bonds. This lowers the activation energy, allowing many more collisions to be successful at the exhaust temperature.
反应为：2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)。铂提供表面使反应物分子吸附并削弱化学键，从而降低活化能，让更多的碰撞在尾气温度下就能成功发生。
A mechanic tests the converter by measuring the CO concentration before and after. Before: 4.2% CO; after: 0.08% CO. The efficiency is over 98%. However, leaded petrol poisons the catalyst, blocking its active sites and reducing the rate significantly.
机械师通过测量前后的一氧化碳浓度来检测转化器：前 4.2%，后 0.08%，效率超过 98%。但含铅汽油会使催化剂中毒，堵住活性位点，使速率大幅下降。
Plotting volume of CO₂ produced against time after adding a fresh catalyst shows a steep initial slope, which levels off as reactants are used up. This demonstrates the rate changing over time.
加入新催化剂后，以生成的二氧化碳体积对时间作图，初始斜率很陡，随着反应物的消耗曲线趋于平缓，这说明了速率随时间的变化。
7. Exothermic and Endothermic Reactions – Hot & Cold Packs | 放热与吸热反应——冷热敷袋
A physiotherapist uses an instant cold pack containing ammonium nitrate and a water pouch. When squeezed, the ammonium nitrate dissolves, and the pack becomes very cold. A different hot pack uses iron powder that oxidises rapidly when exposed to air.
理疗师使用内含硝酸铵和水的冷敷袋。挤压后硝酸铵溶解，敷袋迅速变冷。另一种热敷袋使用铁粉，接触空气后快速氧化放热。
rature drops. The hot pack reaction: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) is exothermic, releasing heat.
NH₄NO₃(s) + 水 → NH₄⁺(aq) + NO₃⁻(aq) 是吸热过程。破坏离子晶格需要从周围吸收能量，因此温度下降。热敷袋反应：4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) 为放热反应，释放热量。
to 11.5 °C. Using Q = mcΔT, the heat absorbed is approximately 50 × 4.18 × 8.5 ≈ 1.78 × 10³ J.
在实验室里，学生将 8.0 g NH₄NO₃ 加入 50 cm³ 水中测量温度变化，从 20.0 °C 降到 11.5 °C。利用 Q = mcΔT，吸收的热量约为 50 × 4.18 × 8.5 ≈ 1.78 × 10³ J。
.10 mol = +17.8 kJ/mol (endothermic). The sign convention is important: + for endothermic, – for exothermic.
使用的 NH₄NO₃ 摩尔数：8.0 ÷ 80 = 0.10 mol。摩尔焓变 = +1.78×10³ J ÷ 0.10 mol = +17.8 kJ/mol（吸热）。正确使用正负号很重要：吸热为正，放热为负。
8. Fractional Distillation & Cracking of Crude Oil | 原油的分馏与裂化
An oil refinery processes crude oil into useful fractions. A distillation column separates hydrocarbons based on boiling points. Small molecules like propane (C₃H₈) leave at the top; long-chain alkanes such as C₁₆H₃₄ condense lower down.
炼油厂将原油加工成有用的馏分。分馏塔根据沸点分离烃类。小分子如丙烷 (C₃H₈) 在塔顶离开，长链烷烃如 C₁₆H₃₄ 在较低处冷凝。
ide catalyst: C₁₆H₃₄ → C₈H₁₈ + 2C₂H₄ + C₄H₈. This produces shorter alkanes (petrol) and alkenes (ethene, butene).
长链烷烃因黏度高、不易燃而用途较少。炼油厂在高温下用氧化铝催化剂裂化 C₁₆H₃₄：C₁₆H₃₄ → C₈H₁₈ + 2C₂H₄ + C₄H₈。生成较短的烷烃（汽油）和烯烃（乙烯、丁烯）。
Alkenes are tested by adding bromine water. Orange colour changes to colourless, indicating an addition reaction has occurred. Ethene can be polymerised to make poly(ethene) – a plastic used for carrier bags.
用溴水检验烯烃，橙色变为无色，证明发生了加成反应。乙烯可以聚合制成聚乙烯——用于制造购物袋的塑料。
A refinery process engineer calculates that cracking 1 tonne of long-chain alkane yields 0.72 tonnes of petrol-range alkanes and 0.26 tonnes of alkenes. The remaining mass is gas and loss. This illustrates atom economy, a key green chemistry principle.
工艺工程师计算，裂化 1 吨长链烷烃可得到 0.72 吨汽油范围内的烷烃和 0.26 吨烯烃，其余为气体和损耗。这体现了原子经济性这一绿色化学重要原则。
9. Acid Rain – Causes and Chemical Remedies | 酸雨——成因与化学治理
Power stations burning coal release sulfur dioxide (SO₂) and nitrogen oxides (NO and NO₂). These gases dissolve in rainwater, forming sulfurous acid (H₂SO₃), sulfuric acid (H₂SO₄), and nitric acid (HNO₃). Acid rain damages limestone buildings and freshwater ecosystems.
燃煤发电厂排放二氧化硫 (SO₂) 和氮氧化物 (NO、NO₂)。这些气体溶于雨水，生成亚硫酸 (H₂SO₃)、硫酸 (H₂SO₄) 和硝酸 (HNO₃)。酸雨会侵蚀石灰岩建筑和淡水生态系统。
aSO₄(aq) + H₂O(l) + CO₂(g). Carbonate rocks fizz and weather away. Lakes become too acidic for fish; limestone powder (lime) is sometimes added to neutralise them.
石灰石 (CaCO₃) 与酸雨的反应：CaCO₃(s) + H₂SO₄(aq) → CaSO₄(aq) + H₂O(l) + CO₂(g)。碳酸盐岩石起泡并被侵蚀。湖泊酸化到鱼无法生存时，有时投放石灰石粉末（石灰）进行中和。
exhaust: CaCO₃ + SO₂ → CaSO₃ + CO₂, then oxidised to CaSO₄ (gypsum). This removes up to 95% of SO₂.
电厂安装烟气脱硫装置，向废气中喷淋石灰石浆液：CaCO₃ + SO₂ → CaSO₃ + CO₂，再氧化为 CaSO₄（石膏）。可去除高达 95% 的 SO₂。
₂ + 6H₂O. This end-of-pipe technology reduces the nitrogen oxides that contribute to acid deposition.
汽车尾气中通过氨选择性催化还原 NOₓ：4NO + 4NH₃ + O₂ → 4N₂ + 6H₂O。这种末端处理技术可减少形成酸沉降的氮氧化物。
10. Limestone Cycle – From Quarry to Cement | 石灰石循环——从采石场到水泥
bon dioxide: CaCO₃(s) → CaO(s) + CO₂(g). The process requires high temperatures around 900–1000 °C.
石灰石 (CaCO₃) 被开采并在窑中加热生产生石灰 (CaO)。这个热分解释放二氧化碳：CaCO₃(s) → CaO(s) + CO₂(g)。过程需要 900–1000 °C 的高温。
laked lime. This is used to neutralise acidic soils and in mortar. Slaked lime is slightly soluble in water, forming limewater [Ca(OH)₂(aq)] which turns milky when CO₂ is bubbled through.
生石灰与水剧烈反应称为熟化：CaO(s) + H₂O(l) → Ca(OH)₂(s)，生成熟石灰。用于中和酸性土壤和制作砂浆。熟石灰微溶于水，形成石灰水 [Ca(OH)₂(aq)]，通入二氧化碳变浑浊。
Mortar is a mixture of slaked lime, sand and water. It sets by absorbing CO₂ from the air, forming calcium carbonate again. Cement is produced by heating limestone with clay; when mixed with sand, gravel and water it makes concrete.
砂浆是熟石灰、沙子和水的混合物。它吸收空气中的二氧化碳而凝固，重新生成碳酸钙。水泥则由石灰石与黏土共热制得，与沙、石子和水混合制成混凝土。
× (56/100) = 1.12 tonnes. Only 1.0 tonne is actually collected due to losses, giving a yield of 89%.
采石场经营者想计算 2.0 吨石灰石可制得多少生石灰。CaCO₃ → CaO + CO₂。Mᵣ CaCO₃ = 100，CaO = 56。CaO 质量 = 2.0 t × (56/100) = 1.12 吨。实际只收集到 1.0 吨，损失导致产率为 89%。
11. Blast Furnace Iron Extraction | 高炉炼铁
Iron is extracted from haematite (Fe₂O₃) in a blast furnace. Coke (carbon), limestone and iron ore are added at the top. Hot air is blasted in near the bottom. A series of reactions produce molten iron and slag.
铁从赤铁矿 (Fe₂O₃) 中通过高炉冶炼提取。焦炭（碳）、石灰石和铁矿石从顶部加入，热空气从底部吹入。一系列反应生成铁水和炉渣。
(s) → 2CO(g). Carbon monoxide is the reducing agent that removes oxygen from the iron ore.
焦炭先燃烧生成 CO₂：C(s) + O₂(g) → CO₂(g)。然后 CO₂ 与更多焦炭反应生成一氧化碳：CO₂(g) + C(s) → 2CO(g)。一氧化碳是还原剂，将铁矿石中的氧脱去。
ron sinks to the bottom and is tapped off. Limestone decomposes to CaO, which reacts with silica (SiO₂) impurity to form calcium silicate (CaSiO₃) slag.
氧化铁的还原：Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g)。铁水下沉并在底部流出。石灰石分解为 CaO，与杂质二氧化硅 (SiO₂) 反应生成硅酸钙 (CaSiO₃) 炉渣。
The slag is less dense than iron, floats on top, and prevents the iron from re-oxidising. It is tapped separately and used for road building. Modern blast furnaces recycle hot gases to pre-heat incoming air, saving energy.
炉渣密度比铁小，浮在表面，防止铁重新氧化。炉渣单独排出，用于筑路。现代高炉回收热废气来预热进风，从而节约能源。
A plant produces 500 tonnes of iron per day. The ore contains 80% 0 × (160/112) = 714 tonnes. Total ore = 714 ÷ 0.80 = 893 tonnes (allowing for impurities).
一家工厂每天生产 500 吨铁。矿石含 80% Fe₂O₃。计算所需矿石质量：所需 Fe₂O₃ = 500 × (160/112) = 714 吨。总矿石 = 714 ÷ 0.80 = 893 吨（考虑杂质）。
12. Chemical Analysis – Testing for Ions in an Unknown Salt | 化学分析——检验未知盐中的离子
A laboratory technician receives an unlabelled white crystalline solid. A series of tests must identify the cation and anion. The solid is dissolved in distilled water for analysis.
实验室技术员收到一瓶无标签的白色晶体固体，需要通过一系列测试来鉴定其中的阳离子和阴离子。固体溶于蒸馏水后进行分析。
Flame test: a clean platinum wire dipped in the solution gives a lilac flame, indicating the presence of potassium ions (K⁺). Sodium ions would give a persistent yellow, and calcium a brick-red colour.
焰色试验：干净的铂丝蘸取溶液，产生淡紫色火焰，表明存在钾离子 (K⁺)。钠离子产生持久的黄色，钙离子产生砖红色。
Addition of dilute sodium hydroxide solution to the sample produces a white precipitate that dissolves in excess NaOH. This confirms aluminium ions (Al³⁺) present, as Al(OH)₃ is amphoteric. However, the lilac flame points to potassium, so the salt may be a mixture. The technician re-dissolves the original solid and adds NaOH again – no precipitate forms, ruling out Al³⁺. The lilac flame persists, and a separate test with nitric acid and silver nitrate gives no precipitate, excluding halides.
加入稀氢氧化钠溶液，有白色沉淀生成并在过量 NaOH 中溶解，这证实铝离子 (Al³⁺) 的存在，因为氢氧化铝是两性的。但是淡紫色火焰指向钾，所以该盐可能是混合物。技术员重新溶解原固体并再次加入 NaOH，没有沉淀产生，排除了 Al³⁺。淡紫色火焰持续存在，另一个用硝酸和硝酸银的测试无沉淀产生，排除了卤离子。
Adding dilute hydrochloric acid followed by barium chloride solution produces a white precipitate, proving sulfate ions (SO₄²⁻). Therefore, the salt is potassium sulfate, K₂SO₄. The Al³⁺ test was an error due to contaminated glassware.
加入稀盐酸后加入氯化钡溶液，产生白色沉淀，证明有硫酸根离子 (SO₄²⁻)。因此该盐是硫酸钾 K₂SO₄。之前 Al³⁺ 的测试是因玻璃器皿污染造成的错误。
hite barium sulfate is insoluble and filtered, dried and weighed to confirm stoichiometry.
为确认，技术员用硝酸钡进行沉淀反应：Ba(NO₃)₂(aq) + K₂SO₄(aq) → BaSO₄(s) + 2KNO₃(aq)。白色的硫酸钡不溶，经过滤、干燥并称重以验证化学计量。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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