📚 Case Study Practice Drills | 案例分析实战演练
Case studies bridge the gap between textbook theory and the messy, exciting world of real chemistry. In this revision workout, you will step into the shoes of a forensic scientist, an environmental chemist, a process engineer and more. Each scenario challenges you to apply Year 9 OCR knowledge to solve a practical puzzle. Read the English description, then the corresponding Chinese explanation, and test yourself before scrolling on.
案例分析在课本理论与真实化学世界之间架起桥梁。在这份复习训练中，你会化身法证科学家、环境化学家、工艺工程师等角色。每个场景都要求你运用九年级 OCR 课程知识解决实际问题。先阅读英文描述，再阅读对应的中文解析，在继续往下翻之前考考自己。
1. Identifying an Unknown Metal | 识别未知金属
A lab technician discovers three unlabelled silvery metal blocks. She places a small piece of each into separate test tubes of dilute hydrochloric acid. Metal A fizzes violently and the test tube becomes hot. Metal B produces a steady stream of bubbles. Metal C shows only a slow, sparse fizzing. She also holds a lit splint at the mouth of each tube and hears a 'squeaky pop', confirming the gas is hydrogen.
一位实验员发现三块未标识的银白色金属块。她分别取一小块放入装有稀盐酸的三支试管中。金属 A 剧烈冒泡，试管发热。金属 B 持续、稳定地产生气泡。金属 C 只缓慢、零星冒泡。她还用点燃的木条靠近每支试管口，听见“噗”的一声，证明气体是氢气。
Based on the reactivity series (potassium, sodium, calcium, magnesium, aluminium, zinc, iron, copper), can you deduce which metal is magnesium, which is zinc and which is likely to be iron? Explain why the reaction rates differ.
根据金属活动性顺序（钾、钠、钙、镁、铝、锌、铁、铜），你能推断出哪一块是镁，哪一块是锌，哪一块可能是铁吗？解释为什么反应速率不同。
Magnesium sits higher in the reactivity series than zinc and iron, so it reacts most vigorously with acids, producing heat and rapid bubbling. Zinc reacts moderately, and iron—being the least reactive of the three—fizzes only slowly. Therefore, metal A is magnesium, metal B is zinc, and metal C is iron. The difference in reactivity arises from how easily each metal atom loses electrons to form positive ions.
镁在活动性顺序中位于锌和铁之上，因此与酸反应最剧烈，产生热量并迅速起泡。锌反应温和，而铁是三者中最不活泼的，仅缓慢冒泡。因此，金属 A 是镁，金属 B 是锌，金属 C 是铁。反应性差异源于金属原子失去电子形成阳离子的难易程度不同。
2. Acid Rain and Limestone Buildings | 酸雨与石灰石建筑
An old cathedral built from limestone (mainly calcium carbonate, CaCO₃) has weathered badly over the last century. Environmental chemists measure the local rainfall pH at 4.2, indicating acid rain containing sulfuric acid (H₂SO₄). Samples of crumbling stonework reveal a white, powdery substance and tiny crystals of gypsum (CaSO₄).
一座用石灰石（主要成分为碳酸钙，CaCO₃）建造的古老教堂在过去一个世纪中严重风化。环境化学家测得当地雨水 pH 为 4.2，表明酸雨中含有硫酸（H₂SO₄）。从剥落的石件取样后，发现一种白色粉末状物质和微小的石膏（CaSO₄）晶体。
Write the word and symbol equations for the reaction between calcium carbonate and sulfuric acid. Use the equation to explain how acid rain damages limestone structures.
写出碳酸钙与硫酸反应的文字方程式和化学方程式。利用方程式解释酸雨如何损坏石灰石建筑。
id reacts with insoluble calcium carbonate, converting it into calcium sulfate which is slightly soluble in water. Over time, repeated rainfall washes away the gypsum, eroding the stone and carving grooves into statues and pillars.
文字方程式：碳酸钙 + 硫酸 → 硫酸钙 + 水 + 二氧化碳。化学方程式：CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂。酸与不溶的碳酸钙反应，将其转化为微溶于水的硫酸钙。随着反复降雨，石膏被冲刷掉，石料受到侵蚀，在雕像和石柱上刻出沟槽。
3. Purifying Village Pond Water | 净化村庄池塘水
A remote village relies on a pond that has become cloudy with mud, twigs and an oily film. Your task is to design a low-cost purification system that produces clear, drinkable water. You have access to beakers, filter paper, charcoal, an electric heater and a distillation apparatus.
一个偏远村庄依赖的池塘水变得浑浊不堪，混杂着泥浆、树枝和一层油膜。你的任务是设计一种低成本净化系统，产出清澈可饮用的水。你可使用的器材包括烧杯、滤纸、活性炭、电热器和蒸馏装置。
Outline a step-by-step method. Explain which stage removes insoluble solids, which removes dissolved organic compounds and colour, and which guarantees sterility.
拟一份分步骤操作法。解释哪一步去除不溶性固体，哪一步去除溶解的有机物和颜色，哪一步确保无菌。
Step 1: Filtration – pour the pond water through filter paper to trap mud, twigs and suspended particles. Step 2: Adsorption – pass the filtrate through a column of activated charcoal, which removes dissolved organic molecules, oil traces and colour. Step 3: Distillation – boil the clear water and condense the steam into a clean receiver. Distillation kills microorganisms and removes dissolved salts, yielding pure water. After distillation, a small amount of chlorine bleach can be added to prevent recontamination.
步骤 1：过滤——将池塘水倒入滤纸，截留泥浆、树枝和悬浮颗粒。步骤 2：吸附——让滤液通过活性炭柱，去除溶解的有机物、油迹和颜色。步骤 3：蒸馏——将清澈的水加热煮沸，并将蒸汽冷凝至干净的接收器中。蒸馏能杀灭微生物并去除溶解盐，得到纯水。蒸馏后，可加入少量含氯漂白剂以防止再污染。
4. Distinguishing Pure Substances and Mixtures | 鉴别纯净物与混合物
A chemist is given two white crystalline solids, X and Y, both thought to be naphthalene. Using a heating apparatus, she records the melting behaviour of each. Solid X melts sharply at exactly 80°C. Solid Y begins to soften at 71°C and becomes fully liquid by 78°C.
一位化学家拿到两种白色晶体固体 X 和 Y，两者都被认为是萘。她使用加热装置记录各自的熔化过程。固体 X 在 80°C 时精确、急剧熔化。固体 Y 在 71°C 时开始软化，到 78°C 时才完全变为液体。
Explain which solid is a pure substance and which is an impure mixture. Refer to the concepts of melting point and melting range. Why is melting point determination a powerful purity test?
解释哪种固体是纯净物，哪种是不纯的混合物。提及熔点和熔距的概念。为什么熔点测定是一种有效的纯度检验手段？
A pure substance has a fixed, sharp melting point because all its particles are identical and experience the same intermolecular forces. Solid X melting precisely at 80°C shows it is pure naphthalene. Impurities disrupt the crystal lattice, lowering the melting point and broadening it into a melting range, as seen with solid Y. Thus, melting point analysis quickly distinguishes pure compounds from contaminated samples.
纯净物具有固定、敏锐的熔点，因为其微粒完全相同，承受的分子间作用力一致。固体 X 在 80°C 精确熔化，表明它是纯萘。杂质会破坏晶格结构，降低熔点并使其扩展为一个熔距，正如固体 Y 所表现的那样。因此，熔点分析能迅速区分纯净化合物和受污染的样品。
5. Designing a Reaction Rate Experiment | 设计反应速率实验
A student investigates how temperature affects the rate of reaction between marble chips (calcium carbonate) and hydrochloric acid. She measures the time needed to collect 50 cm³ of carbon dioxide gas. The data collected at five temperatures are shown in the table below.
一名学生探究温度如何影响大理石碎片（碳酸钙）与盐酸的反应速率。她测量收集 50 cm³ 二氧化碳气体所需的时间。下表为在五个温度下收集的数据。
| Temperature (°C) | Time to collect 50 cm³ CO₂ (s) |
|---|---|
| 15 | 120 |
| 25 | 60 |
| 35 | 30 |
| 45 | 15 |
| 55 | 8 |
Identify the independent, dependent and at least two controlled variables in this investigation. Using particle collision theory, explain the trend shown in the data.
指出本实验中的自变量、因变量以及至少两个控制变量。运用粒子碰撞理论解释数据呈现的趋势。
Independent variable: temperature of the acid. Dependent variable: time to collect 50 cm³ of CO₂ (which indicates reaction rate). Controlled variables: mass and surface area of marble chips, volume and concentration of hydrochloric acid. Particle collision theory explains that as temperature increases, particles gain kinetic energy, move faster and collide more frequently and more energetically. A larger proportion of collisions exceed the activation energy, so the product forms faster, shortening the collection time.
自变量：酸的浓度温度。因变量：收集 50 cm³ CO₂ 的时间（表征反应速率）。控制变量：大理石碎片的质量和表面积、盐酸的体积和浓度。粒子碰撞理论解释，随着温度升高，粒子获得动能，运动加快，碰撞更频繁且更有能量。超过活化能的碰撞比例增大，因此产物生成更快，收集时间缩短。
6. Chemistry in Firefighting | 消防中的化学
A kitchen fire breaks out when a pan of cooking oil overheats. A householder instinctively throws a cup of water onto the flames, and the fire instantly expands into a dangerous fireball. Understanding the fire triangle (fuel, heat, oxygen) helps firefighters choose the correct extinguishing method.
厨房里一锅食用油过热着火。住户本能地向火焰泼了一杯水，火势瞬间膨胀成危险的火球。理解火三角（可燃物、热量、氧气）有助于消防员选择正确的灭火方式。
Explain why water should never be used on an oil fire. Describe how a carbon dioxide fire extinguisher puts out such a fire, referring to the fire triangle. Write the word equation for the reaction that occurs between oil and oxygen.
解释为何绝不能用水扑灭油火。联系火三角，描述二氧化碳灭火器如何扑灭此类火灾。写出油与氧气反应的文字方程式。
y vaporises into steam, violently expanding and throwing burning oil droplets into the air— kets the fire, removing the oxygen component of the fire triangle, while also cooling the fuel slightly. The combustion reaction is: oil + oxygen → carbon dioxide + water (plus energy).
水比油密度大，会沉到底部并瞬间汽化成水蒸气，猛烈膨胀并把燃烧的油滴抛向空中——形成火球。水也无法隔绝燃烧油面的氧气。二氧化碳灭火器通过排挤氧气来工作：较重的气体覆盖火焰，消除火三角中的氧气要素，同时轻微冷却燃料。燃烧反应为：油 + 氧气 → 二氧化碳 + 水（加上能量）。
7. Soil pH and Plant Growth | 土壤 pH 和植物生长
A farmer wants to grow blueberries, which thrive in acidic soil with a pH between 4.5 and 5.5. She tests her soil using a universal indicator and finds the pH is 7.2. She also tests the soil after heavy liming and observes a pH of 8.0 in some areas.
一位农民想种植蓝莓，这类植物在 pH 为 4.5 至 5.5 的酸性土壤中才能繁茂生长。她用通用指示剂测试土壤，发现 pH 为 7.2。她还检测了曾经大量施用石灰的区域，测得部分土壤 pH 为 8.0。
Advise the farmer on a chemical that could lower the soil pH and one that could raise it. Write an ionic equation for the neutralisation reaction that occurs when an acid is added to alkaline soil. Explain why maintaining the correct pH matters for nutrient uptake.
向农民推荐一种能降低土壤 pH 的化学品和一种能提高它的化学品。写出酸加入碱性土壤时所发生中和反应的离子方程式。解释为何保持正确 pH 对营养吸收至关重要。
, agricultural lime (calcium carbonate) is used. The neutralisation ionically is: H⁺(aq) + OH⁻(aq) → H₂O(l). Soil pH affects the solubility of mineral nutrients; in alkaline soils, iron and manganese become less available, causing yellowing of leaves. Acid-loving plants like blueberries require acidic conditions to unlock these micronutrients.
若要降低 pH（使土壤更酸），农民可添加硫磺粉或硫酸亚铁，它们会氧化并水解产生酸。若要提高 pH，则使用农用石灰（碳酸钙）。中和反应的离子方程式为：H⁺(aq) + OH⁻(aq) → H₂O(l)。土壤 pH 影响矿物营养素的可溶性；在碱性土壤中，铁和锰变得难被吸收，导致叶片黄化。喜酸植物如蓝莓需要酸性环境才能释放这些微量营养元素。
8. Extracting Copper from Its Ore | 从矿石中提取铜
In a laboratory simulation of ancient smelting, a student heats a mixture of copper(II) oxide (CuO) and powdered carbon in a crucible. After strong heating, a reddish-brown solid appears, and a gas is evolved that turns limewater milky.
在一项模拟古代熔炼的实验中，一名学生将氧化铜（CuO）粉末与碳粉的混合物在坩埚中加热。强热之后，出现红棕色固体，同时逸出一种使石灰水变浑浊的气体。
Identify the reddish-brown solid and the gas. Write the balanced symbol equation for the reaction. Explain, in terms of oxygen loss and gain, why this is a redox reaction. Suggest why carbon was chosen as a reducing agent.
鉴别红棕色固体和气体。写出该反应的配平化学方程式。从得失氧的角度解释为何这是一个氧化还原反应。说明为何选用碳作还原剂。
he gas is carbon dioxide. The equation is: 2CuO + C → 2Cu + CO₂. Copper oxide loses oxygen — it is reduced. Carbon gains oxygen — lentiful, and sits above copper in the reactivity series, allowing it to displace copper from its oxide.
红棕色固体是金属铜，气体是二氧化碳。方程式为：2CuO + C → 2Cu + CO₂。氧化铜失去氧——被还原。碳得到氧——被氧化。因为还原和氧化同时发生，该过程属于氧化还原反应。选用碳是因为它廉价、丰富，且在活动性顺序中位于铜之上，能够从铜的氧化物中置换出铜。
9. Fruit Batteries and the Electrochemical Series | 水果电池与电化学序列
A pupil constructs a simple cell by inserting a zinc strip and a copper strip into a lemon. A voltmeter connected to the strips reads 0.95 V. She then repeats the experiment with other fruit and with different metal pairs, such as magnesium–copper and iron–copper.
一名学生将锌片和铜片插入柠檬中，制作了一个简单电池。连接到金属片的电压表读数为 0.95 V。然后她再用其他水果以及不同金属对（如镁-铜、铁-铜）重复实验。
Predict what would happen to the voltmeter reading if she swapped the zinc with magnesium, and then with silver. Order the metals magnesium, zinc, iron, copper and silver according to their reactivity in a fruit cell, and explain the role of the lemon juice.
预测如果她用镁取代锌、再用银取代锌，电压表读数会如何变化。根据水果电池中的活性顺序排列镁、锌、铁、铜、银，并解释柠檬汁的作用。
Magnesium is more reactive than zinc, so a Mg–Cu cell would produce a higher voltage (around 1.6 V). Silver is less reactive than copper, so an Ag–Cu pair generates only a tiny voltage. The reactivity order in the cell: Mg > Zn > Fe > Cu > Ag. Lemon juice contains citric acid, which acts as an electrolyte; it allows ions to migrate between the strips, completing the circuit and enabling electron flow through the external wire.
镁比锌更活泼，因此 Mg–Cu 电池会产生更高电压（
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