📚 Year 8 AQA Chemistry: Case Study Practice Exercises | AQA 8年级化学：案例分析实战演练
Welcome to a focused set of case study exercises designed for Year 8 students following the AQA KS3 course. Each scenario presents a practical situation that tests your understanding of key chemical ideas—separating mixtures, reactions of acids, changes of state, and more. Work through each case, read the explanations carefully, and use the tips to strengthen your scientific thinking.
欢迎来到这套专为学习AQA KS3课程的8年级学生设计的案例分析练习题。每个情境都呈现了一个真实的实际问题，用来检测你对关键化学概念的理解——包括混合物分离、酸的反应、状态变化等等。请你逐个研读案例，仔细阅读解释，并利用这些小贴士来强化你的科学思维。
1. Case Study 1: Separating Sand and Salt | 案例1：分离沙和盐
A student is given a beaker containing a dusty mixture of sand and common salt. They are asked to obtain pure dry sand and pure crystallised salt. The available equipment includes beakers, deionised water, a glass rod, filter paper, a funnel, an evaporating dish, a Bunsen burner, a tripod and a gauze mat.
一名学生拿到一个烧杯，里面装有一些沙子和食盐的粉末状混合物。他需要得到纯净干燥的沙子和纯净的食盐晶体。可用的器材包括烧杯、去离子水、玻璃棒、滤纸、漏斗、蒸发皿、本生灯、三脚架和石棉网。
Task: Design a method that uses differences in physical properties and explain the science behind each step.
任务：设计一个利用物理性质差异的实验方法，并解释每一步背后的科学原理。
Suggested method:
建议方法：
Step 1 – Add about 50 cm³ of deionised water to the mixture and stir thoroughly with the glass rod. Salt (sodium chloride) dissolves rapidly in water, but sand (mainly silicon dioxide) does not dissolve because its particles are held tightly in a giant covalent structure that water molecules cannot pull apart.
步骤1 – 向混合物中加入约50 cm³去离子水，用玻璃棒充分搅拌。食盐（氯化钠）迅速溶于水，而沙子（主要成分二氧化硅）不溶解，因为它的微粒被紧紧地束缚在巨大的共价结构中，水分子无法拉开。
Step 2 – Fold a filter paper, place it inside the funnel, and filter the suspension. The sand stays on the filter paper as the residue, while the salt solution passes through as the filtrate. Filtration works because the tiny holes in the paper allow dissolved ions and water molecules to pass, but trap the much larger sand grains.
步骤2 – 折好滤纸，放入漏斗中，过滤悬浊液。沙子作为残渣留在滤纸上，而食盐溶液作为滤液流下。过滤能够实现是因为滤纸上的小孔允许溶解的离子和水分子通过，却截留了大得多的沙粒。
Step 3 – Rinse the sand on the filter paper with a little distilled water to remove any trapped salt, then dry it on a watch glass or in a warm oven. The sand is now pure.
步骤3 – 用少量蒸馏水冲洗滤纸上的沙子，除去残余的盐，然后将其放在表面皿上或暖烘箱中干燥。沙子此时是纯净的。
Step 4 – Pour the filtrate into an evaporating dish and heat gently over a steam bath or a very low Bunsen flame. As water evaporates, the salt solution becomes more concentrated until crystals begin to form. Turn off the heat when small white crystals appear and allow the rest of the water to evaporate slowly; this produces well‑formed cubic crystals of sodium chloride.
步骤4 – 将滤液倒入蒸发皿中，在水浴上或非常小的本生灯火焰上缓缓加热。随着水蒸发，盐溶液浓度逐渐增大，直至开始出现晶体。当看到细小白色晶体时停止加热，让剩余的水慢慢蒸发，这样可以得到形状完好的氯化钠立方晶体。
Key science point: The separation depends on the large difference in solubility. Sand is insoluble in water, while salt is highly soluble. Filtration separates an insoluble solid from a liquid, and evaporation (or crystallisation) separates a dissolved solid from its solvent.
关键科学点：分离依赖于溶解度的巨大差异。沙子不溶于水，而食盐高度可溶。过滤将不溶性固体从液体中分离出来，而蒸发（或结晶）将已溶解的固体从其溶剂中分离出来。
2. Case Study 2: Chromatography of Inks | 案例2：墨水色谱分析
A crime scene investigator finds a note written with a black felt‑tip pen. She suspects the ink is a mixture of several coloured dyes. She sets up a paper chromatography experiment using a strip of chromatography paper, a pencil line, and a beaker of water (the mobile phase).
一位犯罪现场调查员发现了一张用黑色毡头笔书写的字条。她怀疑墨水是几种有色染料的混合物。她设计了一个纸色谱实验，使用一条色谱纸、铅笔线和一个装有水（流动相）的烧杯。
Task: Explain why the spot of black ink separates into bands of blue, red and yellow as water travels up the paper. Then predict which dye is most soluble in water and which is most attracted to the paper.
任务：解释为什么随着水沿着纸上升，黑色墨点会分离成蓝色、红色和黄色条带。然后预测哪种染料在水中溶解度最大，哪种染料对纸的吸引力最强。
Explanation: A black ink often contains several pigments. When the paper tip is placed in water, the water rises by capillary action. As the water front passes the origin spot, the dyes dissolve and are carried along. Each dye has a different affinity for the stationary phase (the paper fibres) and a different solubility in the mobile phase (water). A dye that is very soluble in water and only weakly attracted to the paper will travel further; one that is strongly held by the paper but less soluble will barely move.
解释：黑色墨水中通常含有多种色素。当纸的末端放入水中时，水因毛细作用上升。当水前沿经过原点墨点时，染料溶解并被携带移动。每种染料对固定相（纸纤维）具有不同的亲和力，同时对流动相（水）有着不同的溶解度。在水中溶解度大而对纸吸引力弱的染料移动较远；被纸强烈吸附但溶解度低的染料几乎不动。
From the observed bands: the yellow band moves the farthest, so the yellow dye is the most soluble in water and the least attracted to the paper. The blue band stays closest to the start line, so blue dye is the most strongly adsorbed onto the paper and least soluble in water.
从观察到的条带来看：黄色条带移动最远，说明黄色染料在水中溶解度最大，对纸的吸附力最弱；蓝色条带距起点最近，说明蓝色染料被纸吸附得最牢固，在水中的溶解度最小。
y spot ÷ distance moved by solvent front. This is a standard way to identify unknown dyes.
计算提示：可为每种物质计算比移值（Rf）：Rf = 斑点的移动距离 ÷ 溶剂前沿的移动距离。这是鉴定未知染料的标准方法。
3. Case Study 3: Acid–Metal Reaction | 案例3：酸与金属的反应
A laboratory technician drops a few pieces of zinc granules into a test tube containing dilute hydrochloric acid. Bubbles of a colourless gas form immediately and the zinc gradually disappears, leaving a clear solution. The test tube feels warm.
一位实验室技术员将几粒锌粒投入盛有稀盐酸的试管中。立刻有大量无色气泡产生，锌粒逐渐消失，剩下清澈的溶液。试管摸起来发热。
Task: Write a word equation and a balanced symbol equation for the reaction. Identify the gas produced and describe its chemical test.
任务：写出反应的文字表达式和配平的符号方程式。鉴定生成的气体并描述其化学检验方法。
Word equation: zinc + hydrochloric acid → zinc chloride + hydrogen
文字表达式：锌 + 盐酸 → 氯化锌 + 氢气
Balanced symbol equation: Zn + 2HCl → ZnCl₂ + H₂
配平的符号方程式：Zn + 2HCl → ZnCl₂ + H₂
The gas is hydrogen. It can be tested by holding a lit splint at the mouth of the test tube – a squeaky pop sound confirms its presence. This happens because hydrogen mixes with air and burns very rapidly, causing a mini explosion.
气体是氢气。检验方法是将一根点燃的木条置于试管口——听到尖锐的爆鸣声即可证实氢气的存在。这是因为氢气与空气混合后急速燃烧，产生微小的爆鸣。
Energy change: The reaction is exothermic, meaning it releases heat to the surroundings. The temperature rise supports this classification.
能量变化：该反应是放热的，即它向环境释放热量。温度的升高支持了这一判断。
Safety note: Hydrogen is highly flammable. Always use small amounts and work in a well‑ventilated lab.
安全提示：氢气高度易燃。必须使用少量，并在通风良好的实验室中操作。
4. Case Study 4: Neutralisation in Everyday Life | 案例4：生活中的中和反应
After a heavy meal, a person suffers from indigestion caused by excess hydrochloric acid in the stomach. They take a tablet containing calcium carbonate (limestone). The fizzing and relief indicate a chemical reaction.
大餐之后，某人因胃酸过多而感到消化不良。他服用了一片含有碳酸钙（石灰石）的药片。嘶嘶冒泡和症状缓解表明发生了化学反应。
Task: Describe how the tablet works using a balanced equation and explain the concept of neutralisation in this context.
任务：用配平的方程式描述药片的作用，并在这一情境中解释中和反应的概念。
Reaction: Calcium carbonate reacts with hydrochloric acid to form calcium chloride, water and carbon dioxide gas:
反应：碳酸钙与盐酸反应生成氯化钙、水和二氧化碳气体：
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
This is a neutralisation reaction because an acid (HCl) and a base (CaCO₃) react to produce a salt (calcium chloride) and neutral substances – water and carbon dioxide. The excess stomach acid is removed, relieving pain.
这是一个中和反应，因为酸（HCl）与碱（CaCO₃）反应生成盐（氯化钙）和中性物质——水和二氧化碳。多余的胃酸被消除，不适得到缓解。
The fizzing is due to the effervescence of carbon dioxide gas. The product calcium chloride is soluble and harmless in small amounts.
嘶嘶声来自二氧化碳气体的冒泡。产物氯化钙可溶，且在少量下无害。
ation: acid + base → salt + water (and sometimes carbon dioxide when a carbonate is used).
中和反应的通式：酸 + 碱 → 盐 + 水（使用碳酸盐时有时还生成二氧化碳）。
5. Case Study 5: Changes of State – Ice to Steam | 案例5：状态变化——从冰到蒸汽
A block of ice at −10 °C is heated steadily in a beaker until it becomes water and eventually boils into steam. A temperature sensor records the temperature every minute.
将一块−10 °C的冰放入烧杯中稳定加热，直至其变成水，最后沸腾成为蒸汽。温度传感器每分钟记录一次温度。
Task: Sketch the shape of the heating curve and use the particle model to explain why the temperature remains constant during melting and boiling, even though energy is still being supplied.
任务：画出加热曲线的形状，并用粒子模型解释为什么熔化和沸腾过程中温度保持不变，尽管能量仍在持续供应。
Particle behaviour: In the solid state, water particles vibrate around fixed positions in a regular lattice. As heat is added, the particles vibrate faster and the temperature rises. At 0 °C, the energy supplied goes into overcoming the attractive forces holding the lattice together; the particles break free into a liquid arrangement. During this phase change, the temperature stays at 0 °C because the added energy is used to increase the potential energy of the particles, not their kinetic energy.
粒子行为：在固态下，水分子在规则的晶格中绕固定位置振动。加热时，粒子振动加剧，温度升高。在0 °C时，输入的能量被用来克服维持晶格的吸引力；粒子挣脱束缚成为液态排列。在此相变过程中，温度保持在0 °C，因为加入的能量用于增大粒子的势能，而非动能。
Similarly, at 100 °C the liquid particles gain enough energy to escape the liquid surface completely and become a gas. Again, the temperature plateaus while the liquid transforms to vapour.
类似地，在100 °C时液体粒子获得足够能量完全逃逸液面，变为气体。在液体转变为蒸汽的过程中，温度再次出现平台。
Key term: Latent heat – the energy absorbed or released during a change of state at constant temperature.
关键术语：潜热——在温度不变的状态变化过程中吸收或释放的能量。
6. Case Study 6: Combustion of Magnesium | 案例6：镁的燃烧
A teacher holds a strip of magnesium ribbon in tongs and ignites it in a Bunsen burner flame. It burns with a brilliant white light and leaves a crumbly white powder. The mass of the powder is measured and found to be greater than the original mass of the ribbon.
一位老师用坩埚钳夹住一条镁带，在本生灯火焰上点燃。镁带发出耀眼的白光，燃烧后留下松脆的白色粉末。称量粉末发现其质量大于原来镁带的质量。
Task: Explain the mass change using a balanced equation and link it to the law of conservation of mass.
任务：用配平的方程式解释质量变化，并将其与质量守恒定律联系起来。
Balanced equation: 2Mg + O₂ → 2MgO
配平方程式：2Mg + O₂ → 2MgO
Explanation: Magnesium atoms combine with oxygen molecules from the air. For every two magnesium atoms, one oxygen molecule (O₂) is added, forming magnesium oxide (MgO). Oxygen has mass, so the product has a greater mass than the original magnesium. If we weigh all reactants (magnesium plus the consumed oxygen) and compare with the product MgO, the total mass is conserved – no atoms are lost or created, only rearranged.
解释：镁原子与空气中的氧分子结合。每两个镁原子加上一个氧分子（O₂）生成氧化镁（MgO）。氧具有质量，因此产物质量大于原来的镁。如果称量所有反应物（镁加上消耗的氧气）并与产物 MgO 比较，总质量守恒——没有原子丢失或创生，只是进行了重新排列。
The bright light is a sign of a highly exothermic reaction; a large amount of heat and light energy is released.
耀眼光芒是剧烈放热反应的标志；大量热能和光能被释放。
Safety: Never look directly at burning magnesium; use blue‑tinted safety goggles.
安全：切勿直视燃烧的镁；使用蓝色护目镜。
7. Case Study 7: Obtaining Pure Water from Seawater | 案例7：从海水中获取纯水
A sailor on a life raft has only seawater and needs fresh drinking water. He builds a simple distillation apparatus using a metal container, a collection cup and a plastic sheet.
一位乘救生筏的水手只有海水，急需淡水饮用。他用金属容器、收集杯和塑料薄膜搭建了一个简易蒸馏装置。
Task: Explain how distillation purifies water, referring to boiling points and the condensation process. Why does the distilled water taste flat?
任务：解释蒸馏如何净化水，涉及沸点和冷凝过程。为什么蒸馏水尝起来平淡无味？
Principle: Seawater is a solution of various salts (mainly sodium chloride) dissolved in water. Water boils at 100 °C (at sea level), while the dissolved salts have much higher boiling points (e.g., sodium chloride boils at 1465 °C). When the seawater is heated, water evaporates and becomes steam, leaving the salts behind. The steam then hits a cooler surface, condenses, and drips as pure liquid water. Distillation thus separates the solvent from non‑volatile solutes.
原理：海水是多种盐（主要是氯化钠）溶于水形成的溶液。水在100 °C沸腾（海平面），而溶解的盐沸点高得多（如氯化钠沸点1465 °C）。加热海水时，水蒸发成为蒸汽，盐则留在后面。蒸汽遇到较冷的表面冷凝，液化成纯水落下。蒸馏因此将溶剂与不挥发的溶质分离开来。
Distilled water tastes flat because it lacks the dissolved minerals and ions that give natural drinking water its characteristic taste.
蒸馏水尝起来平淡，是因为它缺乏天然饮用水中的溶解矿物质和离子，这些赋予水特有的味道。
In the lab, simple distillation uses a round‑bottom flask, a condenser with cold water running through the jacket, and a receiving beaker. This technique is safe and effective for producing pure water.
在实验室中，简易蒸馏使用圆底烧瓶、带有冷水夹套的冷凝管和接收烧杯。这项技术安全有效，可制取纯水。
8. Case Study 8: Rusting of Iron | 案例8：铁的生锈
A gardener notices that an iron garden fork left outdoors for several months has developed a reddish‑brown flaky coating, while a stainless steel trowel kept nearby remains shiny.
一位园丁发现，放在户外数月的铁制花园叉上出现了红褐色的片状锈层，而旁边的一把不锈钢小铲却依然光亮。
Task: State the conditions required for iron to rust, write a word equation for rusting, and suggest two methods to prevent rusting on the fork, explaining how they work.
任务：说出铁生锈所需的条件，写出铁生锈的文字表达式，并提出两种防止叉子生锈的方法，解释其原理。
Conditions: Iron rusts when it is exposed to both oxygen (from air) and water at the same time. Salt (e.g., from sea spray) and acids accelerate the process. Pure, dry air or pure, air‑free water will not cause rusting.
条件：铁同时接触氧气（来自空气）和水时才会生锈。盐（如海水飞沫）和酸会加速锈蚀。纯净的干燥空气或无空气的水不会导致生锈。
Word equation: iron + oxygen + water → hydrated iron(III) oxide (rust)
文字表达式：铁 + 氧气 + 水 → 水合氧化铁（铁锈）
Rust is a soft, porous solid that flakes away, exposing fresh iron underneath, so rusting is a continuous destructive process.
铁锈是一种松软多孔的固体，会剥落，露出下面的新鲜铁，因此生锈是一个持续的破坏过程。
Prevention methods:
- Painting or oiling: A layer of paint or oil creates a barrier that keeps oxygen and moisture from reaching the iron surface. This breaks the contact required for rusting.
- Galvanising (coating with zinc): Zinc acts as a sacrificial protection. Even if the coating is scratched, zinc, being more reactive than iron, corrodes in preference to the iron, so the iron remains protected.
预防方法：
- 涂漆或涂油：漆层或油层形成屏障，阻止氧气和湿气接触铁表面，从而切断生锈所需的接触。
- 镀锌（涂覆锌层）：锌作为牺牲保护层。即使涂层被刮伤，锌比铁更活泼，会优先腐蚀，从而保护铁不被锈蚀。
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