📚 Case Study Practice for Year 9 AQA Chemistry | Year 9 AQA 化学：案例分析实战演练
Year 9 AQA Chemistry introduces fundamental concepts that explain the world around us, from the behaviour of particles to chemical reactions. This article presents a series of real-world case studies designed to help you apply your knowledge, practise exam-style analysis, and deepen your understanding. Each case study is followed by a detailed breakdown that mirrors how you should think in both classwork and assessments.
九年级 AQA 化学课程引入了解释我们周围世界的基本概念，从粒子行为到化学反应。本文通过一系列真实世界的案例分析，帮助你应用所学知识，练习考试风格的分析，并加深理解。每个案例后都附有详细的分解说明，模拟你在课堂作业和考试中应有的思维方式。
1. Case Study: Separating Sand, Salt and Water | 案例：分离沙、盐和水
A student accidentally mixed sand, table salt and water in a beaker. She needs to recover dry sand, pure salt and pure water. Describe a sequence of separation techniques she can use, explaining the principle behind each step.
一名学生不小心将沙子、食盐和水混合在一个烧杯中。她需要回收干燥的沙子、纯净的盐和纯水。请描述她可采用的一系列分离技术，并解释每一步的原理。
The mixture is first filtered through filter paper. Sand particles are too large to pass through the tiny pores, so they are collected as residue on the paper. The filtrate contains dissolved salt and water. This is a physical separation based on particle size.
混合物首先通过滤纸进行过滤。沙粒太大无法通过微小的孔隙，因此作为残渣留在滤纸上。滤液中含有溶解的盐和水。这是一种基于粒子大小的物理分离。
To obtain pure salt, the filtrate is heated gently in an evaporating basin. Water evaporates, leaving behind salt crystals. The water vapour can be condensed back into liquid by cooling, giving pure water. This uses evaporation and condensation, which rely on differences in boiling points.
要得到纯盐，可将滤液在蒸发皿中小火加热。水蒸发后留下盐的晶体。水蒸气可以通过冷却重新冷凝为液体，得到纯水。这利用了蒸发和冷凝，依据是沸点的差异。
2. Case Study: Heating Curve of Salty Water | 案例：盐水的加热曲线
A lab group heated a sample of seawater and recorded the temperature every 30 seconds. The graph showed a steady rise, then a plateau around 100 °C, but a second plateau at a higher temperature. What does each plateau represent, and why is the boiling point not a sharp single line?
一个实验小组加热了一份海水样品，每 30 秒记录一次温度。图像显示温度稳步上升，随后在 100 °C 附近出现平台，但在更高温度处出现第二个平台。每个平台代表什么？为什么沸点不是一条清晰单一的线？
The first plateau occurs because water boils at 100 °C and the energy supplied is used to overcome attractive forces between particles during the liquid-to-gas phase change, so temperature stays constant. The boiling point of pure water is 100 °C.
第一个平台的出现是因为水在 100 °C 沸腾，提供的能量用来克服液态转气态时粒子间的吸引力，因此温度保持恒定。纯水的沸点是 100 °C。
The second plateau appears because dissolved salts raise the boiling point of the solution. As water boils away, the solution becomes more concentrated, causing the boiling point to increase gradually. This is a colligative property: adding a solute elevates the boiling point.
第二个平台的出现是因为溶解的盐会升高溶液的沸点。随着水的沸腾蒸发，溶液变得更浓，导致沸点逐渐升高。这是一种依数性：加入溶质会升高沸点。
3. Case Study: Burning Magnesium and Conservation of Mass | 案例：燃烧镁与质量守恒
A student weighed a strip of magnesium ribbon, burned it in a crucible with a lid, and reweighed the product. The mass appeared to increase. However, when the lid was lifted during burning, some white smoke escaped. Explain why the mass of the solid product increased and how the law of conservation of mass still holds.
一名学生称量了一条镁带，在带盖的坩埚中燃烧后再称量产物。质量似乎增加了。然而，如果在燃烧过程中打开盖子，一些白烟会逸出。解释为什么固体产物的质量增加了，以及质量守恒定律为何仍然成立。
The balanced equation is 2Mg + O₂ → 2MgO. Magnesium atoms combine with oxygen atoms from the air, so the product magnesium oxide contains the mass of both magnesium and oxygen. Therefore, the mass of solid left in the crucible is greater than the original mass of the metal alone.
该反应的配平方程式为 2Mg + O₂ → 2MgO。镁原子与空气中的氧原子结合，因此产物氧化镁包含了镁和氧两者的质量。所以坩埚中剩余固体的质量大于原来金属单独的质量。
The total mass of the closed system (magnesium + oxygen) would remain constant if no gas escapes. When white smoke escapes, the apparent mass increase may be less than expected, but the mass of the whole universe of reactants and products is conserved. The law states that atoms are neither created nor destroyed.
如果系统密闭且没有气体逸出，总质量（镁 + 氧气）将保持不变。当白烟逸出时，表观质量增加可能小于预期，但所有反应物与产物的总质量是守恒的。定律指出原子既不能被创造也不能被消灭。
4. Case Study: Antacid Tablets and Neutralisation | 案例：抗酸片与中和反应
A person suffering from heartburn takes an antacid tablet containing calcium carbonate. The stomach contains excess hydrochloric acid. Write a word and symbol equation for the neutralisation that occurs and explain why this relieves the pain.
一位胃灼热的患者服用了含碳酸钙的抗酸片。胃中含有过量的盐酸。写出发生的中和反应的字方程式和符号方程式，并解释为何能缓解疼痛。
nate + hydrochloric acid → calcium chloride + water + carbon dioxide. The symbol equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. This is a neutralisation reaction between an acid and a carbonate, producing a salt, water and carbon dioxide gas.
字方程式：碳酸钙 + 盐酸 → 氯化钙 + 水 + 二氧化碳。符号方程式为 CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂。这是一个酸与碳酸盐之间的中和反应，生成盐、水和二氧化碳气体。
The excess acid in the stomach causes discomfort. Calcium carbonate reacts with the acid, lowering the concentration of H⁺ ions and raising the pH toward neutral. The resulting pH closer to 7 relieves the burning sensation by removing the irritant acid.
胃中过剩的酸引起不适。碳酸钙与酸反应，降低了氢离子的浓度，使 pH 向中性升高。由此 pH 更接近 7，通过清除刺激性酸性物质来缓解灼烧感。
5. Case Study: Rusting of Iron Nails | 案例：铁钉的生锈
Five iron nails were placed in different conditions: dry air, boiled water with oil layer, salt water, open air, and air with calcium chloride drying agent. Only the nail in open air and the one in salt water rusted heavily. Explain why rusting occurred only in these tubes and suggest how to prevent rusting.
五根铁钉分别置于不同条件下：干燥空气、有水层覆盖的煮沸水、盐水、敞开的空气以及有氯化钙干燥剂的空气。只有敞开的空气和盐水中的铁钉严重生锈。解释为何仅在上述试管中生锈，并提出防锈建议。
Rusting requires both oxygen and water. In dry air, there is no water; in boiled water with oil, oxygen is removed and the oil layer prevents further oxygen dissolving. Salt water accelerates rusting because ions in salt act as an electrolyte, facilitating the electrochemical corrosion of iron.
生锈需要同时存在氧气和水。干燥空气中无水；煮沸水加一层油则除去了氧气并阻止氧气进一步溶解。盐水加速生锈，因为盐中的离子充当电解质，促进了铁的电化学腐蚀。
The nail in open air has access to moisture and oxygen, so rust forms. Prevention methods include painting, oiling, galvanising (zinc coating), or using sacrificial protection. These create a barrier or use a more reactive metal to corrode preferentially.
敞开的空气中的铁钉能接触到水和氧气，因此形成锈。防锈方法包括涂漆、涂油、镀锌或使用牺牲保护。这些措施或者形成屏障层，或者利用更活泼的金属优先被腐蚀。
6. Case Study: Periodic Table Detective | 案例：元素周期表侦探
Element X is a shiny solid that conducts electricity, melts at 660 °C, and reacts slowly with dilute hydrochloric acid to produce a colourless gas. It is in Group 13. Use periodic table trends to identify X and predict three properties of the element below it in the group.
元素 X 是一种有光泽的导电固体，熔点 660 °C，与稀盐酸缓慢反应产生无色气体。它位于第 13 族。利用周期表的变化规律识别 X，并预测其同族中下一元素的三项性质。
X is aluminium (Al). Aluminium is metallic, conducts electricity, and reacts with acids to produce hydrogen gas: 2Al + 6HCl → 2AlCl₃ + 3H₂. Its position in Group 13 places it between metalloids and metals with a relatively low density and protective oxide layer.
X 是铝 (Al)。铝是金属，能导电，与酸反应生成氢气：2Al + 6HCl → 2AlCl₃ + 3H₂。它在第 13 族中的位置介于类金属与金属之间，密度相对较低并具有保护性氧化膜。
The element directly below aluminium is gallium (Ga). Based on group trends, gallium is expected to be a metal with a lower melting point (29.8 °C, so it can melt in your hand), higher density, and similar reactivity with acids, but more reactive than aluminium due to increased atomic size and easier loss of outer electrons.
铝正下方的元素是镓 (Ga)。根据族内变化规律，预期镓是一种熔点更低的金属（29.8 °C，可在手中熔化），密度更高，与酸反应性类似，但由于原子半径增大、外层电子更易失去，其活泼性比铝更强。
7. Case Study: Acid Rain and Marble Statues | 案例：酸雨与大理石雕像
Marble, mainly calcium carbonate, has been used for centuries in sculptures and buildings. In areas with heavy acid rain, marble statues show signs of weathering and loss of detail. Explain the chemical reaction causing this damage and suggest why limestone areas suffer similar erosion.
大理石的主要成分是碳酸钙，数百年来一直用于雕塑和建筑。在酸雨严重的地区，大理石雕像出现风化迹象和细部缺失。解释导致这种损毁的化学反应，并说明为何石灰岩区域也遭受类似的侵蚀。
Acid rain contains sulfuric acid (from SO₂ emissions) and nitric acid (from NOₓ). These acids react with calcium carbonate: CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂. The product calcium sulfate is slightly soluble and can be washed away, eroding the surface.
酸雨含有硫酸（来自二氧化硫排放）和硝酸（来自氮氧化物）。这些酸与碳酸钙反应：CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂。产物硫酸钙微溶，可被冲刷走，从而侵蚀表面。
Limestone is also largely calcium carbonate, so it is similarly attacked. The reaction releases carbon dioxide gas and dissolves the rock, leading to features such as caves and limestone pavements. The same chemistry explains why limestone buildings are particularly vulnerable to acid deposition.
石灰岩同样主要由碳酸钙组成，因此也会遭受类似的攻击。反应释放二氧化碳气体并溶解岩石，形成溶洞和石灰岩路面等地貌。相同的化学原理也解释了为何石灰岩建筑特别容易受到酸性沉降物的损害。
8. Case Study: Solar Still for Clean Water | 案例：太阳能蒸馏器获取净水
In a survival scenario, a makeshift solar still can be constructed using a plastic sheet, a container and a small pit. Explain how this device produces drinkable water from muddy or salty water by applying the concepts of evaporation and condensation.
在求生情境下，可以用一块塑料布、一个容器和一个小坑搭建简易太阳能蒸馏器。解释该装置如何利用蒸发和冷凝的原理从泥水或盐水中获取饮用水。
Solar radiation heats the impure water, raising the kinetic energy of water particles. Energetic particles at the surface overcome attractive forces and escape as vapour, leaving behind dissolved salts, mud and microorganisms. This is evaporation, which separates water from non-volatile impurities.
太阳辐射加热不纯的水，增加了水分子的动能。表面高能粒子克服相互作用力，变成水蒸气逸出，留下溶解的盐、泥土和微生物。这就是蒸发，能将水与不挥发的杂质分离。
The water vapour rises, hits the cooler plastic sheet, and condenses into liquid droplets. These droplets run down into the collection container. The process is essentially distillation powered by the sun, producing distilled water that is safe to drink because all solutes have been removed.
水蒸气上升，接触较冷的塑料布，凝结成液态水滴。这些水滴流进收集容器。该过程本质上是利用太阳能进行的蒸馏，能产出安全的饮用水，因为所有溶质已被去除。
9. Case Study: Writing and Balancing Equations Challenge | 案例：书写与配平化学方程式挑战
A student reacted lithium with water and observed bubbles of gas and a solution that turned universal indicator purple. Write the word equation and a balanced symbol equation for this reaction. Explain how you balanced it and identify the alkaline product.
一名学生让锂与水反应，观察到气泡产生，且溶液使通用指示剂变为紫色。写出该反应的字方程式和配平的符号方程式。解释如何配平，并指出碱性产物。
m hydroxide + hydrogen. The unbalanced symbol is Li + H₂O → LiOH + H₂. To balance, place a '2' in front of LiOH and then adjust H₂O to 2: 2Li + 2H₂O → 2LiOH + H₂. Now atoms are conserved: 2 Li, 4 H, 2 O on both sides.
字方程式：锂 + 水 → 氢氧化锂 + 氢气。未配平的符号式为 Li + H₂O → LiOH + H₂。为了配平，先在 LiOH 前加系数 2，再调整水为 2：2Li + 2H₂O → 2LiOH + H₂。现在原子数守恒：两侧均为 2 个锂、4 个氢、2 个氧。
The alkaline product is lithium hydroxide, which dissociates in water to give OH⁻ ions, hence the purple colour with universal indicator. The gas is hydrogen, which burns with a squeaky pop. Balancing confirms the Law of Conservation of Mass at the particle level.
碱性产物是氢氧化锂，它在水中解离出氢氧根离子，使通用指示剂呈紫色。气体是氢气，燃烧时有爆鸣声。配平方程式从粒子层面验证了质量守恒定律。
10. Case Study: Exothermic and Endothermic Applications | 案例：放热与吸热反应的应用
Hand warmers often contain iron powder, salt and activated charcoal, while instant cold packs use ammonium nitrate and water. Classify each process as exothermic or endothermic and explain the energy transfer involved in making or breaking chemical bonds.
暖手宝通常含有铁粉、盐和活性炭，而速冷冰袋则使用硝酸铵和水。将每种过程分类为放热或吸热，并解释在化学键的形成或断裂中涉及的能量转移。
The hand warmer works by the oxidation of iron: 4Fe + 3O₂ + 6H₂O → 4Fe(OH)₃. This is exothermic because the energy released when new bonds form in iron hydroxide is greater than the energy absorbed to break bonds in reactants. The excess energy heats the surroundings.
暖手宝通过铁粉的氧化工作：4Fe + 3O₂ + 6H₂O → 4Fe(OH)₃。这是放热过程，因为形成氢氧化铁中新键所释放的能量大于断裂反应物中化学键所吸收的能量。多余的热量加热了周围环境。
The cold pack uses dissolving ammonium nitrate in water: NH₄NO₃(s) + water → NH₄⁺(aq) + NO₃⁻(aq). This is endothermic; the energy needed to overcome ionic bonds in the solid lattice is greater than the energy released when ions are hydrated, so heat is taken from the surroundings, causing the pack to feel cold.
速冷冰袋利用硝酸铵溶于水：NH₄NO₃(s) + water → NH₄⁺(aq) + NO₃⁻(aq)。这是吸热过程；克服固体晶格中离子键所需的能量大于离子水合时释放的能量，因此从周围环境吸热，使冰袋变冷。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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