📚 Mastering Case Studies in Year 9 CCEA Chemistry | 化学案例分析实战演练
Case studies in Year 9 CCEA Chemistry are not just about recalling facts – they require you to apply your knowledge to real-world or experimental situations. This article presents step-by-step case study practice covering key topics from the CCEA Key Stage 3 specification. Each case is designed to strengthen your analytical thinking and problem-solving skills, helping you interpret data, recognise patterns and explain chemical phenomena clearly.
化学案例分析实战演练不仅考查知识记忆，更要求你将所学知识应用到真实或实验情境中。本文围绕 CCEA 九年级化学大纲的核心主题，提供逐步分析的案例练习。每个案例都旨在强化你的分析思维和问题解决能力，帮助你解读数据、识别规律并清晰解释化学现象。
1. Identifying Unknown Substances Using Physical Properties | 利用物理性质鉴别未知物质
A laboratory technician discovers three unlabelled containers of grey solids: one is aluminium, one is sulfur and one is graphite. She records the melting point, electrical conductivity and reaction with dilute hydrochloric acid. Aluminium has a melting point of 660 °C, conducts electricity and reacts with acid to produce bubbles. Sulfur melts at 115 °C, does not conduct electricity and shows no reaction with acid. Graphite conducts electricity but has an extremely high melting point (>3500 °C) and does not react with acid. Which solid is in each container?
一位实验室技术员发现了三个未贴标签的灰色固体：一个装有铝，一个是硫，另一个是石墨。她记录了熔点、导电性以及与稀盐酸的反应情况。铝的熔点为 660℃，能导电并且与酸反应产生气泡。硫在 115℃ 熔化，不导电，与酸不反应。石墨能导电但熔点极高（>3500℃），且不与酸反应。哪个容器中装的是哪种固体？
Begin by using conductivity as the key distinguishing test: two solids conduct, one does not. Sulfur is a non-metal and a poor electrical conductor, so the insulating solid is sulfur. Next, compare the two conductors: aluminium and graphite. Graphite has an exceptionally high melting point, while aluminium melts at 660 °C. If the melting point is measured and far exceeds 2000 °C, the sample is graphite. Alternatively, reaction with acid separates them – aluminium reacts vigorously, graphite does not. Using two tests removes all ambiguity.
首先利用导电性作为关键区分手段：两种固体能导电，一种不能。硫是非金属，导电性差，因此不导电的固体是硫。接下来比较两种导电固体：铝和石墨。石墨的熔点极高，而铝在 660℃ 熔化。如果测得的熔点远高于 2000℃，则该样品是石墨。另外，与酸的反应也能区分它们——铝与酸剧烈反应，石墨不与酸反应。结合两种测试即可完全消除歧义。
2. Conservation of Mass in a Precipitate Reaction | 沉淀反应中的质量守恒
A student mixes 50 cm³ of silver nitrate solution with 50 cm³ of sodium chloride solution in an open beaker. A white precipitate of silver chloride forms. The total mass before mixing was 127.4 g. After the reaction, the mass recorded is still 127.4 g. The student concludes that no chemical change occurred. Is this conclusion correct? Explain using the principle of conservation of mass and the observations.
一名学生在一个敞口烧杯中将 50 cm³ 硝酸银溶液与 50 cm³ 氯化钠溶液混合。生成了白色的氯化银沉淀。混合前总质量为 127.4 g。反应后测量质量仍为 127.4 g。该学生得出结论：没有发生化学变化。这个结论正确吗？请利用质量守恒定律和观察到的现象进行解释。
No, the conclusion is not correct. The formation of a white precipitate is direct visual evidence of a new substance being produced, which indicates a chemical reaction. The total mass remained the same because, in a closed system or an open beaker where no gas escapes, mass is conserved. The atoms of silver, nitrogen, oxygen, sodium and chlorine are simply rearranged into new products. Since no gas was lost to the surroundings, the total mass measured before and after the reaction is identical, perfectly illustrating the law of conservation of mass.
这个结论不正确。白色沉淀的生成是产生新物质的直接视觉证据，表明发生了化学反应。总质量保持不变是因为在一个封闭系统中或没有气体逸出的敞口烧杯中，质量是守恒的。银、氮、氧、钠和氯的原子仅仅是重新组合成了新物质。由于没有气体散逸到周围环境中，反应前后测量的总质量相同，这恰好体现了质量守恒定律。
3. Analysing Concentration Effects on Reaction Rate | 分析浓度对反应速率的影响
In a laboratory investigation, three different concentrations of hydrochloric acid (0.5 mol/dm³, 1.0 mol/dm³, 1.5 mol/dm³) were reacted with marble chips. The volume of carbon dioxide gas produced was measured every 30 seconds. The results are shown in the table below.
在一次实验探究中，用三种不同浓度的盐酸（0.5 mol/dm³、1.0 mol/dm³、1.5 mol/dm³）与大理石碎片反应。每隔 30 秒测量生成二氧化碳气体的体积。结果如下表所示。
| Time (seconds) | 0.5 mol/dm³ CO₂ (cm³) | 1.0 mol/dm³ CO₂ (cm³) | 1.5 mol/dm³ CO₂ (cm³) |
| 0 | 0 | 0 | 0 |
| 30 | 12 | 24 | 36 |
| 60 | 20 | 40 | 58 |
| 90 | 26 | 50 | 68 |
| 120 | 30 | 57 | 72 |
| 150 | 33 | 62 | 73 |
Using the particle collision theory, explain why the initial rate of reaction is higher for the 1.5 mol/dm³ acid. Also, why does the rate eventually slow down for all concentrations?
利用粒子碰撞理论，解释为什么 1.5 mol/dm³ 的酸初始反应速率更快。同时，为什么所有浓度的反应速率最终都会减慢？
A higher concentration means there are more hydrochloric acid particles per unit volume. This increases the frequency of successful collisions between reactant particles, so the initial rate of carbon dioxide production is highest for the 1.5 mol/dm³ acid. Over time, the marble chips are used up and the concentration of acid decreases, reducing the number of effective collisions. Consequently, the rate of reaction slows down and eventually stops when the limiting reactant is consumed.
更高的浓度意味着单位体积内有更多的盐酸粒子。这增加了反应物粒子之间有效碰撞的频率，因此 1.5 mol/dm³ 的酸产生二氧化碳的初始速率最快。随着时间的推移，大理石碎片被消耗，酸的浓度下降，有效碰撞的次数减少。因此，反应速率减慢，当限制性反应物被耗尽时反应最终停止。
4. Identifying an Unknown Gas from Test Results | 依据检测结果鉴别未知气体
A student electrolyses water and collects the gas at the cathode. The gas burns with a pale blue flame and makes a popping sound with a lit splint. A second gas collected at the anode relights a glowing splint. Name the two gases and write the word equation for the decomposition of water. Explain how the atomic structure of water supports the observation that the volume of gas at the cathode is twice that at the anode.
一名学生电解水并收集阴极产生的气体。该气体燃烧时呈淡蓝色火焰，用点燃的小木条检验会发出“噗”的爆鸣声。在阳极收集到的另一种气体能使带火星的木条复燃。请命名这两种气体，写出水分解的文字表达式。解释水的原子结构如何支持阴极气体体积是阳极气体体积两倍的观察结果。
ater → hydrogen + oxygen. A water molecule, H₂O, contains two hydrogen atoms and one oxygen atom. During electrolysis, water decomposes into twice as many hydrogen molecules as oxygen molecules. Because equal volumes of gases contain the same number of molecules at the same temperature and pressure, the volume of hydrogen collected is double the volume of oxygen.
阴极产生的气体是氢气（用点燃的木条检验会发出爆鸣声），阳极产生的气体是氧气（使带火星的木条复燃）。文字表达式为：水 → 氢气 + 氧气。水分子（H₂O）含有两个氢原子和一个氧原子。在电解过程中，水分解产生的氢分子数量是氧分子数量的两倍。由于同温同压下等体积的气体含有相同数目的分子，因此收集到氢气的体积是氧气体积的两倍。
5. Choosing the Best Fuel Using Energy Transfer Data | 利用能量转化数据选择最佳燃料
A group of pupils investigate the energy released when four different alcohols are burned. They use a spirit burner and a copper calorimeter containing 100 g of water. The temperature rise is recorded for each fuel while keeping the mass of alcohol burned constant at 0.5 g. Results: methanol raised the temperature by 12 °C, ethanol by 16 °C, propanol by 21 °C, and butanol by 25 °C. A pupil claims butanol is the best fuel. Do you agree? Justify your answer by calculating the energy transferred for butanol and discussing other practical factors.
一组学生研究燃烧四种不同醇时所释放的能量。他们使用酒精灯和装有 100 g 水的铜制量热器。在燃烧的醇质量固定为 0.5 g 的情况下记录每种燃料的温升。结果：甲醇升温 12℃，乙醇升温 16℃，丙醇升温 21℃，丁醇升温 25℃。一名学生称丁醇是最好的燃料。你同意这个结论吗？请计算丁醇转移的能量并讨论其他实际因素，以证明你的答案。
ly 0.5 g of butanol was burned, the energy released per gram is 10,500 ÷ 0.5 = 21,000 J/g. Yes, in terms of energy density, butanol appears the best among the tested fuels because it releases the most energy per gram. However, other factors matter: butanol may produce more soot, be more expensive, or have a higher toxicity. A fuel must also be safe, readily available and burn cleanly for practical use.
转移到水中的能量可以用 Q = m × c × ΔT 估算，其中 m = 100 g，c = 4.2 J/g°C，丁醇的 ΔT = 25℃。Q = 100 × 4.2 × 25 = 10,500 J。因为只燃烧了 0.5 g 丁醇，每克释放的能量为 10,500 ÷ 0.5 = 21,000 J/g。是的，就能量密度而言，丁醇在被测试的燃料中似乎是最好的，因为它每克释放的能量最多。然而，还需考虑其他因素：丁醇可能产生更多的烟灰、价格更高或毒性更大。在实际使用中，燃料还必须安全、容易获得并且燃烧清洁。
6. Separating a Mixture Using Solubility Differences | 利用溶解性差异分离混合物
A mixture contains sand, salt and iron filings. Describe a sequence of separation techniques to obtain dry samples of each component. Name each method and explain why it is suitable for that particular substance.
一种混合物含有沙子、食盐和铁屑。请描述一系列分离技术，以获得每种组分的干燥样品。说出每种方法的名称，并解释为什么该方法适用于该特定物质。
First, use a magnet to remove the iron filings. Iron is magnetic, so placing a magnet wrapped in paper over the mixture will attract the filings, leaving sand and salt. Next, add water to the remaining mixture and stir. Salt is soluble, sand is insoluble. Filter the mixture; sand remains as residue on the filter paper, and salt solution passes through as filtrate. Rinse and dry the sand. Finally, evaporate the water from the salt solution by heating it gently. The water turns to vapour, leaving solid salt crystals behind. This sequence exploits magnetism, solubility and the physical change of state from liquid to gas.
首先，用磁铁吸出铁屑。铁具有磁性，因此将用纸包裹的磁铁放在混合物上方就能吸走铁屑，留下沙子和食盐。接着，向剩余的混合物中加水并搅拌。食盐可溶于水，沙子不溶。将混合物过滤；沙子作为残渣留在滤纸上，食盐溶液作为滤液通过滤纸。将沙子洗涤并干燥。最后，通过温和加热蒸发掉食盐溶液中的水。水变成水蒸气，留下固态的食盐晶体。这一系列操作依次利用了磁性、溶解性以及液体到气体的物理状态变化。
7. Acid Rain Formation and Its Environmental Impact | 酸雨的形成及其环境影响
Case study: A power station burns coal containing sulfur impurities. The waste gases pass through a tall chimney. The local lake downstream of the prevailing wind has become more acidic, and limestone statues in the town centre show signs of flaking. Explain the chemical processes from the combustion of sulfur to the damage observed. Write balanced word equations for the formation of sulfuric acid and the reaction of acid rain with limestone (calcium carbonate). Suggest one way to reduce acid rain.
案例分析：一座发电站燃烧含有硫杂质的煤。废气通过高烟囱排放。位于主风向的下游的一个本地湖泊变得更加酸化，镇中心的石灰石雕像出现了剥落迹象。请解释从硫燃烧到观察到受损的化学过程。写出硫酸形成的文字表达式以及酸雨与石灰石（碳酸钙）反应的文字表达式。提出一种减少酸雨的方法。
oxygen to form sulfuric acid (sulfur dioxide + water + oxygen → sulfuric acid). Acid rain falls into the lake, lowering its pH and harming aquatic life. Limestone statues are made of calcium carbonate. Sulfuric acid reacts with calcium carbonate to produce calcium sulfate, water and carbon dioxide (calcium carbonate + sulfuric acid → calcium sulfate + water + carbon dioxide). The calcium sulfate is soluble and washes away, causing flaking. To reduce acid rain, the power station can install scrubbers or use flue gas desulfurization, or switch to low-sulfur fuels.
燃烧含硫煤会产生二氧化硫（硫 + 氧气 → 二氧化硫）。SO₂气体上升进入大气，与水和氧气反应形成硫酸（二氧化硫 + 水 + 氧气 → 硫酸）。酸雨降入湖中，降低其 pH 值并危害水生生物。石灰石雕像由碳酸钙制成。硫酸与碳酸钙反应生成硫酸钙、水和二氧化碳（碳酸钙 + 硫酸 → 硫酸钙 + 水 + 二氧化碳）。硫酸钙可溶，会被雨水冲走，导致雕像剥落。为减少酸雨，发电站可以安装洗涤器或进行烟气脱硫，也可以改用低硫燃料。
8. Interpreting Chromatography Results | 解读色谱分析结果
A forensic scientist analyses a dye found on a piece of evidence. She spots the evidence dye and three known dyes (A, B, C) on a chromatography paper. The solvent front moved 10.0 cm. Dye A spot moved 8.5 cm, dye B moved 3.2 cm, dye C moved 6.0 cm. The evidence dye separated into two spots: one moved 3.2 cm and the other 8.5 cm. Calculate the Rf values for the two components of the evidence sample and identify which known dyes are present. Explain why chromatography is a good technique for this analysis.
一位法证科学家分析证据上发现的一种染料。她在色谱纸上点样了证据染料和三种已知染料（A、B、C）。溶剂前沿移动了 10.0 cm。染料 A 斑点移动了 8.5 cm，染料 B 移动了 3.2 cm，染料 C 移动了 6.0 cm。证据染料分离为两个斑点：一个移动了 3.2 cm，另一个移动了 8.5 cm。计算证据样品两个组分的 Rf 值，并鉴定出其中含有哪些已知染料。解释为什么色谱法适用于这种分析。
= 8.5 ÷ 10.0 = 0.85. Comparing with known dyes: dye A (Rf = 0.85), dye B (Rf = 0.32), dye C (Rf = 0.60). Therefore, the evidence dye is a mixture containing dye A and dye B. Chromatography is excellent for this because it separates dissolved substances based on their different solubilities and attractions to the stationary phase, producing distinct spots that can be matched by Rf values. It requires only a tiny sample and gives reliable identification.
Rf = 斑点移动的距离 ÷ 溶剂前沿移动的距离。对于移动 3.2 cm 的组分，Rf = 3.2 ÷ 10.0 = 0.32。对于移动 8.5 cm 的组分，Rf = 8.5 ÷ 10.0 = 0.85。与已知染料对比：染料 A（Rf = 0.85）、染料 B（Rf = 0.32）、染料 C（Rf = 0.60）。因此，证据染料是一种混合物，含有染料 A 和染料 B。色谱法非常适合这种分析，因为它能根据溶解度和对固定相的亲和力差异将溶解的物质分离，产生不同的斑点，再通过 Rf 值进行比对。它只需要极少量样品，并能提供可靠的鉴定结果。
9. Exothermic and Endothermic Processes in Everyday Life | 日常生活中的放热与吸热过程
A student investigates temperature changes when dissolving different salts in water. She adds 5 g of ammonium nitrate to water and observes the temperature drops from 20 °C to 10 °C. With calcium chloride, the temperature rises from 20 °C to 35 °C. Classify each process as exothermic or endothermic, and explain what happens to the energy of the surroundings in each case. Give another everyday example of an exothermic and an endothermic reaction.
一名学生研究不同盐溶于水时的温度变化。她在水中加入 5 g 硝酸铵，观察到温度从 20℃ 降至 10℃。加入氯化钙后，温度从 20℃ 升至 35℃。请将每个过程归类为放热或吸热过程，并解释每种情况下周围环境的能量变化。再分别举出一个日常生活中放热反应和吸热反应的例子。
Ammonium nitrate dissolving absorbs energy from the surroundings, so it is an endothermic process – the temperature of the solution decreases. Calcium chloride dissolving releases energy to the surroundings, making it an exothermic process – the temperature increases. In an endothermic process, the system takes in heat, cooling the immediate environment; in an exothermic process, the system gives out heat, warming the environment. Everyday exothermic examples include combustion (burning natural gas) and respiration. An endothermic example is the thermal decomposition of calcium carbonate or the reaction of citric acid with sodium hydrogencarbonate in some cold packs.
硝酸铵溶解时从周围吸收能量，因此是一个吸热过程——溶液温度降低。氯化钙溶解时向周围释放能量，因此是一个放热过程——溶液温度升高。在吸热过程中，体系吸收热量，使周围环境变冷；在放热过程中，体系释放热量，使环境变暖。日常放热反应的例子包括燃烧（天然气燃烧）和呼吸作用。吸热反应的例子有碳酸钙的热分解或者某些冷敷包中柠檬酸与碳酸氢钠的反应。
10. Evaluating a Model of the Atom from Experimental Evidence | 依据实验证据评价原子模型
Year 9 pupils study the evolution of the atomic model. Rutherford's gold foil experiment fired alpha particles at a thin gold sheet. Most particles passed straight through, but a few were deflected at large angles and a very tiny fraction bounced straight back. Explain what each observation tells us about the structure of the atom. Also, state one limitation of the Bohr model that was improved by the discovery of electron clouds.
九年级学生学习原子模型的发展历程。卢瑟福的金箔实验用 α 粒子轰击薄金片。大多数粒子径直穿过，但少数粒子发生大角度偏转，极少数粒子反弹回来。请解释每一项观察结果揭示了原子的哪些结构特征。同时指出波尔模型的一个局限性，以及电子云的发现如何改进了它。
Most particles passing through indicates that the atom is mostly empty space. The occasional large deflections mean that alpha particles occasionally approach a small, dense, positively charged region – the nucleus. The particles that bounced back show that the nucleus is extremely dense and contains most of the atom’s mass. A limitation of the Bohr model is that electrons are depicted in fixed circular orbits at set distances from the nucleus, while the modern electron cloud model describes electrons moving in regions of high probability rather than precise paths. This explained chemical bonding and spectra more accurately.
大多数粒子穿过说明原子内部大部分是空的。偶尔发生的大角度偏转表明 α 粒子偶尔靠近了一个很小、致密且带正电的区域——原子核。反弹回来的粒子证明原子核极其致密，集中了原子的大部分质量。波尔模型的一个局限在于它把电子描绘成在固定圆形轨道上以确定距离绕核运动，而现代电子云模型则将电子描述为在概率较高的区域运动，而非精确的路径。这更准确地解释了化学键和光谱现象。
11. Investigating Displacement Reactions Using the Reactivity Series | 利用活动性顺序研究置换反应
A student places a piece of zinc metal into copper(II) sulfate solution. The blue colour fades and a brown solid coats the zinc. Write the word equation and explain why this reaction occurs, using the reactivity series (potassium, sodium, calcium, magnesium, aluminium, zinc, iron, lead, copper, silver, gold). Predict whether magnesium would react with zinc sulfate solution and justify your answer.
一名学生将一片锌金属放入硫酸铜(II)溶液中。蓝色逐渐褪去，锌表面覆盖了一层棕色固体。请写出文字表达式，并利用金属活动性顺序（钾、钠、钙、镁、铝、锌、铁、铅、铜、银、金）解释为什么该反应能够发生。预测镁是否会与硫酸锌溶液反应并说明理由。
he brown solid is copper metal. Magnesium is above zinc in the reactivity series, meaning it is even more reactive. Therefore, magnesium would react with zinc sulfate solution, displacing zinc: magnesium + zinc sulfate → magnesium sulfate + zinc.
文字表达式：锌 + 硫酸铜(II) → 硫酸锌 + 铜。锌比铜更活泼，因此能把铜离子从溶液中置换出来。棕色固体是金属铜。镁在活动性顺序中排在锌的前面，意味着它更活泼。因此，镁会与硫酸锌溶液反应，置换出锌：镁 + 硫酸锌 → 硫酸镁 + 锌。
12. Applying pH Scale and Neutralisation in Context | pH 值与中和反应的实际应用
A farmer tests soil from two fields. Field X has a pH of 4.5; Field Y has a pH of 7.2. She wants to grow wheat, which grows best at pH 6.0–7.0. Explain which field’s soil is too acidic and name a substance she could add to raise the pH. Write a general word equation for the neutralisation reaction. Explain why adding too much of the substance could harm crop growth.
一位农夫测试了两块田地的土壤。X 地块的 pH 值为 4.5；Y 地块的 pH 值为 7.2。她想种植小麦，最适宜小麦生长的 pH 范围是 6.0–7.0。请解释哪块地的土壤酸性过强，并说出她可以添加的一种能提高 pH 值的物质。写出中和反应的一般文字表达式。解释为什么添加过多该物质会损害作物生长。
estone). The general neutralisation reaction is: acid + base → salt + water. If too much base is added, the soil pH will rise above 7.0, becoming too alkaline. Most plants absorb nutrients best when the pH is close to neutral; high alkalinity can lock up essential minerals, making them unavailable to the roots and harming growth.
X 地块的 pH 值为 4.5，对小麦而言酸性过强。为了提高 pH 值，农夫可以添加碱性物质，如氧化钙（生石灰）或碳酸钙（石灰石）。中和反应的一般文字表达式为：酸 + 碱 → 盐 + 水。如果添加过多的碱，土壤 pH 就会升高至超过 7.0，变得碱性过强。大多数植物在接近中性时吸收养分的能力最佳；过高的碱性会固定必需矿物质，使植物根部无法吸收，从而损害生长。
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