📚 Year 7 WJEC Chemistry: Cross-disciplinary Integrated Question Practice | 跨学科综合题型训练
Cross-disciplinary questions in Year 7 chemistry help you see how chemical ideas connect with maths, physics, biology and geography. This article presents ten typical integrated question types, each demonstrating how to combine knowledge from different subjects to solve problems. Practising these will strengthen your analytical skills and prepare you for real-world scientific thinking.
七年级化学中的跨学科题目帮助你理解化学概念如何与数学、物理、生物和地理相联系。本文呈现十种典型的综合题型，每种题型都展示如何结合不同学科的知识来解决问题。练习这些题目将增强你的分析能力，并为你进行现实世界中的科学思考做好准备。
1. Mass Change Calculations in Reactions | 化学反应中的质量变化计算
When a chemical reaction produces a gas, the mass of the apparatus may appear to change. This question type blends chemistry with maths by asking you to calculate the missing mass and then explain the result using the conservation of mass. You must subtract one mass from another and then recall that atoms are never lost or gained in a chemical change.
当化学反应产生气体时，装置的质量可能看起来发生了变化。此类题目将化学与数学融合，要求你计算缺失的质量，然后用质量守恒定律解释结果。你需要用一个质量减去另一个质量，然后记住在化学变化中原子既不会消失也不会增加。
Example: A student heats 10.0 g of calcium carbonate in an open crucible. The solid residue left is 5.6 g. (a) Calculate the mass of carbon dioxide produced. (b) Explain why the final mass is lower.
例题： 一名学生在敞口坩埚中加热 10.0 g 碳酸钙。剩下的固体残渣为 5.6 g。(a) 计算产生的二氧化碳的质量。(b) 解释为什么最终质量变小了。
Answer: .4 g. (b) Carbon dioxide gas escaped into the air, so the solid left had a smaller mass. The total mass of the product gas and solid still equals the initial mass, following the law of conservation of mass.
解答： (a) 二氧化碳的质量 = 10.0 g − 5.6 g = 4.4 g。(b) 二氧化碳气体逸散到空气中，因此剩余固体的质量较小。产生的气体和固体的总质量仍然等于初始质量，这遵循了质量守恒定律。
2. States of Matter and Particle Model | 物态变化与粒子模型
This type of question asks you to explain changes such as melting, boiling or condensation using the particle model, which is a key idea in both chemistry and physics. You may need to draw or describe how the arrangement and movement of particles alter, while emphasising that the mass stays the same because the number of particles does not change.
这类题目要求你用粒子模型解释熔化、沸腾或冷凝等变化，粒子模型是化学和物理中的核心概念。你可能需要绘制或描述粒子的排列和运动如何改变，同时要强调质量保持不变，因为粒子数量没有改变。
Example: An ice cube with a mass of 20 g is left in a beaker until it melts completely. (a) Describe how the spacing and movement of water particles change. (b) Predict the mass of liquid water. Explain your answer.
例题： 一块质量为 20 g 的冰放在烧杯中直至完全融化。(a) 描述水粒子的间距和运动发生了怎样的变化。(b) 预测液态水的质量。解释你的答案。
Answer: (a) In ice, particles are held in fixed positions, close together and vibrating. As it melts, particles gain energy, move faster and are no longer in fixed positions; they can slide past one another. (b) The mass will still be 20 g, because the number of particles is unchanged and each particle still has the same mass.
解答： (a) 在冰中，粒子固定在位置上，排列紧密并振动。融化时，粒子获得能量，运动加快，不再位于固定位置，它们可以彼此滑动。(b) 质量仍然是 20 g，因为粒子的数量没有改变，并且每个粒子的质量不变。
3. Photosynthesis and Respiration Word Equations | 光合作用与呼吸作用的文字方程式
Biochemical processes such as photosynthesis and respiration involve chemical reactions that can be written as word equations. These questions link chemistry to biology by testing your ability to identify reactants and products, recognise energy changes and see how the two equations are opposites. You may also be asked to relate them to the carbon cycle.
光合作用和呼吸作用等生化过程涉及可以用文字方程式表示的化学反应。这类题目通过检测你识别反应物和产物、辨认能量变化并看出两个方程式互为逆过程的能力，将化学与生物联系在一起。你还可能需要将它们与碳循环关联起来。
Example: + ________ (+ energy). Then explain why these reactions are important for the balance of carbon dioxide in the atmosphere.
例题： 补全下列文字方程式：光合作用：________ + 水 → 葡萄糖 + ________。呼吸作用：葡萄糖 + ________ → 二氧化碳 + ________ (+ 能量)。然后解释这些反应对大气中二氧化碳平衡为何重要。
Answer: Photosynthesis: carbon dioxide + water → glucose + oxygen. Respiration: glucose + oxygen → carbon dioxide + water (+ energy). Photosynthesis removes CO₂ from the air, while respiration returns it, helping to keep the level of carbon dioxide fairly stable in nature.
解答： 光合作用：二氧化碳 + 水 → 葡萄糖 + 氧气。呼吸作用：葡萄糖 + 氧气 → 二氧化碳 + 水 (+ 能量)。光合作用从空气中移除 CO₂，而呼吸作用将其归还，这有助于保持自然界中二氧化碳的含量基本稳定。
4. Rock Weathering and Acid Rain | 岩石风化与酸雨
This integrated topic combines chemistry and geography. Acid rain, containing sulfuric and nitric acids, reacts with rocks and buildings made of limestone (calcium carbonate). You need to write word or simple symbol equations, explain how human activities cause acid rain and describe the long-term effects on landscapes and statues.
这个综合主题将化学与地理结合在一起。含有硫酸和硝酸的酸雨与石灰岩（碳酸钙）构成的岩石和建筑物发生反应。你需要写出文字或简单的符号方程式，解释人类活动如何导致酸雨，并描述对地貌和雕像的长期影响。
Example: Acid rain can damage marble statues. The main reaction is between sulfuric acid and calcium carbonate. (a) Write the word equation for this reaction. (b) Suggest why acid rain is more common in areas with many coal-fired power stations.
例题： 酸雨会损坏大理石雕像。主要反应是硫酸与碳酸钙之间的反应。(a) 写出该反应的文字方程式。(b) 说说为什么在有许多燃煤电站的地区酸雨更常见。
Answer: (a) calcium carbonate + sulfuric acid → calcium sulfate + carbon dioxide + water. (b) Burning coal releases sulfur dioxide, which reacts with water in clouds to form acid rain. More coal-fired power stations mean more sulfur dioxide, leading to more acidic rainfall.
解答： (a) 碳酸钙 + 硫酸 → 硫酸钙 + 二氧化碳 + 水。(b) 煤的燃烧释放二氧化硫，二氧化硫与云层中的水反应生成酸雨。燃煤电站越多，意味着二氧化硫越多，导致降雨酸性更强。
5. Water Purification and the Water Cycle | 水净化与水循环
Producing clean drinking water involves separation techniques such as filtration, sedimentation and chlorination, which are chemical and physical methods. Paired with geography, questions can ask you to map the water cycle onto a treatment plant or to explain why water changes state through evaporation and condensation before being collected.
生产清洁饮用水涉及过滤、沉降和氯化等分离技术，这些是化学和物理方法。与地理学结合，题目可能要求你将水循环映射到水处理厂上，或者解释为什么水在被收集之前会通过蒸发和冷凝发生物态变化。
Example: A town draws water from a river. Describe the steps needed to make this water safe to drink, naming any separation techniques. Also explain how the water cycle naturally evaporates and condenses water before it enters the river.
例题： 一个小镇从河流中取水。描述使这些水可以安全饮用的步骤，说出所用的分离技术。并解释在水进入河流之前，水循环是如何自然地蒸发和冷凝水的。
Answer: First, sedimentation allows large particles to settle. Then filtration through sand and gravel removes smaller suspended solids. Finally, chlorination kills harmful microorganisms. In nature, water evaporates from oceans, leaves behind impurities, and condenses to form clouds; when it rains, the water enters rivers, relatively pure but picks up soil and minerals.
解答： 首先，沉降使大颗粒沉淀。然后通过沙子和砾石的过滤去除较小悬浮固体。最后，氯化杀死有害微生物。在自然界中，水从海洋蒸发，留下杂质，然后冷凝形成云；降雨时，水进入河流，相对纯净但会携带土壤和矿物质。
6. Density and Material Identification | 密度与物质鉴定
en match it to a table of known metals or plastics. The formula ρ = m / V must be applied, and you need to identify the substance based on its characteristic density.
密度将化学与物理和数学联系起来。你可能会得到一个未知固体的质量和体积，要求计算其密度，然后与已知金属或塑料的表格相匹配。必须应用公式 ρ = m / V，并且根据其特征密度识别该物质。
ρ = m ÷ V
Example: g/cm³, Copper = 8.9 g/cm³. (a) Calculate the density of the block. (b) Identify the metal.
例题： 一个金属块的质量为 89 g，体积为 10 cm³。使用密度值：铝 = 2.7 g/cm³，铁 = 7.9 g/cm³，铜 = 8.9 g/cm³。(a) 计算该金属块的密度。(b) 鉴定该金属。
Answer: ivide; 10 cm³ = 8.9 g/cm³. (b) The density matches copper exactly, so the metal is copper.
解答： (a) 密度 = 89 g ÷ 10 cm³ = 8.9 g/cm³。(b) 该密度与铜完全匹配，因此该金属是铜。
7. Ores and Metal Extraction | 矿石与金属提取
Many metals are found in the Earth's crust as ores, which are compounds. Extracting metals involves reduction reactions that can be linked to geography (mining locations) and economics. In this question type, you might balance a simple equation or calculate the mass of ore needed to obtain a certain mass of metal, using the idea of percentage composition.
许多金属以矿石（即化合物）的形式存在于地壳中。金属的提取涉及还原反应，可以与地理（采矿位置）和经济学联系。在这类题目中，你可能需要配平简单的方程式，或利用百分含量的概念，计算获得一定质量金属所需的矿石质量。
Example: Haematite is an ore containing iron oxide. The simplified reaction is: iron oxide + carbon → iron + carbon dioxide. If an ore sample contains 70% iron by mass, what mass of ore would be needed to produce 140 kg of pure iron?
例题： 赤铁矿是一种含有氧化铁的矿石。简化反应为：氧化铁 + 碳 → 铁 + 二氧化碳。如果一个矿石样品含有质量分数为 70% 的铁，那么生产 140 kg 纯铁需要多少质量的矿石？
Answer: The ore is 70% ning 100 kg of ore contains 70 kg of iron. To get 140 kg of iron, you need 140 ÷ 70% helps explain the type of reaction but the calculation is a proportion problem combining chemistry and maths.
解答： 矿石含铁 70%，意味着 100 kg 矿石含 70 kg 铁。要得到 140 kg 铁，需要 140 ÷ 70% = 200 kg 矿石。化学方程式有助于说明反应类型，但计算是一个结合化学与数学的比例问题。
8. pH and the Digestive System | pH 与消化系统
The pH scale, used in chemistry to classify acids and alkalis, is critical in biology too. The stomach has a strong acidic environment (pH 1–2) for digestion, while the small intestine is alkaline. Questions may require you to interpret universal indicator colours, predict where enzymes work best and link these to pH values.
化学中用于分类酸和碱的 pH 标度在生物学中也至关重要。胃内有强酸性环境（pH 1–2）用于消化，而小肠则是碱性的。题目可能要求你解读通用指示剂的颜色，预测酶在哪里工作最好，并将这些与 pH 值联系起来。
Example: A student tests stomach juice with universal indicator and gets a red colour. (a) Estimate the pH. (b) Explain why the stomach needs this pH. (c) Later, food enters the small intestine where bile raises the pH to about 8. Name one chemical substance made by the liver that is responsible.
例题： 一名学生用通用指示剂检测胃液，得到红色。(a) 估计其 pH 值。(b) 解释为什么胃需要这样的 pH 值。(c) 之后，食物进入小肠，胆汁将 pH 值升高到约 8。说出一种由肝脏制造并负责此作用的化学物质的名称。
Answer: (a) pH 1–2, strongly acidic. (b) This acidic pH helps kill harmful bacteria swallowed with food and provides the best condition for the enzyme pepsin to break down proteins. (c) Bile (or bile salts) neutralises stomach acid and makes the conditions alkaline for enzymes in the small intestine.
解答： (a) pH 1–2，强酸性。(b) 这种酸性 pH 有助于杀死随食物吞入的有害细菌，并为胃蛋白酶分解蛋白质提供最佳条件。(c) 胆汁（或胆盐）中和胃酸，使其变为碱性，为小肠中的酶提供条件。
9. Interpreting Graphs of Reaction Rate | 解读反应速率图
Graphs showing how volume of gas or mass changes over time are common in chemistry investigations. You need mathematical skills to read axes, calculate the gradient or compare the steepness of lines. This reveals which reaction is faster or when the reaction stops, linking data analysis from maths with chemical reaction concepts.
显示气体体积或质量随时间变化的图表在化学探究中很常见。你需要数学技能来读取坐标、计算梯度或比较线条的陡度。这可以揭示哪个反应更快或反应何时停止，将数学的数据分析与化学反应概念联系起来。
Example: A student adds zinc to two different concentrations of hydrochloric acid. The graph plots volume of H₂ gas against time. Curve A is steeper and levels off earlier than curve B. (a) Which curve shows the faster reaction? (b) Explain why the reaction stops.
例题： 一位学生将锌加入两个不同浓度的盐酸中。该图绘制了 H₂ 气体体积随时间的变化。曲线 A 比曲线 B 更陡，且较早达到平顶。(a) 哪条曲线表示反应更快？(b) 解释反应为什么会停止。
Answer: (a) Curve A shows a faster reaction because its initial slope is steeper, meaning more gas is produced in the same time. (b) The reaction stops because one of the reactants (likely the zinc or the acid) has been completely used up; no more product can form.
解答： (a) 曲线 A 表示反应更快，因为其初始斜率更陡，意味着在相同时间内产生了更多气体。(b) 反应停止是因为其中一种反应物（可能是锌或酸）已被完全消耗；无法再形成产物。
10. Exothermic and Endothermic Energy Changes | 放热与吸热能量变化
Some reactions give out heat (exothermic), while others absorb heat (endothermic). These concepts appear in both chemistry and physics. Graphs of temperature against time let you see temperature rises or falls, and you must interpret them to classify the change. Real-life examples include burning fuels, neutralisation and dissolving salts.
有些反应释放热量（放热），而另一些则吸收热量（吸热）。这些概念同时出现在化学和物理中。温度-时间图可以让你看到温度上升或下降，你必须解读它们来判断变化的类型。现实生活中的例子包括燃烧燃料、中和反应和溶解盐。
Example: A beaker of sodium hydroxide solution and a beaker of ammonium nitrate solution are each measured over 5 minutes after the solute is added to water. In beaker 1, temperature rises from 20 °C to 28 °C. In beaker 2, temperature falls from 20 °C to 12 °C. (a) Identify which is exothermic and which is endothermic. (b) Give one everyday use of an exothermic reaction.
例题： 在两个烧杯中，将氢氧化钠溶液和硝酸铵溶液分别在溶质加入水中后测量 5 分钟。烧杯 1 中，温度从 20 °C 上升到 28 °C。烧杯 2 中，温度从 20 °C 下降到 12 °C。(a) 判断哪个是放热的，哪个是吸热的。(b) 给出放热反应的一个日常用途。
Answer: (a) Beaker 1 shows a temperature rise → exothermic, because energy is released to the surroundings. Beaker 2 shows a temperature drop → endothermic, because energy is absorbed from the surroundings. (b) Self-heating cans for coffee or hand warmers use exothermic reactions.
解答： (a) 烧杯 1 温度上升 → 放热，因为能量释放到周围环境中。烧杯 2 温度下降 → 吸热，因为能量从周围环境中吸收。(b) 自热咖啡罐或暖手宝都利用了放热反应。
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