📚 Year 8 Edexcel Science: Interdisciplinary Mixed Question Practice | 八年级爱德思科学：跨学科综合题型训练
Welcome to this interdisciplinary mixed question practice for Year 8 Edexcel Science. This resource brings together essential skills from biology, chemistry and physics through carefully designed exam-style tasks. You will practise identifying variables, interpreting graphs, balancing equations, performing calculations, applying the particle model, analysing ecosystems, exploring acids and alkalis, and evaluating experimental methods. Each section presents a typical question followed by a step-by-step solution, with paired English and Chinese text to support learners in bilingual classrooms. Work through each example to build confidence in tackling the integrated challenges often found in Year 8 assessments.
欢迎参加八年级爱德思科学跨学科综合题型训练。本资源通过精心设计的考试风格题目，将生物学、化学和物理学中的关键技能融为一体。您将练习识别变量、解释图表、配平方程式、进行计算、应用粒子模型、分析生态系统、探索酸与碱，并评估实验方法。每个部分先给出典型问题，再展示分步解答，并以中英配对的方式呈现，以帮助双语课堂中的学习者。逐一完成例题，您将更自信地应对八年级测评中常见的综合挑战。
1. Multiple-Choice Questions on Cells and Particles | 细胞与粒子的选择题
Multiple-choice questions often test your ability to distinguish between scientific ideas from different disciplines. Read the question carefully and eliminate answers you know are wrong.
选择题常常考查你区分不同学科科学概念的能力。仔细阅读题目，并排除你认为错误的选项。
Question: Which of the following statements about living things and matter is correct?
问题：下列关于生物和物质的陈述哪一项是正确的？
- A. All cells have a cell wall.
- A. 所有细胞都有细胞壁。
- B. Atoms in a liquid are arranged in a regular pattern.
- B. 液体中的原子呈规则排列。
- C. Animal cells store genetic material inside a nucleus.
- C. 动物细胞将遗传物质储存在细胞核内。
- D. Particles in a gas are tightly packed together.
- D. 气体中的粒子紧密堆积在一起。
Correct answer: C. Animal cells are eukaryotic and contain a true nucleus where DNA is held. Option A is false because animal cells lack a cell wall. Option B is incorrect because liquids have a disordered, random arrangement of particles. Option D is wrong because gas particles are widely spaced and move freely.
正确答案：C。动物细胞是真核细胞，含有真正的细胞核，DNA 储存其中。选项 A 错误，因为动物细胞没有细胞壁。选项 B 不正确，因为液体中粒子排列无序、随机。选项 D 错误，因为气体粒子间距很大并自由运动。
This question links cell biology (Year 8 topic) with the particle model from chemistry and physics. Always check each option against what you know about both topics.
这道题将细胞生物学（八年级主题）与化学和物理中的粒子模型联系起来。始终将每个选项与您对这两个主题的了解进行核对。
2. Identifying Independent, Dependent and Control Variables | 识别自变量、因变量和控制变量
In a scientific investigation, being able to name the variables is a core skill. The independent variable is what you change, the dependent variable is what you measure, and control variables are factors you keep constant to make the test fair.
在科学探究中，能够说出变量名称是一项核心技能。自变量是你改变的变量，因变量是你测量的变量，控制变量是你为了使测试公平而保持恒定的因素。
Question: A student investigates how light intensity affects the rate of photosynthesis in pondweed. She places a lamp at different distances from the beaker and counts the number of oxygen bubbles produced per minute. Identify the independent variable, the dependent variable and two control variables in this experiment.
问题：一名学生研究光照强度如何影响水草的光合作用速率。她将灯放在离烧杯不同距离的地方，并数出每分钟产生的氧气泡数。指出该实验中的自变量、因变量和两个控制变量。
Independent variable: light intensity (varied by changing the distance of the lamp). Dependent variable: rate of photosynthesis (measured by the number of oxygen bubbles per minute). Control variables could include: temperature of the water, concentration of carbon dioxide, the same piece of pondweed, or the waiting time before counting bubbles. Keeping these constant ensures that only the light intensity affects the results.
自变量：光照强度（通过改变灯的距离来改变）。因变量：光合作用速率（以每分钟产生的氧气泡数来衡量）。控制变量可以包括：水温、二氧化碳浓度、使用同一根水草，或数气泡前的等待时间。保持这些因素不变可确保只有光照强度影响结果。
Many practical questions in Edexcel Year 8 science require you to suggest how to control variables. Think about everything that could accidentally change the outcome and explain how you would keep it the same.
在八年级爱德思科学的许多实验问题中，你需要提出如何控制变量。想想所有可能意外改变结果的因素，并说明你将如何使它们保持不变。
3. Interpreting Graphs: Temperature Change During a Neutralisation Reaction | 图线解释：中和反应中的温度变化
Graphs are a universal language in science. You need to read axis labels, recognise trends and link them to scientific explanations. A common graph shows how temperature changes over time during a chemical reaction.
图表是科学中的通用语言。你需要读坐标轴标签、识别趋势并将其与科学解释联系起来。常见的曲线图展示化学反应过程中温度随时间的变化。
Question: The graph below represents temperature against time when a student adds sodium hydroxide solution to hydrochloric acid. The temperature rises from 20 °C to a maximum of 28 °C in the first 3 minutes, then gradually falls back to room temperature. (a) Name the type of reaction that gives out heat. (b) Explain why the temperature rises. (c) Suggest why the temperature falls after 3 minutes.
问题：下图展示了学生将氢氧化钠溶液加入盐酸时温度随时间的变化。前 3 分钟温度从 20 °C 升高到最高 28 °C，然后逐渐降回室温。(a) 指出放出热量的反应类型。(b) 解释温度为什么会升高。(c) 提出 3 分钟后温度下降的原因。
(a) The reaction is exothermic. Neutralisation reactions between acids and alkalis are exothermic. (b) The temperature rises because the forming of new bonds in the products (salt and water) releases more energy than is taken in to break bonds in the reactants, so thermal energy is transferred to the surroundings. (c) After the reaction is complete, no more chemical energy is released; the hot solution loses heat to the cooler air by thermal energy transfer, so the temperature falls towards room temperature.
(a) 该反应为放热反应。酸与碱之间的中和反应是放热的。(b) 温度升高是因为生成物（盐和水）中新键形成时释放的能量大于反应物断键时吸收的能量，因此热能传递给周围环境。(c) 反应结束后不再有化学能释放；热溶液通过热能传递将热量散失到较冷的空气中，因此温度回落至室温。
When describing a graph, always quote numbers from the axes, use scientific vocabulary such as ‘exothermic’ and ‘thermal energy transfer’, and link your explanation to the underlying chemistry and physics.
在描述图表时，要始终引用坐标轴上的数据，使用“放热”和“热能传递”等科学词汇，并将你的解释与背后的化学和物理原理联系起来。
4. Balancing Chemical Equations and Conservation of Mass | 配平化学方程式与质量守恒
Balancing equations is a key chemistry skill that also reinforces the physics idea of conservation of mass. Atoms cannot be created or destroyed, so the total mass of reactants equals the total mass of products.
配平方程式是一项关键的化学技能，同时也巩固了物理学中质量守恒的概念。原子不能被创造或消灭，因此反应物的总质量等于生成物的总质量。
Question: Magnesium burns in oxygen to form magnesium oxide. The word equation is: magnesium + oxygen → magnesium oxide. The symbol equation is: Mg + O₂ → MgO. (a) Balance the symbol equation. (b) 12 g of magnesium are burnt completely. Use the idea of conservation of mass to calculate the mass of oxygen needed. (Relative atomic masses: Mg = 24, O = 16.)
问题：镁在氧气中燃烧生成氧化镁。文字方程式为：镁 + 氧气 → 氧化镁。符号方程式为：Mg + O₂ → MgO。(a) 配平该符号方程式。(b) 完全燃烧 12 g 镁。利用质量守恒的思想，计算所需氧气的质量。（相对原子质量：Mg = 24，O = 16。）
(a) To balance, we need 2 magnesium atoms reacting with 1 oxygen molecule to produce 2 units of magnesium oxide. The balanced equation is: 2Mg + O₂ → 2MgO. (b) From the balanced equation, 2 moles of Mg react with 1 mole of O₂. The mass ratio of Mg to O₂ is (2 × 24) : (2 × 16) = 48 : 32, which simplifies to 24 : 16. So every 24 g of Mg needs 16 g of oxygen. Using proportion, 12 g of Mg needs (12 ÷ 24) × 16 g = 8 g of oxygen. The total mass of reactants (12 g + 8 g) equals the total mass of product, which is 20 g of magnesium oxide, demonstrating conservation of mass.
(a) 配平后，需要 2 个镁原子与 1 个氧分子反应生成 2 个氧化镁单元。配平的方程式为：2Mg + O₂ → 2MgO。(b) 从配平方程式可知，2 mol 镁与 1 mol 氧气反应。镁与氧气的质量比为 (2 × 24) : (2 × 16) = 48 : 32，化简为 24 : 16。因此每 24 g 镁需要 16 g 氧气。按比例计算，12 g 镁需氧气 (12 ÷ 24) × 16 g = 8 g。反应物总质量 (12 g + 8 g) 等于生成物 20 g 氧化镁的质量，体现了质量守恒。
Always check your balanced equation by counting atoms on both sides. Even if you are not yet using moles fully, you can use simple ratio to solve mass problems in Year 8.
始终通过数两边原子数来检查配平的方程式。即便你在八年级还未完全使用摩尔，你也可以用简单的比例来解决质量计算问题。
5. Calculating Speed, Distance and Time | 计算速度、距离和时间
Physics calculations often appear in interdisciplinary contexts, such as describing the movement of organisms or particles. The relationship between speed, distance and time is fundamental.
物理计算常出现在跨学科情境中，例如描述生物体或粒子的运动。速度、距离和时间之间的关系是基础。
Question: A peregrine falcon dives a distance of 360 metres in 4 seconds to catch its prey. (a) Calculate its average speed in m/s. (b) Convert the speed to km/h. (c) The falcon weighs 1.2 kg. Explain whether its size or speed gives it more kinetic energy.
问题：一只游隼在 4 秒内俯冲了 360 米去捕捉猎物。(a) 计算其平均速度，单位为 m/s。(b) 将速度换算为 km/h。(c) 该游隼重 1.2 kg。解释是它的体积还是速度使它具有更大的动能。
(a) Speed = distance ÷ time = 360 m ÷ 4 s = 90 m/s. (b) To convert m/s to km/h, multiply by 3.6: 90 × 3.6 = 324 km/h. So the falcon reaches a speed of 324 km/h. (c) Kinetic energy depends on both mass and speed, but it is proportional to speed squared (KE = ½ × m × v²). A high speed (90 m/s) squared gives a large value, so the extremely fast dive dominates the kinetic energy rather than the modest mass of 1.2 kg. This answers a cross-disciplinary question linking physics of motion with biology of a predator.
(a) 速度 = 距离 ÷ 时间 = 360 m ÷ 4 s = 90 m/s。(b) 将 m/s 转换为 km/h，乘以 3.6：90 × 3.6 = 324 km/h。因此游隼达到 324 km/h 的速度。(c) 动能同时取决于质量和速度，但与速度的平方成正比（KE = ½ × m × v²）。高速（90 m/s）的平方会得到很大的值，因此极高的俯冲速度对动能起主导作用，而非其不大的 1.2 kg 质量。这道题将运动的物理知识与捕食者的生物知识联系起来。
6. Particle Model and Gas Pressure | 粒子模型与气体压强
The particle model explains behaviour in all states of matter. When we heat a gas, the particles move faster, hit the container walls more often and with greater force, increasing pressure. This concept bridges chemistry and physics.
粒子模型解释了所有物质状态的行为。当我们加热气体时，粒子运动加快，更频繁且更有力地撞击容器壁，使压强增大。这一概念连接了化学和物理学。
Question: A sealed syringe contains air at room temperature. When the syringe is placed in hot water, the plunger moves outward. Use the particle model to explain why this happens and what would occur if the syringe could not expand.
问题：一支密封的注射器在室温下装有一些空气。当把注射器放入热水中时，活塞向外移动。用粒子模型解释为什么会发生这种现象，并说明如果注射器无法膨胀会发生什么。
When the air inside the syringe is heated, the gas particles gain kinetic energy. They move faster, collide with the inner walls (including the plunger) more frequently and with greater force per collision. This raises the pressure inside the fixed volume, pushing the plunger outwards until the pressure inside balances the atmospheric pressure outside. If the plunger were locked so that the syringe could not expand, the faster-moving particles would simply collide harder with the rigid walls, causing a continued increase in pressure. If heated enough, the syringe could burst. This shows how temperature, particle motion and pressure are linked.
当注射器内的空气被加热时，气体粒子获得动能。它们运动得更快，与内壁（包括活塞）碰撞更频繁，且每次碰撞的力量更大。这使固定体积内的压强升高，将活塞向外推，直到内部压强与外部的气压相平衡。如果活塞被锁住，注射器无法膨胀，运动更快的粒子只会更强烈地撞击刚性壁，导致压强持续升高。如果加热到足够程度，注射器可能爆裂。这展示了温度、粒子运动和压强之间的关联。
In Year 8 exams, you are often asked to apply the particle model to both physical changes and simple chemical reactions. Remember to mention speed, collisions and force when explaining gas behaviour.
在八年级考试中，你经常需要应用粒子模型来解释物理变化和简单的化学反应。记住在解释气体行为时要提及速度、碰撞和作用力。
7. Food Chains and Energy Transfer in Ecosystems | 食物链与生态系统中的能量传递
Food chains show the flow of energy through an ecosystem. Only about 10% of energy is transferred from one trophic level to the next; the rest is lost as heat through respiration, uneaten parts and waste. This is a key crossover between biology and physics (energy).
食物链展示了能量在生态系统中的流动。只有大约 10% 的能量从一个营养级传递到下一级；其余的能量通过呼吸作用以热的形式散失，或因未被食用部分和废物而流失。这是生物学与物理学（能量）之间的关键交叉点。
Question: In a grassland food chain, grass traps 15,000 kJ of solar energy by photosynthesis. The food chain is: grass → grasshopper → frog → snake. Assume 10% of energy is transferred at each stage. (a) Calculate the amount of energy available to the snake. (b) Explain why so little energy reaches the top predator. Use ideas about respiration, movement and heat.
问题：在草原食物链中，草通过光合作用捕获了 15,000 kJ 的太阳能。食物链为：草 → 蚱蜢 → 青蛙 → 蛇。假设每个阶段传递 10% 的能量。(a) 计算蛇能够获得的能量值。(b) 解释为什么到达顶级捕食者的能量如此之少。请利用呼吸作用、运动和热量等概念。
(a) Energy passed from grass to grasshopper: 15,000 kJ × 0.1 = 1,500 kJ. Grasshopper to frog: 1,500 kJ × 0.1 = 150 kJ. Frog to snake: 150 kJ × 0.1 = 15 kJ. The snake obtains just 15 kJ from the original 15,000 kJ. (b) Most energy is lost at each level because organisms use energy for movement, growth and staying warm (respiration releases thermal energy that cannot be transferred). Also, not every part of an organism is eaten, and some energy is lost in undigested waste. This inefficiency limits the length of food chains.
(a) 从草传递到蚱蜢的能量：15,000 kJ × 0.1 = 1,500 kJ
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