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Interdisciplinary Skills in Year 8 Chemistry | 八年级化学跨学科综合题型训练

2026-07-18 06:22 · 本文部分内容由AI辅助创作

📚 Interdisciplinary Skills in Year 8 Chemistry | 八年级化学跨学科综合题型训练

Cambridge Lower Secondary Science often tests your ability to link chemistry with physics, biology, and earth science. This article provides integrated question practice that mirrors real exam-style challenges. You will strengthen both your scientific knowledge and your skill in applying it across different contexts.

剑桥初中科学考试经常考查你将化学与物理、生物、地球科学联系起来的能力。本文提供模拟真实考题的跨学科综合练习，帮助你在巩固科学知识的同时，提升在不同情境中应用知识的能力。

1. The Particle Model and Physics: Density | 粒子模型与物理：密度

A student places an unknown metal object on a balance; its mass is 78 g. She then lowers it into a measuring cylinder containing 50 cm³ of water. The water level rises to 80 cm³. Calculate the density of the metal and use the particle model to explain why the metal sinks in water.

一名学生将一块未知金属放在天平上，质量为78 g。然后她将其浸入装有50 cm³水的量筒中，水位上升至80 cm³。计算该金属的密度，并利用粒子模型解释金属为何会沉入水中。

First, find the volume of the metal: 80 cm³ – s a density of 1.0 g/cm³. In the metal, particles are packed closely in a regular arrangement, giving it a much higher mass in the same volume. In water, particles are slightly further apart. The metal sinks because its density is greater than that of water — a idea that combines the particle theory of matter with the physics of buoyancy.

首先，求金属体积：80 cm³ – 50 cm³ = 30 cm³。密度 = 质量 ÷ 体积，因此 ρ = 78 g ÷ 30 cm³ = 2.6 g/cm³。水的密度为1.0 g/cm³。在金属中，粒子紧密排列，规则有序，使相同体积内的质量更大。在水中，粒子间距稍大。金属下沉是因为其密度大于水的密度——这个结论融合了物质粒子理论与物理中的浮力原理。

2. Chemical Equations in Biology: Respiration and Photosynthesis | 生物中的化学方程式：呼吸与光合作用

The same six-carbon sugar molecule, glucose, is central to both energy release in cells and energy capture in plants. Complete the table below by writing the balanced symbol equation for each process and then explain how they form a cycle in nature.

同一种六碳糖分子——葡萄糖，既是细胞释放能量的核心，也是植物捕获能量的枢纽。完成下表中每个过程的配平化学方程式，并解释它们如何在自然界形成一个循环。

Process Word Equation Symbol Equation
Aerobic respiration glucose + oxygen → carbon dioxide + water C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Photosynthesis carbon dioxide + water → glucose + oxygen 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Respiration releases carbon dioxide and water, the very reactants needed for photosynthesis. Photosynthesis, in turn, produces glucose and oxygen, which are used in respiration. This interdependence shows how biological processes are driven by chemical transformations, linking biology to the chemistry of reversible-like, life-sustaining cycles.

呼吸作用释放二氧化碳和水，这些恰好是光合作用所需的反应物。光合作用又生成葡萄糖和氧气，供呼吸使用。这种相互依存关系表明，生物过程由化学变化驱动，将生物学与维持生命的可逆循环式的化学紧密相连。

3. Acids and Bases in Geography: Acid Rain and Rock Weathering | 地理中的酸与碱：酸雨与岩石风化

Marble, a metamorphic rock, is mainly calcium carbonate (CaCO₃). Modelling acid rain with dilute sulfuric acid (H₂SO₄) gives a vigorous reaction. Write a balanced chemical equation for the reaction and explain how this accelerates chemical weathering of statues.

大理石是一种变质岩，主要成分是碳酸钙(CaCO₃)。用稀硫酸(H₂SO₄)模拟酸雨，会发生剧烈反应。写出该反应的配平化学方程式，并解释此反应如何加速雕像的化学风化。

CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Calcium carbonate reacts with sulfuric acid to form calcium sulfate, water and carbon dioxide gas. In geography, this is a classic example of chemical weathering: acid rain (a cocktail of sulfuric and nitric acids from pollution) dissolves limestone and marble over time. The solid rock is transformed into soluble or crumbly products, eroding monuments. This question mixes rock cycle knowledge with neutralisation chemistry.

碳酸钙与硫酸反应生成硫酸钙、水和二氧化碳气体。在地理学科中，这是化学风化的典型例子：酸雨（污染产生的硫酸和硝酸混合物）会逐渐溶解石灰石和大理石。坚硬的岩石转变成可溶性或疏松的产物，侵蚀历史遗迹。这道题将岩石循环知识与酸碱中和化学融合在一起。

4. Combustion and Physics: Exothermic Energy Calculations | 燃烧与物理：放热能量计算

culate the energy released by the wax per gram and per mole (Mᵣ ≈ 352). How does this experiment demonstrate conversion of chemical energy to thermal energy?

燃烧1.0 g烛蜡（假定为C₂₅H₅₂）使100 g水温升高32 °C。水的比热容为4.2 J/g°C。计算每克蜡和每摩尔蜡释放的能量（相对分子质量约352）。该实验如何展示化学能转化为热能？

3.44 kJ/g. Energy per mole = 13 440 J/g × 352 g/mol = 4 730 880 J/mol ≈ 4 730 kJ/mol. The chemical potential energy stored in the wax' bonds is released as heat during combustion, demonstrating the first law of thermodynamics — a perfect integration of chemistry and physics.

水吸收的能量 = 质量 × 比热容 × 温度变化 = 100 g × 4.2 J/g°C × 32 °C = 13 440 J。每克蜡释放能量 = 13 440 J / 1.0 g = 13.44 kJ/g。每摩尔能量 = 13 440 J/g × 352 g/mol = 4 730 880 J/mol ≈ 4 730 kJ/mol。储存在蜡分子键中的化学势能在燃烧时以热能形式释放，体现了热力学第一定律——这是化学与物理的完美结合。

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5. Separating Mixtures and Environmental Science: Water Purification | 混合物分离与环境科学：水净化

Design a sequence of separation techniques to obtain pure water from a sample of muddy, salty river water. Your method must remove insoluble solids, dissolved salts and any microbes. Explain the scientific principle behind each step, linking to particle size and boiling point differences.

设计一套分离技术，从一份浑浊的咸河水样品中获得纯水。你的方法必须去除不溶性固体、溶解盐和任何微生物。解释每一步背后的科学原理，并将其与粒子大小和沸点差异联系起来。

This problem combines material separation techniques from chemistry with environmental engineering and public health concepts, highlighting how understanding physical properties can provide safe drinking water.

这道题将化学中的物质分离技术与环境工程和公共卫生概念相结合，突显了理解物理性质如何提供安全饮用水。

6. Metals and Geology: Extraction and Reactivity | 金属与地质：冶炼与活泼性

Iron is extracted from its ore, haematite (Fe₂O₃), in a blast furnace using carbon monoxide. Write the balanced equation for the reduction of iron(III) oxide and relate the process to the reactivity series. Explain why aluminium, a more reactive metal, cannot be extracted in the same way.

铁通过高炉中用一氧化碳还原其矿石赤铁矿(Fe₂O₃)来提取。写出还原氧化铁的配平方程式，并将该过程与金属活泼性顺序关联起来。解释为什么更活泼的金属铝不能用同样的方法提取。

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Carbon monoxide removes oxygen from iron oxide because carbon is more reactive than iron. In the reactivity series, metals above carbon require electrolysis for extraction. Aluminium is above carbon, so its oxide (Al₂O₃) must be electrolysed in molten cryolite — a method that relies on the physics of electric current driving chemical decomposition. This integrates geology, chemistry of reduction, and the industrial physics of electrolysis.

一氧化碳从氧化铁中夺走氧，因为碳比铁更活泼。在金属活泼性顺序中，位置高于碳的金属需要用电解法提取。铝的位置高于碳，因此其氧化物(Al₂O₃)必须在熔融冰晶石中电解——这种方法依赖电流驱动化学分解的物理学原理。该题综合了地质学、氧化还原化学和电解工业的物理学。

7. The Carbon Cycle and Climate: Chemistry Meets Geography | 碳循环与气候：化学与地理交汇

Explain how the combustion of fossil fuels and the decay of organic matter both return CO₂ to the atmosphere. Include balanced equations for one example of each. Why is this understanding essential for both chemists and climate scientists?

解释化石燃料的燃烧和有机物的腐烂如何都将二氧化碳归还到大气中。各举一例并写出配平方程式。为什么这种理解对化学家和气候科学家都至关重要？

on on dead plants). Both processes oxidise carbon-based compounds, releasing CO₂. In geography, the enhanced greenhouse effect is explained by the extra CO₂ from human activities trapping infrared radiation. Chemistry provides the molecular mechanism, while geography analyses the global impacts — a true interdisciplinary link.

燃烧：CH₄ + 2O₂ → CO₂ + 2H₂O（燃烧天然气）。腐烂：C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O（微生物对死植物的呼吸作用）。这两个过程都氧化碳基化合物，释放CO₂。在地理中，人为活动增加CO₂导致红外辐射被截留，从而强化温室效应。化学提供了分子层面的机制，而地理分析全球影响——这是真正的跨学科联系。

8. Mathematics in Chemistry: Calculating Reacting Masses | 化学中的数学：计算反应质量

Mg + O₂ → 2MgO. If 4.8 g of magnesium is ignited, what mass of magnesium oxide is produced? (Aᵣ: Mg = 24, O = 16). Show your working and explain the law of conservation of mass.

镁在氧气中燃烧：2Mg + O₂ → 2MgO。如果点燃4.8 g镁，生成多少克氧化镁？（相对原子质量：Mg = 24, O = 16）。写出计算过程，并解释质量守恒定律。

s (4.8 g × 5/3) = 8.0 g MgO. The total mass before and after reaction remains constant because atoms are only rearranged, not created or destroyed — key principle linking stoichiometric calculations to the quantitative nature of chemistry.

MgO的相对分子质量 = 24 + 16 = 40。由方程式，2 mol Mg (2×24=48 g) 生成2 mol MgO (2×40=80 g)。Mg与MgO的质量比为48:80=3:5。因此4.8 g Mg生成 (4.8 g × 5/3) = 8.0 g MgO。反应前后总质量保持不变，因为原子只是重新排列，不产生也不消失——这一关键原则将化学计量计算与化学的定量本质联系起来。

9. Material Science: Polymers and Innovative Design | 材料科学：聚合物与创新设计

Compare the properties and bonding of a thermoplastic like poly(ethene) with a thermosetting plastic like Bakelite. For each, suggest a practical use based on its chemical structure and describe why cross-linking changes the material's response to heat.

比较热塑性塑料如聚乙烯和热固性塑料如电木的性质与键合。对每一种塑料，根据其化学结构提出一个实际用途，并说明交联为何会改变材料对热的反应。