AS AQA International AS Chemistry CH01 Unit 1 Example Responses | AS AQA 国际化学 CH01 第一单元示例答案

📚 AS AQA International AS Chemistry CH01 Unit 1 Example Responses | AS AQA 国际化学 CH01 第一单元示例答案

This article walks through example responses for AQA International AS Chemistry Unit 1 (CH01). It focuses on how examiners award marks, how to structure definitions, calculations and explanations, and how to avoid common errors that often appear in candidate answers.

本文梳理 AQA 国际 AS 化学第一单元（CH01）的示例答案，重点讲解考官如何给分、如何组织定义题、计算题和解释题，以及如何避免考生答案中常见的失分点。

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1. How to Use Model Responses | 如何使用示例答案

Model answers are not just "correct facts"; they show how command words control the depth of response. For "state", a short accurate phrase is enough. For "explain", you must link cause to consequence using scientific ideas such as charge, orbital energy or collision frequency.

示例答案不只是 "正确事实"；它们体现题干指令词如何控制答题深度。"state" 只需给出简短准确的说法；"explain" 则必须用电荷、轨道能量或碰撞频率等科学概念把因果连接起来。

In CH01, typical command words include: define, state, calculate, explain, suggest and compare. Each requires a different style of answer. The table below summarises what examiners expect.

CH01 中常见的指令词包括：define、state、calculate、explain、suggest 和 compare。每个词要求不同的答题方式。下表总结了考官期望。

Command word	Expected response	Example
Define	Exact meaning, often with key conditions	Define first ionisation energy
Calculate	Clear steps, units and significant figures	Calculate the empirical formula
Explain	Use scientific reasons, not just restate the question	Explain why aluminium has a lower first ionisation energy than magnesium
Compare	Give similarities and differences	Compare the bonding in sodium chloride and iodine

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2. Atomic Structure and Mass Spectrometry | 原子结构与质谱

Relative atomic mass (Aᵣ) is the weighted mean mass of an atom of an element relative to 1/12th the mass of one carbon-12 atom. It has no units because it is a ratio.

相对原子质量（Aᵣ）是指一种元素的一个原子的加权平均质量与一个碳-12 原子质量的 1/12 的比值。它是一个比值，因此没有单位。

A mass spectrometer gives the relative abundance of isotopes. The main stages are ionisation, acceleration, deflection in a magnetic field and detection. Heavier ions and ions with smaller charge are deflected less.

质谱仪可以给出同位素的相对丰度。其主要步骤是电离、加速、在磁场中偏转和检测。质量越大、电荷越小的离子偏转越少。

Aᵣ = Σ (isotope mass × abundance) ÷ 100

Example: naturally occurring copper contains 69.2% copper-63 and 30.8% copper-65. Aᵣ = (63 × 69.2 + 65 × 30.8) ÷ 100 = 63.6.

示例：天然铜含有 69.2% 的铜-63 和 30.8% 的铜-65。Aᵣ = (63 × 69.2 + 65 × 30.8) ÷ 100 = 63.6。

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3. Ionisation Energy Trends | 电离能趋势

First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous +1 ions. The equation is written with state symbols.

第一电离能是指从一摩尔气态原子中移走一摩尔电子，形成一摩尔气态 +1 离子所需的能量。书写方程式时必须标注状态符号。

X(g) → X⁺(g) + e⁻

Across Period 3 from sodium to argon, first ionisation energy generally increases because nuclear charge increases while shielding remains similar. However, aluminium has a lower value than magnesium because the outer electron in Al is in a higher-energy 3p orbital, while Mg loses from 3s. Sulfur has a lower value than phosphorus because paired p electrons repel each other, making one electron easier to remove.

从钠到氩的第三周期中，第一电离能总体上升，因为核电荷增加而屏蔽作用变化不大。但铝的电离能低于镁，因为铝的外层电子位于能量更高的 3p 轨道，而镁失去的是 3s 电子。硫的电离能低于磷，因为硫的 p 轨道中存在成对电子排斥，使其中一个电子更容易被移走。

Element	Na	Mg	Al	Si	P	S	Cl	Ar
IE / kJ mol⁻¹	496	738	578	789	1012	1000	1251	1521

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4. Amount of Substance and Empirical Formula | 物质的量与经验式

nks mass, concentration and gas volume through the equations n = m/M, n = cV and pV = nRT.

物质的量以摩尔为单位。一摩尔含有 6.02 × 10²³ 个粒子。摩尔通过 n = m/M、n = cV 和 pV = nRT 把质量、浓度和气体体积联系起来。

Example: a compound contains 40.0% carbon, 6.7% hydrogen and 53.3% ividing by the smallest value gives a ratio of 1 : 2 : 1, so the empirical formula is CH₂O.

示例：某化合物按质量含碳 40.0%、氢 6.7%、氧 53.3%。假设取 100 g，n(C) = 40.0 ÷ 12.0 = 3.33 mol，n(H) = 6.7 ÷ 1.0 = 6.7 mol，n(O) = 53.3 ÷ 16.0 = 3.33 mol。除以最小值得到比例 1 : 2 : 1，因此经验式为 CH₂O。

als if R = 8.31 J K⁻¹ mol⁻¹ is used. A common error is using °C or kPa without converting.

气体计算中，如果使用 R = 8.31 J K⁻¹ mol⁻¹，温度必须转换为开尔文，压力必须转换为帕斯卡。常见错误是直接使用摄氏度或千帕而没有换算。

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5. Bonding and Structure | 化学键与结构

Sodium chloride is an ionic compound with a giant ionic lattice. It conducts electricity when molten or in solution because ions are free to move and carry charge, but it does not conduct as a solid because ions are fixed in the lattice.

氯化钠是离子化合物，具有巨型离子晶格。熔融或溶于水时能导电，因为离子可以自由移动并携带电荷；固态时不能导电，因为离子被固定在晶格中。

Iodine is a simple molecular covalent substance. It has a low melting point because only weak intermolecular forces need to be overcome between molecules, even though the covalent bonds within each I₂ molecule are strong.

碘是简单分子共价物质。它的熔点较低，因为分子间只需克服微弱的分子间作用力，而每个 I₂ 分子内部的共价键虽然很强但不会断裂。

Type	Particles	Forces broken on melting	Electrical conductivity
Ionic	Positive and negative ions	Strong electrostatic forces	Conducts when molten or dissolved
Simple molecular	Molecules	Weak intermolecular forces	Does not conduct
Metallic	Positive ions and delocalised electrons	Metallic bonds	Conducts as solid and liquid

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6. Electronegativity and Intermolecular Forces | 电负性与分子间作用力

Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond. In hydrogen fluoride, fluorine is more electronegative than hydrogen, so the bond is polar: H δ⁺ — F δ⁻.

电负性是指原子吸引共价键中成键电子对的能力。在氟化氢中，氟的电负性大于氢，因此该键具有极性：H δ⁺ — F δ⁻。

Hydrogen fluoride has an unexpectedly high boiling point because molecules form hydrogen bonds. Hydrogen bonding only occurs when H is bonded to N, O or F, and there is a lone pair on the electronegative atom. Hydrogen chloride, hydrogen bromide and hydrogen iodide have dipole-dipole and London forces, which increase with molecular size and electron number.

氟化氢的沸点异常高，因为分子间形成氢键。氢键只有在 H 与 N、O 或 F 成键，且电负性原子上有孤对电子时才会出现。氯化氢、溴化氢和碘化氢之间存在偶极-偶极力和伦敦力，且这些作用随分子大小和电子数增加而增强。

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7. Hess's Law and Enthalpy Changes | 赫斯定律与焓变

Hess's law states that the total enthalpy change for a reaction is independent of the route taken. This allows unknown enthalpy changes to be calculated from formation or combustion data.

赫斯定律指出，一个反应的总焓变与反应路径无关。这使得我们可以利用生成焓或燃烧焓数据计算未知的焓变。

For formation data, the enthalpy change of reaction is the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. Elements in their standard states have zero enthalpy of formation.

对于生成焓数据，反应的焓变等于生成物标准生成焓之和减去反应物标准生成焓之和。处于标准状态的单质其生成焓为零。

ΔHᵣ = Σ ΔHf° (products) − Σ ΔHf° (reactants)

−394 kJ mol⁻¹, H₂O = −286 kJ mol⁻¹. ΔH = [(−394) + 2(−286)] − [(−75) + 0] = −891 kJ mol⁻¹.

示例：甲烷燃烧反应 CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)，使用生成焓值 CH₄ = −75 kJ mol⁻¹、CO₂ = −394 kJ mol⁻¹、H₂O = −286 kJ mol⁻¹。ΔH = [(−394) + 2(−286)] − [(−75) + 0] = −891 kJ mol⁻¹。

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8. Maxwell–Boltzmann Distribution and Kinetics | 麦克斯韦–玻尔兹曼分布与动力学

The Maxwell–Boltzmann distribution shows the spread of molecular energies in a gas at a given temperature. Only particles with energy equal to or greater than the activation energy, Eₐ, can react when they collide with the correct orientation.

麦克斯韦–玻尔兹曼分布显示气体在一定温度下分子能量的分布。只有能量等于或大于活化能 Eₐ 的粒子，在取向正确时碰撞才能发生反应。

When temperature increases, the distribution flattens and shifts to the right. A much larger proportion of particles now have energy greater than or equal to Eₐ, so the frequency of successful collisions increases and the rate increases significantly.

当温度升高时，分布曲线变得平坦并向右移动。此时能量大于或等于 Eₐ 的粒子比例大大增加，因此有效碰撞频率增加，反应速率显著提高。

A catalyst provides an alternative reaction route with a lower activation energy. This does not

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