AS AQA Chemistry Paper 1 June 2022: Full Paper Walkthrough & Revision Guide | AS AQA 化学 2022年6月试卷1 全面解析与复习指南

📚 AS AQA Chemistry Paper 1 June 2022: Full Paper Walkthrough & Revision Guide | AS AQA 化学 2022年6月试卷1 全面解析与复习指南

The June 2022 AQA AS Chemistry Paper 1 (7404/1) was the first full sitting of the new specification after the pandemic, covering physical and inorganic chemistry alongside practical skills. This paper tested students on atomic structure, amount of substance, bonding, energetics, kinetics, equilibria, redox chemistry, and the periodic table. In this comprehensive walkthrough, we will analyse the key question types, revisit critical concepts, and highlight the pitfalls that cost marks across the paper.

2022年6月AQA AS化学试卷1（7404/1）是疫情后新大纲的首次完整考试，涵盖物理化学、无机化学及实验技能。本试卷考查了原子结构、物质的量、化学键、能量学、动力学、化学平衡、氧化还原反应和元素周期表。在本次全面解析中，我们将逐一分析主要题型，回顾关键概念，并指出该试卷中常见的失分陷阱。

1. Paper Structure & Marking | 试卷结构与评分

The paper is 1 hour 30 minutes long and carries 80 marks, representing 50% of the total AS Chemistry qualification. Section A consists of 40 multiple-choice questions worth 1 mark each, while Section B contains structured and extended-response questions worth approximately 40 marks. Students must answer all questions in both sections, and a data sheet containing the periodic table and physical constants is provided.

本试卷时长1小时30分钟，满分80分，占AS化学总成绩的50%。A部分包含40道单项选择题，每题1分；B部分为结构化和扩展回答题，合计约40分。考生必须回答两个部分的全部题目，考试会提供包含元素周期表和物理常数的数据表。

The June 2022 paper was notable for its emphasis on application of knowledge to unfamiliar contexts. The mean marks for Section A were slightly lower than previous years, suggesting the multiple-choice questions demanded deeper conceptual understanding rather than simple recall.

2022年6月的试卷特别注重将知识应用于陌生情境。A部分的平均分略低于往年，说明选择题更强调深层概念理解而非简单记忆。

Section A: 40 multiple-choice questions, 1 mark each | A部分：40道选择题，每题1分
Section B: Structured questions, ~40 marks | B部分：结构化问题，约40分
Allow 2 minutes per MCQ and 12 minutes for each 8-mark extended question | 选择题每题2分钟，8分扩展题每题12分钟

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

The first section of the paper tested fundamental atomic theory. Students were required to state the relative masses and charges of protons, neutrons and electrons — a question that proved straightforward but essential. The key values are: proton (mass 1, charge +1), neutron (mass 1, charge 0), and electron (mass 1/1840, charge −1).

试卷第一部分考查了基础原子理论。题目要求学生写出质子、中子和电子的相对质量与电荷——这看似简单却至关重要。关键数值为：质子（质量1，电荷+1），中子（质量1，电荷0），电子（质量1/1840，电荷−1）。

A more challenging question involved mass spectrometry data. Given the isotopic abundance of chlorine, students had to calculate the relative atomic mass (Aᵣ). The spectrum of Cl₂ shows peaks at m/z values of 70, 72 and 74, corresponding to ³⁵Cl–³⁵Cl, ³⁵Cl–³⁷Cl and ³⁷Cl–³⁷Cl respectively, with intensity ratios of 9:6:1.

更具挑战性的题目涉及质谱数据。给定氯的同位素丰度，学生需要计算相对原子质量（Aᵣ）。Cl₂的质谱在m/z值为70、72和74处出现峰，分别对应³⁵Cl–³⁵Cl、³⁵Cl–³⁷Cl和³⁷Cl–³⁷Cl，强度比为9:6:1。

The formula for Aᵣ is: Aᵣ = Σ(isotopic mass × fractional abundance) | Aᵣ的计算公式为：Aᵣ = Σ(同位素质量 × 丰度分数)

Students frequently confused the molecular ion peaks of Cl₂ with atomic peaks. The accelerator voltage in a mass spectrometer determines the kinetic energy of ions, not their mass — a common misconception in multiple-choice questions.

学生经常将Cl₂的分子离子峰与原子峰混淆。质谱仪中的加速电压决定离子的动能而非质量——这是选择题中常见的误解点。

First ionisation energy: energy required to remove 1 mole of electrons from 1 mole of gaseous atoms | 第一电离能：从1摩尔气态原子中移除1摩尔电子所需的能量
Second ionisation energy is always larger because the electron is removed from a smaller, more positively charged ion | 第二电离能总是更大，因为电子是从更小、正电荷更高的离子中移出的
Mass spectrometry can determine relative atomic mass, relative molecular mass and fragmentation patterns | 质谱可测定相对原子质量、相对分子质量及碎片模式

3. Amount of Substance: The Mole | 物质的量：摩尔

No AQA chemistry paper is complete without rigorous mole calculations, and June 2022 was no exception. Students were asked to calculate the number of moles in a given mass of anhydrous sodium carbonate, using the equation n = m/M. For a 2.65 g sample of Na₂CO₃ (M = 106.0 g mol⁻¹), the correct answer is n = 2.65 ÷ 106.0 = 0.0250 mol.

AQA化学试卷一定包含严格的摩尔计算，2022年6月也不例外。题目要求计算给定质量的无水碳酸钠中的物质量，使用公式n = m/M。对于2.65 g的Na₂CO₃样品（M = 106.0 g mol⁻¹），正确答案为n = 2.65 ÷ 106.0 = 0.0250 mol。

The ideal gas equation pV = nRT appeared in a calculation involving the volume of gas evolved from a reaction. Students were reminded that the gas constant R = 8.31 J K⁻¹ mol⁻¹, and that pressure must be converted from kPa to Pa before substitution. A common error was using temperature in degrees Celsius rather than kelvin, yielding answers in error by a factor of 273.

理想气体方程pV = nRT出现在一个涉及反应产生气体体积的计算题中。题目提醒学生气体常数R = 8.31 J K⁻¹ mol⁻¹，且压力必须从kPa转换为Pa才能代入。一个常见错误是使用摄氏温度而非开尔文温度，导致答案偏差273倍。

pV = nRT | where p in Pa, V in m³, n in mol, T in K | p的单位为Pa，V的单位为m³，n的单位为mol，T的单位为K

For empirical formula questions, the June 2022 paper presented a compound containing C, H and O. Students had to divide each mass by its relative atomic mass, then divide by the smallest value to obtain the simplest whole-number ratio. Hydrated salts asked students to determine the value of x in CuSO₄·xH₂O from the mass loss on heating.

对于实验式问题，2022年6月试卷给出了一种含C、H、O的化合物。学生必须将各元素质量除以各自的相对原子质量，再除以最小值以获得最简整数比。水合盐题目要求学生通过加热后的质量损失来确定CuSO₄·xH₂O中的x值。

Molar volume at room temperature and pressure: 24.0 dm³ mol⁻¹ | 室温常压下的摩尔体积：24.0 dm³ mol⁻¹
Concentration formula: c = n/V, where V is in dm³ | 浓度公式：c = n/V，其中V以dm³为单位
Titration calculations require working down to 4 significant figures | 滴定计算需精确到4位有效数字

4. Bonding & Intermolecular Forces | 化学键与分子间作用力

The bonding section tested students’ ability to draw dot-and-cross diagrams. The June 2022 paper required the diagram for magnesium oxide (MgO), showing the transfer of two electrons from magnesium to oxygen. Crucially, the charges (Mg²⁺ and O²⁻) and the completed outer shells of both ions must be clearly labelled. Unlabelled charges cost students marks in this question.

化学键部分考查了学生绘制电子点叉图的能力。2022年6月试卷要求画出氧化镁（MgO）的电子图，显示两个电子从镁转移到氧。关键在于必须清晰标注两个离子的电荷（Mg²⁺和O²⁻）及完整的外层电子壳。未标注电荷是本题的主要失分点。

Electronegativity questions asked students to explain why the O–H bond is polar. The answer requires reference to the dipole formed when oxygen, being more electronegative than hydrogen, attracts the bonding electron pair more strongly. Students also had to identify the strongest intermolecular force present in water — hydrogen bonding — and explain how it affects boiling point.

电负性题目要求学生解释为什么O–H键是极性的。回答需要指出：氧的电负性大于氢，更强烈地吸引成键电子对，从而形成偶极。学生还需识别水中存在的最强分子间作用力——氢键——并解释它如何影响沸点。

The distinction between permanent dipole–dipole forces and London (dispersion) forces was tested in a comparison question. London forces arise from instantaneous dipoles caused by fluctuating electron distributions, whereas permanent dipole–dipole forces arise between polar molecules. Students who wrote “Van der Waals’ forces” without further classification failed to gain full marks.

永久偶极–偶极作用力与伦敦（色散）力的区别在比较题中考查。伦敦力源于电子分布波动引起的瞬时偶极；而永久偶极–偶极作用力存在于极性分子之间。仅写”范德华力”而未进一步分类的学生不能得满分。

Hydrogen bonds occur between N–H, O–H and F–H bonds | 氢键存在于N–H、O–H和F–H键之间
Metallic bonding: electrostatic attraction between positive ions and delocalised electrons | 金属键：正离子与离域电子之间的静电引力
Giant covalent structures (diamond, graphite, silicon dioxide) have high melting points due to strong covalent bonds | 巨型共价结构（金刚石、石墨、二氧化硅）因强共价键而具有高熔点

5. Energetics & Hess’s Law | 能量学与赫斯定律

Calorimetry was assessed through a practical-based question. Students were given data from a combustion experiment of methanol and asked to calculate the enthalpy change of combustion using q = mcΔT. With 200 cm³ of water and a temperature rise of 15.5 °C, the heat transferred is:

量热法通过一道实验情境题进行考查。题目给出甲醇燃烧实验的数据，要求学生使用q = mcΔT计算燃烧焓变。对于200 cm³水和15.5 °C的温升，传递的热量为：

q = m × c × ΔT = 200 g × 4.18 J g⁻¹ K⁻¹ × 15.5 K = 12,958 J = 12.96 kJ

Students then divided by the number of moles of methanol burnt (0.0250 mol) to obtain ΔH = −518 kJ mol⁻¹. The negative sign is essential because combustion is exothermic. The question also required students to suggest why the experimental value is less exothermic than the data-book value: heat is lost to the surroundings, incomplete combustion, and non-standard conditions.

学生随后除以燃烧的甲醇物质量（0.0250 mol）得到ΔH = −518 kJ mol⁻¹。负号至关重要，因为燃烧是放热反应。题目还要求学生解释为什么实验值比数据手册值放热更少：热量散失到环境中、燃烧不完全、以及非标准条件。

Hess’s law was tested with a Born–Haber style cycle for the formation of magnesium oxide. The key principle is that the enthalpy change of a reaction is independent of the route taken, allowing students to calculate lattice enthalpy or enthalpy of formation via bond-breaking and bond-forming pathways.

赫斯定律通过一个类似Born–Haber循环的题目来考查氧化镁的生成。核心原理是反应焓变与路径无关，学生可以通过断键和成键路径来计算晶格焓或生成焓。

When dealing with bond dissociation enthalpies, students must remember that bond breaking is endothermic (positive) and bond making is exothermic (negative). The equation used is:

在处理键解离焓时，学生必须记住断键是吸热（正值），成键是放热（负值）。所用公式为：

ΔH = Σ(bond enthalpies of reactants) − Σ(bond enthalpies of products) | ΔH = Σ(反应物键焓) − Σ(生成物键焓)

Standard enthalpy of formation: formation of 1 mole of compound from its elements in standard states | 标准生成焓：在标准状态下从单质生成1摩尔化合物
Standard enthalpy of combustion: complete combustion of 1 mole of substance in oxygen | 标准燃烧焓：1摩尔物质在氧气中完全燃烧
Always quote units (kJ mol⁻¹) and signs in enthalpy answers | 焓变答案务必注明单位（kJ mol⁻¹）和正负号

6. Kinetics: Collision Theory | 动力学：碰撞理论

Kinetics questions on the June 2022 paper focused on factors affecting reaction rate. Students were asked to explain, using collision theory, why increasing the concentration of a reactant increases the rate of reaction. The expected answer: at higher concentrations, there are more particles per unit volume, so collisions between reactant particles occur more frequently, leading to a higher frequency of successful collisions.

2022年6月试卷的动力学问题聚焦于影响反应速率的因素。题目要求学生用碰撞理论解释为什么增加反应物浓度会提高反应速率。预期答案为：浓度越高，单位体积内粒子数越多，反应物粒子之间的碰撞更加频繁，从而有效碰撞的频率更高。

A graph question showed the Maxwell–Boltzmann distribution at two temperatures. Students had to recognise that at the higher temperature, the curve shifts to the right and becomes lower and flatter, while the total area under the curve remains constant because it represents the total number of molecules. The fraction of molecules with kinetic energy equal to or greater than the activation energy (Eₐ) increases markedly.

一道图形题展示了两种温度下的Maxwell–Boltzmann分布。学生必须认识到：温度升高时，曲线右移且变得更低更平缓，但曲线下总面积保持不变，因为面积代表分子总数。动能等于或超过活化能（Eₐ）的分子比例显著增加。

The paper also asked students to interpret the effect of a catalyst. A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the proportion of molecules with sufficient energy to react. Common errors included stating that catalysts are “used up” or that they increase the kinetic energy of particles — both incorrect, as catalysts lower Eₐ rather than changing particle velocities.

试卷还要求学生解释催化剂的作用。催化剂提供了具有更低活化能的替代反应路径，增加了具有足够能量反应的分子比例。常见错误包括说催化剂”被消耗”或增加粒子动能——两者都不正确，因为催化剂降低的是Eₐ而非改变粒子速度。

Maxwell–Boltzmann curve always starts at the origin and approaches the x-axis asymptotically | Maxwell–Boltzmann曲线总从原点开始并渐近逼近x轴
Activation energy: the minimum energy required for a collision to result in reaction | 活化能：碰撞导致反应所需的最小能量
Catalysts are specific: different reactions require different catalysts | 催化剂具有选择性：不同反应需要不同催化剂

7. Chemical Equilibrium | 化学平衡

The equilibrium section of the paper examined Le Chatelier’s principle in the context of the industrially important reaction for the production of ethanol:

试卷的平衡部分在工业上重要的乙醇合成反应情境中考查勒夏特列原理：

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) ΔH = −46 kJ mol⁻¹

Students were asked to predict the effect of increasing pressure on the position of equilibrium. Since there are 2 moles of gas on the left and 1 mole on the right, increasing pressure shifts the equilibrium to the right, favouring the side with fewer gas molecules — in this case, ethanol production. This shifts the equilibrium to the right in the forward direction.

题目要求学生预测增加压力对平衡位置的影响。由于左侧有2摩尔气体，右侧有1摩尔气体，增加压力会使平衡向右侧移动，即向气体分子数较少的一侧移动——本例中为乙醇生成方向。平衡向正反应方向移动。

Temperature effects were also tested. Because the forward reaction is exothermic (ΔH = −46 kJ mol⁻¹), a decrease in temperature favours the forward reaction, increasing the yield of ethanol. However, students also needed to recognise the industrial trade-off: lower temperatures reduce the rate of reaction, so an optimum temperature of around 300 °C is used in practice with an iron catalyst.

温度的影响也受到考查。因为正反应为放热反应（ΔH = −46 kJ mol⁻¹），降低温度有利于正反应，提高乙醇产率。然而，学生还需要认识到工业上的权衡：低温会降低反应速率，因此实际工业生产中约300 °C使用铁催化剂作为最佳温度。

An equilibrium constant (Kc) calculation required students to set up an ICE table for the Haber process. For the equation N₂ + 3H₂ ⇌ 2NH₃, given equilibrium concentrations of [N₂] = 0.25 mol dm⁻³, [H₂] = 0.75 mol dm⁻³ and [NH₃] = 0.50 mol dm⁻³:

一道平衡常数（Kc）计算题要求学生为哈伯法建立ICE表格。对于方程N₂ + 3H₂ ⇌ 2NH₃，给定平衡浓度[N₂] = 0.25 mol dm⁻³，[H₂] = 0.75 mol dm⁻³，[NH₃] = 0.50 mol dm⁻³：

Kc = [NH₃]² ÷ ([N₂] × [H₂]³) = (0.50)² ÷ (0.25 × 0.75³) = 0.25 ÷ 0.1055 = 2.37 mol⁻² dm⁶

Students frequently forgot to square or cube the concentrations according to the stoichiometric coefficients, or omitted the units of Kc. The units must be derived from the equilibrium expression: mol⁻² dm⁶ in this case.

学生经常忘记根据化学计量系数对浓度取平方或立方，或省略Kc的单位。单位必须根据平衡表达式推导：本例中为mol⁻² dm⁶。

Only temperature changes the value of Kc | 只有温度会改变Kc的值
Adding a catalyst does not affect the position of equilibrium or Kc | 加入催化剂不影响平衡位置或Kc
Kc represents the equilibrium constant of the forward reaction as written | Kc表示正向反应（按书写方程）的平衡常数

8. Redox Reactions & Oxidation States | 氧化还原反应与氧化态

Redox chemistry appeared in both sections of the June 2022 paper. Students were required to assign oxidation states to elements in a range of species, including MnO₄⁻, Cr₂O₇²⁻ and SO₂. The rules are: the oxidation state of a free element is zero; in ions, the sum equals the ionic charge; and the sum in a neutral molecule is zero.

氧化还原化学在2022年6月试卷的两个部分中均出现。学生需要为多种物质中的元素指定氧化态，包括MnO₄⁻、Cr₂O₇²⁻和SO₂。规则为：游离元素氧化态为零；离子中氧化态之和等于离子电荷；中性分子中氧化态之和为零。

For MnO₄⁻: oxygen is −2 each, and with four oxygens the total is −8. To give an overall charge of −1, manganese must be +7. Similarly, in Cr₂O₇²⁻, seven oxygens contribute −14, so each chromium is +6. Students are advised to write out these calculations in the exam.

对于MnO₄⁻：每个氧为−2，四个氧总计−8。要使总电荷为−1，锰必须为+7。同理，在Cr₂O₇²⁻中，七个氧贡献−14，因此每个铬为+6。建议考生在考试中写出这些计算过程。

Balancing redox equations by the oxidation-state method was a high-tariff question. Taking the reaction between iodide ions and acidified dichromate:

通过氧化态方法配平氧化还原方程是高分值题目。以碘离子与酸化重铬酸盐之间的反应为例：

Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O

The half-equation method involves balancing atoms and charges in each half separately. For the oxidation half: 2I⁻ → I₂ + 2e⁻. For the reduction half: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Multiplying the oxidation half by 3 and adding yields the balanced overall equation with 6 electrons cancelled.

半反应法需要分别配平每个半反应的原子和电荷。氧化半反应：2I⁻ → I₂ + 2e⁻。还原半反应：Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O。将氧化半反应乘以3并相加，抵消6个电子，即可得到配平的总体方程。

Students also had to identify the oxidising agent and reducing agent in a given reaction. The oxidising agent is the species that is itself reduced (gains electrons), while the reducing agent is the species that is oxidised (loses electrons).

学生还需在给定反应中识别氧化剂和还原剂。氧化剂是自身被还原（获得电子）的物质，而还原剂是被氧化（失去电子）的物质。

Oxidation is loss of electrons; reduction is gain of electrons (OIL RIG) | 氧化是失电子；还原是得电子
In acidified potassium manganate, the colour change is purple (MnO₄⁻) to colourless (Mn²⁺) | 酸性高锰酸钾中，颜色变化为紫色（MnO₄⁻）到无色（Mn²⁺）
Oxidation state of oxygen is −2 in most compounds, except peroxides (−1) and OF₂ (+2) | 大多数化合物中氧的氧化态为−2，过氧化物为−1，OF₂中为+2

9. Periodicity & Group 2 Chemistry | 元素周期律与第二主族

The final content-heavy section covered periodic trends and the chemistry of Group 2 metals. Students were asked to explain the trend in atomic radius across Period 3. The expected answer: as the nuclear charge increases across the period, the electrons are added to the same principal energy level (n = 3), so the effective nuclear charge experienced by outer electrons increases, pulling them closer to the nucleus and decreasing atomic radius.

最后一个内容密集的部分涵盖周期趋势和第二主族金属化学。题目要求学生解释第三周期原子半径趋势。预期答案为：随着核电荷在周期中增加，电子被添加到同一主能级（n = 3），外层电子感受到的有效核电荷增加，将它们拉向原子核，导致原子半径减小。

First ionisation energy trends were tested with a graph. The general increase across Period 3 is interrupted by two drops: from magnesium to aluminium (because the electron removed from aluminium is in a 3p orbital, which is at a higher energy level than 3s), and from phosphorus to sulfur (because the electron removed from sulfur is paired in a 3p orbital, resulting in extra electron–electron repulsion).

第一电离能趋势通过图表考查。第三周期的总上升趋势被两个下降点中断：从镁到铝（因为铝中被移除的电子在3p轨道，能量高于3s），以及从磷到硫（因为硫中被移除的电子在3p轨道中成对，产生额外的电子–电子排斥）。

Group 2 reactions with water were examined. Magnesium reacts slowly with cold water but more readily with steam, producing magnesium oxide and hydrogen. The solubility trend of Group 2 hydroxides increases down the group; consequently, magnesium hydroxide produces a slightly alkaline solution while barium hydroxide is more soluble and forms a solution with higher pH. This links to their use as antacids.

第二主族与水的反应受到考查。镁与冷水反应缓慢，但与水蒸气反应更剧烈，生成氧化镁和氢气。第二主族氢氧化物的溶解度向下递增；因此，氢氧化镁产生微碱性溶液，而氢氧化钡更易溶，形成pH更高的溶液。这与它们作为抗酸剂的应用相关。

A precipitation question involved adding sodium hydroxide to solutions of Group 2 metal ions. Students needed to recognise that all Group 2 hydroxides are sparingly soluble, appearing as white precipitates. However, calcium hydroxide is more soluble than magnesium hydroxide, so a saturated solution of calcium hydroxide is used in limewater testing for carbon dioxide.

一道沉淀题涉及向第二主族金属离子溶液中加入氢氧化钠。学生需要认识到所有第二主族氢氧化物微溶于水，表现为白色沉淀。然而，氢氧化钙比氢氧化镁溶解度更大，因此氢氧化钙饱和溶液用于石灰水检验二氧化碳。

Group 2 metals are reducing agents: their reactivity increases down the group | 第二主族金属是还原剂：其反应活性向下递增
Thermal stability of Group 2 carbonates increases down the group | 第二主族碳酸盐的热稳定性向下递增
Melting points of Period 3 elements increase from Na to Si, then sharply decrease | 第三周期元素熔点从Na到Si递增，然后急剧下降

10. Exam Technique & Common Pitfalls | 答题技巧与常见错误

Analysis of examiner reports for the June 2022 paper reveals several recurring issues that cost students marks. The most common mistakes in Section A involved careless reading of units, particularly confusion between kJ and J, or between dm³ and cm³. In titration calculations, students must ensure that volumes are converted to dm³ by dividing by 1000 before substitution into concentration equations.

对2022年6月试卷的考官报告分析揭示了几个反复出现导致失分的问题。A部分最常见的错误是对单位的粗心阅读，特别是kJ与J、或dm³与cm³的混淆。在滴定计算中，学生必须确保体积先除以1000换算为dm³，再代入浓度方程。

For extended-response questions (6 marks), the AQA mark schemes typically allocate 1 mark for each discrete correct point. Students should bullet-point their answers to ensure every relevant point is visible to the examiner. Excessive irrelevant information is not penalised directly but wastes precious time.

对于扩展回答题（6分），AQA评分标准通常为每个独立的正确要点分配1分。学生应该用要点形式列出答案，确保每个相关要点都能被考官看到。过多无关信息不会被直接扣分，但会浪费宝贵时间。

In enthalpy calculations, the most common error was omitting the negative sign for exothermic reactions. In equilibrium questions, students frequently stated that concentration changes Kc, which is incorrect — only temperature can change Kc for a given reaction. Always quote units derived from the equilibrium expression.

在焓变计算中，最常见的错误是省略放热反应的负号。在平衡问题中，学生经常错误地说浓度会改变Kc——实际上只有温度能改变给定反应的Kc。务必注明从平衡表达式推导出的单位。

Finally, for multiple-choice questions, the AQA data sheet is copied into the paper, so all constants required will be provided. Students who memorised key values such as Avogadro’s constant (6.022 × 10²³ mol⁻¹) and the gas constant (8.31 J K⁻¹ mol⁻¹) had an advantage, allowing them to spot impossible answer options immediately. The 2022 paper rewarded students who showed all working, since error carried forward (ECF) marks were applied generously where methods were correct.

最后，对于选择题，AQA数据表随试卷提供，所需全部常数均会给出。记忆关键数值（如阿伏加德罗常数6.022 ×

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AS AQA Chemistry Paper 1 June 2022: Full Paper Walkthrough & Revision Guide | AS AQA 化学 2022年6月试卷1 全面解析与复习指南