📚 AS AQA Chemistry June 2022 Insert 1: Data Skills and Core Concepts | AS AQA 化学 2022年6月试卷一附页：数据技能与核心概念
In AQA AS Chemistry Paper 1, the June 2022 Insert 1 is a supporting resource that often contains data tables, graphs, experimental procedures or additional information needed to answer several questions. This article reviews the core concepts and calculation skills that commonly link to insert-based questions, helping you to use the insert efficiently and avoid avoidable errors.
在 AQA AS 化学试卷一中，2022 年 6 月的附页 1 是一份辅助材料，通常包含回答若干题目所需的数据表、图形、实验步骤或补充信息。本文复习与附页题相关的核心概念和计算技能，帮助你高效使用附页并避免常见错误。
1. Understanding the Insert | 理解附页的作用
Insert 1 in AQA AS Chemistry Paper 1 June 2022 usually provides extra material such as a results table, a graph of volume against time, a calorimetry data table, or an experimental setup. You must read the insert with the question paper, paying close attention to units, headings, axis labels and any anomalies in the data.
AQA AS 化学试卷一 2022 年 6 月的附页 1 通常提供额外的材料，例如结果表、体积-时间曲线、量热数据表或实验装置图。你必须将附页与试题结合阅读，仔细关注单位、表头、坐标轴标签以及数据中的任何异常。
- Results tables with initial, final and titre readings
- Graphs showing temperature change or gas volume over time
- Experimental instructions for Group 2, Group 7 or redox reactions
- 包含初读数、末读数和滴定体积的结果表
- 显示温度变化或气体体积随时间变化的图形
- 第 2 族、第 7 族或氧化还原反应的实验操作说明
2. Amount of Substance and Molar Calculations | 物质的量与摩尔计算
Insert-based questions frequently ask you to calculate the amount of substance in moles from mass, concentration or gas volume. The two most important equations are n = m ÷ M and c = n ÷ V. Always convert mass to grams and volume to dm³ before substituting into these equations.
附页题经常要求你根据质量、浓度或气体体积计算物质的量（摩尔）。两个最重要的公式是 n = m ÷ M 和 c = n ÷ V。在代入这些公式之前，始终将质量换算为克，将体积换算为 dm³。
n = m ÷ M and c = n ÷ V
If a results table gives a mean titre in cm³, divide by 1000 to convert to dm³. For example, a mean titre of 24.50 cm³ is 0.02450 dm³. Then multiply concentration in mol dm⁻³ by this volume to find the moles of the titrant.
如果结果表给出的平均滴定体积以 cm³ 为单位，将其除以 1000 转换为 dm³。例如，平均滴定体积 24.50 cm³ 等于 0.02450 dm³。然后用浓度（mol dm⁻³）乘以该体积，即可求得滴定剂的物质的量。
3. Ideal Gas Equation | 理想气体状态方程
When the insert provides the volume, temperature and pressure of a gas, use the ideal gas equation pV = nRT. Temperature must be in kelvin; add 273 to a Celsius value. The gas constant R is 8.31 J K⁻¹ mol⁻¹ when pressure is in pascals and volume is in cubic metres.
当附页提供气体的体积、温度和压强时，使用理想气体状态方程 pV = nRT。温度必须以开尔文为单位；将摄氏温度加 273。当压强以帕斯卡为单位、体积以立方米为单位时，气体常数 R 为 8.31 J K⁻¹ mol⁻¹。
pV = nRT
Be careful with unit conversions. Pressure in kPa must be multiplied by 1000 to give Pa, and volume in cm³ must be divided by 1,000,000 to give m³. If the insert gives volume in dm³, divide by 1000 to obtain m³.
注意单位换算。压强若以 kPa 为单位，需乘以 1000 得到 Pa；体积若以 cm³ 为单位，需除以 1,000,000 得到 m³。如果附页给出的体积单位为 dm³，则除以 1000 得到 m³。
4. Titration Calculations and Uncertainties | 滴定计算与不确定度
Insert tables often show raw burette readings. Concordant titres are those within 0.10 cm³ of each other, and only concordant results should be used to calculate the mean titre. Percentage uncertainty is calculated as (uncertainty ÷ measured value) × 100.
附页表格经常显示原始滴定管读数。一致滴定值是指彼此相差在 0.10 cm³ 以内的读数，只有一致的结果才能用于计算平均滴定体积。百分不确定度按 (不确定度 ÷ 测量值) × 100 计算。
% uncertainty = (uncertainty ÷ titre) × 100
For example, a burette reading has an uncertainty of ±0.05 cm³. If the mean titre is 25.00 cm³, the percentage uncertainty is (0.05 ÷ 25.00) × 100 = 0.20%. For a balance reading of ±0.01 g and a mass of 2.00 g, the percentage uncertainty is 0.50%.
例如，滴定管读数的不确定度为 ±0.05 cm³。如果平均滴定体积为 25.00 cm³，则百分不确定度为 (0.05 ÷ 25.00) × 100 = 0.20%。对于精度为 ±0.01 g 的天平，若称量质量为 2.00 g，则百分不确定度为 0.50%。
5. Group 2 Trends and Reactions | 第 2 族变化规律与反应
Insert 1 may describe Group 2 reactions, such as magnesium or calcium reacting with water or dilute acid. Reactivity increases down Group 2 because the atomic radius increases and the first and second ionisation energies decrease, making it easier to remove electrons.
附页 1 可能描述第 2 族反应，例如镁或钙与水或稀酸反应。第 2 族自上而下反应活性增强，因为原子半径增大，第一和第二电离能降低，使得电子更容易失去。
Solubility trends are also important. Group 2 sulfates become less soluble down the group; barium sulfate is insoluble and forms a white precipitate. Group 2 hydroxides become more soluble down the group, so barium hydroxide is much more soluble than magnesium hydroxide.
溶解性规律也很重要。第 2 族硫酸盐自上而下溶解性降低；硫酸钡不溶并形成白色沉淀。第 2 族氢氧化物自上而下溶解性增大，因此氢氧化钡比氢氧化镁的溶解性大得多。
6. Group 7 Chemistry and Halide Tests | 第 7 族化学与卤化物检验
Group 7 elements become less oxidising down the group. Chlorine displaces bromide and iodide ions; bromine displaces iodide ions but not chloride. These displacement reactions can be used to identify the halogen present in an insert-based question.
第 7 族元素自上而下氧化性减弱。氯能置换溴离子和碘离子；溴能置换碘离子但不能置换氯离子。这些置换反应可用于鉴定附页题中存在的卤素。
Acidified silver nitrate tests give characteristic precipitates: chloride forms a white precipitate, bromide forms a cream precipitate, and iodide forms a yellow precipitate. Adding ammonia solution can distinguish them: silver chloride dissolves in dilute ammonia, silver bromide dissolves only in concentrated ammonia, and silver iodide does not dissolve.
酸化硝酸银检验会产生特征沉淀：氯离子生成白色沉淀，溴离子生成乳白色沉淀，碘离子生成黄色沉淀。加入氨水可加以区分：氯化银溶于稀氨水，溴化银仅溶于浓氨水，碘化银不溶。
7. Enthalpy Changes from Insert Data | 由附页数据计算焓变
Calorimetry data in an insert usually provide the mass of water or solution, the temperature change, and the specific heat capacity. The heat change q is calculated using q = mcΔT, where m is the mass of the solution in grams, c is usually 4.18 J g⁻¹ K⁻¹, and ΔT is the temperature change in K or °C.
附页中的量热数据通常提供水或溶液的质量、温度变化和比热容。热量变化 q 用 q = mcΔT 计算，其中 m 是溶液质量（g），c 通常为 4.18 J g⁻¹ K⁻¹，ΔT 是温度变化（K 或 °C）。
q = mcΔT
Remember that q is in joules; divide by 1000 to convert to kJ. Then divide by the number of moles of the limiting reactant to find ΔH in kJ mol⁻¹. Exothermic reactions have a negative ΔH because heat is released.
记住 q 的单位是焦耳；除以 1000 转换为千焦。然后除以限制反应物的物质的量，得到 ΔH（kJ mol⁻¹）。放热反应的 ΔH 为负值，因为热量被释放。
8. Kinetics and Maxwell-Boltzmann Distribution | 动力学与麦克斯韦-玻尔兹曼分布
If Insert 1 shows a Maxwell-Boltzmann distribution curve, the most probable energy corresponds to the peak, not the mean or average energy. A temperature increase flattens the curve and shifts the peak to the right because more molecules have higher energies.
如果附页 1 显示麦克斯韦-玻尔兹曼分布曲线，最概然能量对应曲线的峰值，而不是平均能量。温度升高使曲线变平，峰值右移，因为更多分子具有较高能量。
A catalyst provides an alternative reaction pathway with a lower activation energy, so a greater proportion of molecules have energy greater than or equal to the activation energy. This increases the rate of reaction without the catalyst being used up.
催化剂提供一条活化能较低的反应途径，因此能量大于或等于活化能的分子比例增大。这加快了反应速率，而催化剂本身不被消耗。
9. Equilibrium and Kc | 平衡与 Kc
If the insert gives equilibrium amounts or concentrations, you may need to calculate the equilibrium constant Kc. The expression includes only aqueous solutions and gases; solids and pure liquids are omitted. For a general reaction aA + bB ⇌ cC + dD, Kc is written as ([C]ᶜ[D]ᵈ) ÷ ([A]ᵃ[B]ᵇ).
如果附页给出平衡量或平衡浓度，你可能需要计算平衡常数 Kc。该表达式只包括溶液和气体；固体和纯液体不写入。对于一般反应 aA + bB ⇌ cC + dD，Kc 写为 ([C]ᶜ[D]ᵈ) ÷ ([A]ᵃ[B]ᵇ)。
Kc = ([C]ᶜ[D]ᵈ) ÷ ([A]ᵃ[B]ᵇ)
Always use equilibrium concentrations in mol dm⁻³. If the question gives initial amounts and an equilibrium amount, set up an ICE table to find the equilibrium amounts, then convert to concentrations before substituting.
始终使用平衡时的浓度（mol dm⁻³）。如果题目给出初始量和某一平衡量，建立 ICE 表求出平衡量，然后转换为浓度再代入。
10. Redox and Oxidation States | 氧化还原与氧化态
Insert-based questions may ask you to work out oxidation states from a table of formulas or an equation. Use the rules: oxygen is usually −2, hydrogen is +1, and the sum of oxidation states equals the overall charge on the ion or molecule.
附页题可能要求你根据分子式或方程式表判断氧化态。使用以下规则：氧通常为 −2，氢为 +1，氧化态的总和等于离子或分子的总电荷。
In redox reactions, the oxidising agent is reduced and the reducing agent is oxidised. To balance a redox equation, combine the two half-equations and ensure the number of electrons lost equals the number gained before adding the half-equations.
在氧化还原反应中，氧化剂被还原，还原剂被氧化。要配平氧化还原方程式，先结合两个半反应，并确保失去的电子数等于得到的电子数，然后再相加。
11. Common Exam Pitfalls in Insert Questions | 附页题的常见失分点
Many students lose marks by using a non-concordant titre, forgetting to convert cm³ to dm³, misreading the insert’s axes or units, or giving ΔH in J mol⁻¹ instead of kJ mol⁻¹. Another common error is using the mass of the wrong substance in a calculation.
许多学生失分的原因包括：使用非一致滴定值、忘记将 cm³ 转换为 dm³、误读附页的坐标轴或单位、或将 ΔH 的单位写成 J mol⁻¹ 而非 kJ mol⁻¹。另一个常见错误是在计算中使用了错误物质的质量。
Always quote final answers to the appropriate number of significant figures, usually the smallest number of significant figures in the data provided. If the insert gives values to three significant figures, your final answer should generally also be to three significant figures.
始终将最终答案保留适当的有效数字，通常取所给数据中最少的有效数字位数。如果附页给出的数值保留三位有效数字，你的最终答案通常也应保留三位有效数字。
12. Revision Checklist for Insert-Based AS Paper 1 | 基于附页的 AS 试卷一复习清单
Before your exam, practise extracting data accurately from tables and graphs. Make sure you can confidently calculate moles, gas volume, enthalpy changes, percentage uncertainty and Kc. Know the trends in Group 2 and Group 7, and be able to explain them in terms of atomic structure.
考试之前，练习从表格和图形中准确提取数据。确保你能够熟练计算物质的量、气体体积、焓变、百分不确定度和 Kc。掌握第 2 族和第 7 族的变化规律，并能从原子结构角度加以解释。
- Read insert headings and units before each question
- Convert cm³ to dm³ for titration and concentration calculations
- Use concordant titres only for the mean titre
- Convert J to kJ and check ΔH sign for exothermic or endothermic reactions
- Use an ICE table for Kc calculations when initial and equilibrium amounts are given
- Link Group 2 and Group 7 observations to the species present in the insert
- 在回答每题前先阅读附页的表头和单位
- 滴定和浓度计算时将 cm³ 转换为 dm³
- 仅使用一致滴定值求平均滴定体积
- 将 J 转换为 kJ，并检查 ΔH 符号与放热或吸
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