📚 AS AQA International Chemistry Unit 1 (CH01): Core Physical & Inorganic Concepts | AS AQA 国际化学第一单元（CH01）：核心物理与无机概念
The AS AQA International Chemistry Unit 1 (CH01) paper tests a wide range of physical and inorganic chemistry concepts, from atomic structure and quantitative chemistry to bonding, energetics, kinetics, equilibrium, redox and the trends of Groups 2 and 7. Mastering the exact definitions, calculation methods and explanations required by the mark scheme is essential for a top grade.
AS AQA 国际化学第一单元（CH01）考试涵盖广泛的物理化学与无机化学概念，从原子结构、定量化学到化学键、能量学、动力学、平衡、氧化还原以及第2族和第7族的趋势。掌握评分标准要求的确切定义、计算方法和解释是取得高分的关键。
1. Atomic Structure and Mass Spectrometry | 原子结构与质谱
In CH01, you must be able to describe the stages of a time-of-flight mass spectrometer: vaporisation, ionisation by electron impact or electrospray, acceleration through an electric field, ion drift through the flight tube, detection, and data analysis. The time taken for an ion to reach the detector depends on its mass-to-charge ratio: lighter or more highly charged ions move faster.
在 CH01 中，你必须能够描述飞行时间质谱仪的各个阶段：气化、通过电子轰击或电喷雾电离、在电场中加速、离子在飞行管中漂移、检测以及数据分析。离子到达检测器所需的时间取决于其质荷比：较轻或电荷较高的离子运动更快。
Relative atomic mass (Aᵣ) is defined as the weighted average mass of an atom of an element relative to 1/12th the mass of a carbon-12 atom. For a sample containing isotopes with masses m₁, m₂ and percentage abundances a₁, a₂, the calculation is:
相对原子质量（Aᵣ）定义为元素的一个原子的加权平均质量相对于碳-12原子质量的 1/12。对于含有质量为 m₁、m₂ 且丰度为 a₁、a₂ 同位素的样品，其计算公式为：
Aᵣ = (m₁ × a₁ + m₂ × a₂) / 100
Always check that your percentage abundances add to 100 before using this formula, and give your final answer to the appropriate number of significant figures.
在使用此公式之前，务必检查丰度百分比总和是否为 100，并将最终答案保留适当数量的有效数字。
2. Electron Configuration and Periodicity | 电子排布与周期性
Electron configurations follow the Aufbau principle, Pauli exclusion principle and Hund’s rule. For example, carbon is 1s² 2s² 2p² and chlorine is 1s² 2s² 2p⁶ 3s² 3p⁵. When writing configurations for ions, add or remove electrons from the highest energy subshell first; transition metal ions lose 4s electrons before 3d.
电子排布遵循构造原理、泡利不相容原理和洪特规则。例如，碳的电子排布为 1s² 2s² 2p²，氯为 1s² 2s² 2p⁶ 3s² 3p⁵。书写离子的电子排布时，首先从最高能量亚层添加或移除电子；过渡金属离子会先失去 4s 电子再失去 3d 电子。
Across Period 3, atomic radius decreases because the nuclear charge increases while shielding remains roughly constant, pulling outer electrons closer. First ionisation energy generally increases across a period, but drops occur between Groups 2 and 3 because the p subshell is higher in energy, and between Groups 5 and 6 because of electron pairing in the p orbital.
在第三周期中，原子半径逐渐减小，因为核电荷增加而屏蔽效应基本保持不变，使外层电子被拉得更近。第一电离能通常沿周期增大，但第2族和第3族之间出现下降，因为 p 亚层能量更高；第5族和第6族之间也出现下降，因为 p 轨道中出现电子配对。
3. Amount of Substance and the Mole | 物质的量与摩尔
The mole is the unit for amount of substance. One mole contains 6.022 × 10²³ specified particles. The number of moles n can be calculated from mass m and molar mass M using n = m / M. For gases at room temperature and pressure, the molar gas volume is often taken as 24.0 dm³ mol⁻¹, but only use this if the question states RTP.
摩尔是物质的量的单位。一摩尔包含 6.022 × 10²³ 个指定粒子。物质的量 n 可通过质量 m 和摩尔质量 M 使用 n = m / M 计算。对于常温常压下的气体，摩尔气体体积通常取 24.0 dm³ mol⁻¹，但只有在题目说明 RTP 时才能使用。
Solution calculations use c = n / V, where c is concentration in mol dm⁻³ and V is volume in dm³. Titration questions require you to combine reacting ratios from a balanced equation with n = c × V to find an unknown concentration or mass.
溶液计算使用 c = n / V，其中 c 为物质的量浓度，单位为 mol dm⁻³，V 为体积，单位为 dm³。滴定题目要求你结合配平方程式中的反应比例与 n = c × V 来求未知浓度或质量。
4. Empirical and Molecular Formulae | 实验式与分子式
The empirical formula gives the simplest whole-number ratio of atoms in a compound. To find it, divide the mass or percentage of each element by its relative atomic mass, then divide each result by the smallest value to obtain a simple ratio. If the ratio is not whole numbers, multiply all values by the same factor.
实验式给出化合物中原子的最简整数比。要确定实验式，将每种元素的质量或百分比除以其相对原子质量，然后将每个结果除以最小值以获得简单比例。如果比例不是整数，则将所有值乘以同一因子。
The molecular formula is a whole-number multiple of the empirical formula. To find the multiple, divide the relative molecular mass by the empirical formula mass. For example, if the empirical formula is CH₂ and Mᵣ is 84.0, the multiple is 84.0 / 14.0 = 6, so the molecular formula is C₆H₁₂.
分子式是实验式的整数倍。要确定倍数，将相对分子质量除以实验式质量。例如，如果实验式为 CH₂，Mᵣ 为 84.0，则倍数为 84.0 / 14.0 = 6，因此分子式为 C₆H₁₂。
5. Ideal Gas Equation | 理想气体状态方程
The ideal gas equation links pressure p, volume V, number of moles n, the gas constant R and temperature T:
理想气体状态方程将压强 p、体积 V、物质的量 n、气体常数 R 和温度 T 联系起来：
pV = nRT
Use p in pascals (Pa), V in cubic metres (m³), n in mol, T in kelvin (K), and R = 8.31 J K⁻¹ mol⁻¹. Convert kPa to Pa by multiplying by 10³, cm³ to m³ by dividing by 10⁶, dm³ to m³ by dividing by 10³, and °C to K by adding 273.
使用压强 p 的单位为帕斯卡（Pa），体积 V 的单位为立方米（m³），n 的单位为 mol，温度 T 的单位为开尔文（K），R = 8.31 J K⁻¹ mol⁻¹。将 kPa 乘以 10³ 转换为 Pa，将 cm³ 除以 10⁶ 转换为 m³，将 dm³ 除以 10³ 转换为 m³，将°C 加 273 转换为 K。
6. Chemical Bonding and Structure | 化学键与结构
Ionic bonding occurs between a metal and a non-metal and is the electrostatic attraction between oppositely charged ions. Ionic compounds tend to be brittle, have high melting points, and conduct electricity when molten or dissolved in water because ions become mobile.
离子键发生在金属和非金属之间，是相反电荷离子之间的静电吸引力。离子化合物通常易碎，熔点高，在熔融或溶于水时可以导电，因为离子可以自由移动。
Covalent bonding is the strong electrostatic attraction between a shared pair of electrons and the nuclei of the bonded atoms. A dative covalent bond forms when one atom supplies both electrons. Metals bond by metallic bonding: the attraction between positive metal ions and a sea of delocalised electrons, which explains their conductivity and malleability.
共价键是共用电子对与成键原子核之间的强静电吸引力。配位共价键在一个原子提供两个电子时形成。金属通过金属键结合：正金属离子与离域电子海之间的吸引力，这解释了金属的导电性和延展性。
7. Intermolecular Forces and Shapes | 分子间作用力与分子形状
Shapes of molecules are predicted by electron-pair repulsion theory: electron pairs around a central atom arrange themselves to minimise repulsion. Common shapes include linear (2 bonding pairs, 180°), trigonal planar (3 pairs, 120°), tetrahedral (4 pairs, 109.5°), trigonal bipyramidal (5 pairs, 90° and 120°) and octahedral (6 pairs, 90°).
分子形状可通过电子对互斥理论预测：中心原子周围的电子对会排列成使排斥力最小的构型。常见形状包括直线形（2 对成键电子对，180°）、平面三角形（3 对，120°）、四面体（4 对，109.5°）、三角双锥（5 对，90° 和 120°）和八面体（6 对，90°）。
Lone pairs repel more strongly than bonding pairs, reducing bond angles. For example, NH₃ has one lone pair and a bond angle of 107°, while H₂O has two lone pairs and a bond angle of 104.5°. Electronegativity differences create polar bonds, and polar molecules can have permanent dipole-dipole forces. Hydrogen bonding occurs when H is bonded to N, O or F; it is the strongest intermolecular force and explains the unusually high boiling point of water.
孤对电子比成键电子对的排斥力更强，会减小键角。例如，NH₃ 有一对孤对电子，键角为 107°，而 H₂O 有两对孤对电子，键角为 104.5°。电负性差异产生极性键，极性分子可具有永久偶极-偶极作用力。当 H 与 N、O 或 F 成键时会产生氢键；它是最强的分子间作用力，解释了水异常高的沸点。
8. Energetics and Calorimetry | 能量学与量热法
Enthalpy change ΔH is negative for exothermic reactions, which release heat to the surroundings, and positive for endothermic reactions, which absorb heat. In calorimetry, the heat change q is calculated from the mass m of the solution, its specific heat capacity c and the temperature change ΔT:
焓变 ΔH 对于放热反应为负值，反应向环境释放热量；对于吸热反应为正值，反应吸收热量。在量热法中，热量变化 q 由溶液质量 m、比热容 c 和温度变化 ΔT 计算：
q = m × c × ΔT
For water, c = 4.18 J g⁻¹ K⁻¹. The enthalpy change per mole is then ΔH = −q / n, where n is the number of moles of the limiting reactant. Remember to include the negative sign for exothermic reactions.
对于水，c = 4.18 J g⁻¹ K⁻¹。每摩尔的焓变为 ΔH = −q / n，其中 n 为限制反应物的物质的量。对于放热反应，记得加上负号。
Hess’s law states that the total enthalpy change for a reaction is independent of the route taken. This allows you to calculate ΔH from enthalpy changes of formation or combustion by constructing a cycle.
赫斯定律指出，反应的总焓变与所采取的路径无关。这使你可以通过构建能量循环，利用生成焓或燃烧焓来计算 ΔH。
9. Kinetics and Maxwell-Boltzmann Distribution | 动力学与麦克斯韦-玻尔兹曼分布
The Maxwell-Boltzmann distribution shows the range of kinetic energies of gas molecules at a given temperature. The area under the curve represents the total number of molecules, and only molecules with energy greater than or equal to the activation energy Eₐ can react.
麦克斯韦-玻尔兹曼分布显示了在给定温度下气体分子动能的范围。曲线下面积代表分子总数，只有能量大于或等于活化能 Eₐ 的分子才能发生反应。
Increasing temperature shifts the distribution to the right and lowers the peak, so a much larger proportion of molecules have energy exceeding Eₐ, which greatly increases reaction rate. Adding a catalyst provides an alternative reaction pathway with lower activation energy, so more molecules can react without changing the temperature.
升高温度使分布曲线向右移动并降低峰值，因此能量超过 Eₐ 的分子比例大大增加，从而显著提高反应速率。加入催化剂提供了活化能较低的另一条反应路径，因此在不改变温度的情况下有更多分子可以反应。
10. Equilibrium and Le Chatelier’s Principle | 平衡与勒夏特列原理
Dynamic equilibrium occurs when the forward and reverse reactions proceed at the same rate, so the concentrations of reactants and products remain constant. The equilibrium constant Kc is written as the concentration of products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their coefficients.
动态平衡发生在正反应和逆反应速率相等时，因此反应物和产物的浓度保持恒定。平衡常数 Kc 表示为产物浓度以其化学计量数为幂的乘积除以反应物浓度以其化学计量数为幂的乘积。
Le Chatelier’s principle states that if a system at equilibrium is disturbed, it shifts in the direction that opposes the change. Increasing pressure favours the side with fewer gas moles; increasing temperature favours the endothermic direction. A catalyst does not affect the position of equilibrium or Kc, only the rate at which equilibrium is reached.
勒夏特列原理指出，如果平衡系统受到扰动，它会向削弱该扰动的方向移动。增大压强有利于气体物质的量较少的一侧；升高温度有利于吸热方向。催化剂不会影响平衡位置或 Kc，只会影响达到平衡的速率。
11. Redox and Oxidation States | 氧化还原与氧化数
Oxidation is loss of electrons, and reduction is gain of electrons; a redox reaction involves both processes. Oxidation states are assigned using rules: elements are 0, oxygen is usually −2, hydrogen is usually +1, and the sum of oxidation states equals the overall charge.
氧化是失去电子，还原是获得电子；氧化还原反应同时包含这两个过程。氧化数通过规则确定：单质为 0，氧通常为 −2，氢通常为 +1，氧化数之和等于总电荷。
In a redox equation, the species oxidised loses electrons and its oxidation state increases, while the species reduced gains electrons and its oxidation state decreases. Half-equations can be combined by balancing electrons, atoms and charge. For example, Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu combine to give Zn + Cu²⁺ → Zn²⁺ + Cu.
在氧化还原方程中，被氧化的物质失去电子，氧化数升高；被还原的物质获得电子，氧化数降低。半反应式可通过平衡电子、原子和电荷来合并。例如，Zn → Zn²⁺ + 2e⁻ 与 Cu²⁺ + 2e⁻ → Cu 合并为 Zn + Cu²⁺ → Zn²⁺ + Cu。
12. Group 2 and Group 7 Trends | 第2族和第7族趋势
Group 2 elements become more reactive down the group because atomic radius increases and ionisation energy decreases, making it easier to lose the two outer electrons. The solubility of Group 2 hydroxides increases down the group, while the solubility of Group 2 sulfates decreases.
第2族元素的反应活性沿族递增，因为原子半径增大且电离能降低，使其更容易失去两个外层电子。第2族氢氧化物的溶解度沿族递增，而第2族硫酸盐的溶解度沿族递减。
Group 7 halogens become less reactive down the group because atomic radius increases and the attraction for an added electron decreases. A more reactive halogen will displace a less reactive halide from solution, for example Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂. This trend is linked to the decreasing oxidising ability of the halogens down the group.
第7族卤素的反应活性沿族递减，因为原子半径增大，对添加电子的吸引力减弱。反应性更强的卤素会从溶液中置换出反应性较弱的卤化物，例如 Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂。这一趋势与卤素沿族氧化能力的递减有关。
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