📚 AQA GCSE Chemistry Quick Reference: Formulae & Theorems | AQA GCSE 化学公式定理速查手册
Welcome to your ultimate quick reference guide for AQA GCSE Chemistry. This handbook consolidates all essential formulas, equations, and key theorems you will need for your Year 11 revision and exams. From mole calculations to energetics, each section is presented with clear paired explanations in both English and Chinese.
欢迎使用 AQA GCSE 化学终极速查指南。本手册汇集了您在 11 年级复习和考试中所需的所有关键公式、方程式和重要定理。从摩尔计算到能量学，每个部分都配有清晰的中英对照解释。
1. Relative Atomic Mass (Aᵣ) & Isotopes | 相对原子质量 (Aᵣ) 与同位素
The relative atomic mass (Ar) is the weighted average mass of the isotopes of an element, relative to 1/12th the mass of a carbon‑12 atom.
相对原子质量 (Ar) 是某元素各同位素质量的加权平均值，相对于碳‑12 原子质量的 1/12。
It is calculated using: Ar = Σ (isotopic mass × % abundance) / 100, or using fractional abundance.
计算公式为：Ar = Σ (同位素质量 × 丰度百分比) / 100，也可以使用小数形式的丰度。
Example: Chlorine has two main isotopes, ³⁵Cl (75%) and ³⁷Cl (25%). Ar = (35 × 75 + 37 × 25) / 100 = 35.5
示例：氯有两种主要同位素 ³⁵Cl (75%) 和 ³⁷Cl (25%)。Ar = (35×75 + 37×25)/100 = 35.5
2. The Mole & Molar Mass (Mᵣ) | 摩尔与摩尔质量 (Mᵣ)
The mole (mol) is the unit for amount of substance. One mole contains 6.02 × 10²³ particles (Avogadro constant).
摩尔 (mol) 是物质的量的单位。1 摩尔含有 6.02 × 10²³ 个微粒 (阿伏加德罗常数)。
Molar mass (Mr) is the mass of one mole of a substance, in g/mol. It is numerically equal to the relative formula mass.
摩尔质量 (Mr) 是一摩尔物质的质量，单位是 g/mol，数值上等于相对式量。
moles = mass (g) / Mr (g/mol)
摩尔 = 质量 (g) / 相对分子质量 (g/mol)
Example: How many moles in 20 g of NaOH? Mr = 40, so moles = 20 / 40 = 0.5 mol
示例：20 g NaOH 是多少摩尔？Mr = 40，故摩尔 = 20/40 = 0.5 mol
3. Concentration Calculations | 浓度计算
Concentration can be expressed in g/dm³ or mol/dm³. Remember 1 dm³ = 1000 cm³.
浓度可以用 g/dm³ 或 mol/dm³ 表示。记住 1 dm³ = 1000 cm³。
Concentration (g/dm³) = mass (g) / volume (dm³)
浓度 (g/dm³) = 质量 (g) / 体积 (dm³)
Concentration (mol/dm³) = moles / volume (dm³)
浓度 (mol/dm³) = 摩尔 / 体积 (dm³)
Example: 5 g of NaCl dissolved in 250 cm³ water. Convert volume: 250 cm³ = 0.25 dm³. Concentration = 5 / 0.25 = 20 g/dm³.
示例：5 g NaCl 溶于 250 cm³ 水。转换体积：250 cm³ = 0.25 dm³，浓度 = 5 / 0.25 = 20 g/dm³。
4. Molar Gas Volume (RTP) | 气体摩尔体积 (常温常压下)
At room temperature and pressure (RTP: 20 °C, 1 atm), one mole of any gas occupies 24 dm³ (or 24 000 cm³).
在常温常压下 (RTP: 20 °C, 1 atm)，1 摩尔任何气体的体积为 24 dm³ (或 24 000 cm³)。
Volume of gas (dm³) = moles × 24
气体体积 (dm³) = 摩尔 × 24
To find moles from a gas volume: moles = volume (dm³) / 24.
由气体体积求摩尔：摩尔 = 体积 (dm³) / 24。
Example: 3 mol of hydrogen occupies 3 × 24 = 72 dm³ at RTP.
示例：3 mol 氢气在 RTP 下占 3 × 24 = 72 dm³。
5. Atom Economy & Percentage Yield | 原子经济性与百分产率
Atom economy measures the efficiency of a reaction in incorporating atoms into the desired product.
原子经济性衡量反应中反应物原子转化为目标产物的效率。
Atom economy = (Mr of desired product / sum of Mr of all products) × 100%
原子经济性 = (目标产物相对分子质量 / 所有产物相对分子质量之和) × 100%
Percentage yield compares the actual mass of product obtained to the maximum theoretical mass.
百分产率 比较实际得到的产物质量与理论最大质量。
Percentage yield = (actual yield / theoretical yield) × 100%
百分产率 = (实际产量 / 理论产量) × 100%
High atom economy reduces waste; high yield indicates an effective procedure.
原子经济性高可减少废物；产率高说明实验效率高。
6. Energy Changes (Calorimetry) | 能量变化 (量热法)
In calorimetry, heat energy transferred to or from the surroundings is calculated using: q = m c ΔT.
在量热法中，传递到周围环境或从周围环境吸收的热量用 q = m c ΔT 计算。
q = m × c × ΔT
q = 质量 (g) × 比热容 (J/g°C) × 温度变化 (°C)
For water, c = 4.2 J/g°C. ΔT = final temperature − initial temperature.
对于水，比热容 c = 4.2 J/g°C。ΔT = 最终温度 − 初始温度。
To find molar enthalpy change (ΔH), divide by moles and express in kJ/mol: ΔH = −q / n (for exothermic, q is heat given out; if you define q as energy absorbed by water, ΔH = −q/n for exothermic).
要计算摩尔焓变 (ΔH)，除以摩尔数并转换为 kJ/mol：ΔH = −q / n (对于放热反应，q 为水吸收的热量时，ΔH 为负值)。
Example: 0.1 mol of fuel heats 500 g water from 20°C to 40°C. q = 500 × 4.2 × 20 = 42 000 J = 42 kJ. ΔH = −42 / 0.1 = −420 kJ/mol.
示例：0.1 mol 燃料使 500 g 水从 20°C 升到 40°C。q = 500×4.2×20 = 42 000 J = 42 kJ。ΔH = −42/0.1 = −420 kJ/mol。
7. Rate of Reaction | 反应速率
The mean rate of reaction measures how quickly a reactant is used up or a product is formed over time.
平均反应速率衡量反应物消耗或产物生成的快慢。
Mean rate = change in amount (mass/volume/moles) / time
平均速率 = 变化量 (质量/体积/摩尔) / 时间
Units can be g/s, cm³/s or mol/s. For an accurate instantaneous rate, draw a tangent on a concentration–time graph.
单位可以是 g/s、cm³/s 或 mol/s。要获得精确的瞬时速率，可在浓度-时间图上画切线。
Collision theory states that particles must collide with sufficient energy (greater than activation energy) and correct orientation to react.
碰撞理论指出，微粒必须发生有效碰撞，碰撞能量需大于活化能且取向正确。
8. Titration Calculations (Neutralisation) | 滴定计算 (中和反应)
Titrations use the neutralisation reaction between an acid and an alkali to determine an unknown concentration.
滴定利用酸碱中和反应来测定未知浓度。
Key method: use a known standard solution, measure exact volumes at the endpoint, then apply mole ratios.
关键方法：使用已知标准溶液，准确量取终点时的体积，然后应用摩尔比。
Steps: (1) nknown = cknown × Vknown (dm³). (2) Use balanced equation to find moles of unknown. (3) cunknown = nunknown / Vunknown (dm³).
步骤：(1) 已知摩尔 = c已知 × V已知 (dm³)。(2) 根据配平的方程式求未知物的摩尔。(3) c未知 = n未知 / V未知 (dm³)。
For a 1:1 acid-alkali reaction, you can also use: caVa = cbVb (only when the mole ratio is 1:1).
对于 1:1 的酸碱反应，也可使用：c酸V酸 = c碱V碱 (仅当摩尔比为 1:1 时)。
9. Empirical & Molecular Formulae | 实验式与分子式
The empirical formula gives the simplest whole‑number ratio of atoms in a compound. The molecular formula shows the actual number of atoms in a molecule.
实验式给出化合物中各原子的最简整数比。分子式显示分子中实际的原子数目。
To find empirical formula: convert element masses (or %) to moles by dividing by Ar, then divide by the smallest number of moles, and round to whole numbers.
求实验式：将各元素的质量 (或百分比) 除以各自的 Ar 得到摩尔数，再除以最小的摩尔数，最后取整。
Molecular formula = (empirical formula) × n, where n = Mr / empirical formula mass.
分子式 = (实验式) × n，其中 n = 相对分子质量 / 实验式质量。
Example: a compound contains 80% carbon, 20% hydrogen by mass. Moles C = 80/12 = 6.67, H = 20/1 = 20. Ratio ≈ 1:3 → empirical formula CH₃. If Mr = 30, then molecular formula is C₂H₆.
示例：某化合物含碳80%，氢20%。C 摩尔 = 80/12 = 6.67，H = 20/1 = 20。比例约1:3 → 实验式 CH₃。若 Mr = 30，则分子式为 C₂H₆。
10. Dynamic Equilibrium & Le Chatelier’s Principle | 动态平衡与勒夏特列原理
In a reversible reaction, dynamic equilibrium is reached when the forward and reverse reactions occur at the same rate, and concentrations remain constant.
在可逆反应中，当正反应与逆反应速率相等且浓度保持恒定时，即达到动态平衡。
Le Chatelier’s principle: if a change in condition (concentration, temperature, pressure) is applied to a system at equilibrium, the position of equilibrium shifts to oppose the change.
勒夏特列原理：如果改变平衡体系的条件 (浓度、温度、压强)，平衡位置会向减弱这种改变的方向移动。
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Increasing reactant concentration → shifts to products. | 增加反应物浓度 → 向产物方向移动。
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Increasing temperature in exothermic reaction → shifts to reactants; endothermic → shifts to products. | 放热反应升温 → 向反应物方向移动；吸热反应升温 → 向产物方向移动。
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Increasing pressure shifts to side with fewer gas moles. | 增大压强，平衡向气体分子数减少的方向移动。
11. Ionic & Half Equations | 离子方程式与半反应
Ionic equations show only the species that actually change during a reaction; spectator ions are omitted.
离子方程式只显示反应中实际发生变化的物种；旁观离子省略不写。
Example: NaOH + HCl → NaCl + H₂O. The ionic equation is: H⁺ + OH⁻
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