📚 Year 11 AQA Chemistry Formula & Theorem Quick Reference | Year 11 AQA 化学公式定理速查手册
Welcome to the Year 11 AQA Chemistry Quick Reference Handbook – your go-to guide for essential formulas, theorems, and key principles. Master these to excel in your GCSE exams. All the content is aligned with the AQA specification, providing bilingual explanations to reinforce understanding.
欢迎使用 11 年级 AQA 化学速查手册——你必备的公式、定理及核心原理指南。掌握这些内容，在 GCSE 考试中脱颖而出。所有内容严格遵循 AQA 考纲，并通过双语解释加深理解。
1. The Mole & Molar Mass | 摩尔与摩尔质量
the SI unit for amount of substance. One mole of any substance contains exactly 6.02 × 1023 elementary entities (Avogadro constant). The mass of one mole of a substance is its molar mass (M), which has the unit g/mol and is numerically equal to the relative atomic mass (Ar) or relative formula mass (Mr).
摩尔（mol）是国际单位制中物质的量的单位。1 摩尔任何物质恰好包含 6.02 × 1023 个基本单元（阿伏伽德罗常数）。1 摩尔物质的质量就是其摩尔质量（M），单位为 g/mol，数值上等于相对原子质量（Ar）或相对式量（Mr）。
n = m / M
Where n = amount of substance (mol), m = mass (g), and M = molar mass (g/mol). Always use a periodic table to find Ar values and sum correctly for compounds.
其中 n = 物质的量（mol），m = 质量（g），M = 摩尔质量（g/mol）。务必查阅元素周期表获取 Ar 值，并对化合物正确求和。
Example: Calculate moles present in 25 g of calcium carbonate (CaCO3). M of CaCO3 = 40 + 12 + (3 × 16) = 100 g/mol. n = 25 ÷ 100 = 0.25 mol.
示例：计算 25 g 碳酸钙 (CaCO3) 所含摩尔数。M (CaCO3) = 40 + 12 + (3 × 16) = 100 g/mol，因此 n = 25 ÷ 100 = 0.25 mol。
| Quantity / 量 | Symbol / 符号 | Unit / 单位 |
|---|---|---|
| Mass / 质量 | m | g |
| Molar mass / 摩尔质量 | M | g/mol |
| Amount / 物质的量 | n | mol |
2. Concentration of Solutions | 溶液浓度
Concentration refers to the amount of solute dissolved in a given volume of solution. In quantitative chemistry, the most useful measure is molar concentration, which has units of mol/dm3 (often written as M). A 1 dm3 volume equals 1000 cm3.
浓度指溶解在给定体积溶液中的溶质的量。在定量化学中，最常用的是摩尔浓度，单位为 mol/dm3（常写作 M）。1 dm3 等于 1000 cm3。
c = n / V
Where c = concentration (mol/dm3), n = amount (mol), and V = volume (dm3). If the volume is given in cm3, convert it to dm3 by dividing by 1000.
其中 c = 浓度（mol/dm3），n = 物质的量（mol），V = 体积（dm3）。若体积以 cm3 给出，需除以 1000 转换为 dm3。
Example: 0.50 mol of sodium chloride is dissolved in 250 cm3 of water. V = 250/1000 = 0.250 dm3. c = 0.50 / 0.250 = 2.0 mol/dm3.
示例：0.50 mol 氯化钠溶于 250 cm3 水中。V = 250/1000 = 0.250 dm3，c = 0.50 / 0.250 = 2.0 mol/dm3。
| Variable / 变量 | Meaning / 含义 | Common unit / 常用单位 |
|---|---|---|
| c | Molar concentration / 摩尔浓度 | mol/dm3 |
| V | Volume of solution / 溶液体积 | dm3 or cm3 |
You can also combine the mole and concentration formulas: m = c × V × M. This directly links mass, concentration, volume and molar mass, which is very useful in titrations.
你也可以合并摩尔与浓度公式：m = c × V × M。该式直接将质量、浓度、体积和摩尔质量联系起来，在滴定计算中格外有用。
3. Gas Volumes at RTP | 常温常压下的气体体积
At room temperature and pressure (RTP: 20 °C and 1 atmosphere), one mole of any gas occupies 24 dm3 (or 24,000 cm3). This is a fixed molar gas volume and can be used to convert between moles and volume for gases only.
在常温常压（RTP：20 °C，1 个大气压）下，1 摩尔任何气体占 24 dm3（或 24,000 cm3）。这是一个固定的气体摩尔体积，仅用于气体与摩尔之间的转换。
V (dm3) = n × 24
V (cm3) = n × 24,000
These equations work only when the gas behaves ideally. Remember that the volume of a gas depends on temperature and pressure, so these relationships hold strictly at RTP.
这两个方程仅适用于气体呈现理想行为时。注意气体体积受温度和压力影响，因此上述关系严格仅在 RTP 下成立。
Example: How much volume does 0.250 mol of carbon dioxide occupy at RTP? V = 0.250 × 24 = 6.0 dm3.
示例：0.250 mol 二氧化碳在 RTP 下占据多大体积？V = 0.250 × 24 = 6.0 dm3。
The mole–gas volume relationship is often used in conjunction with reacting mass calculations. You can find the mass of a gas evolved by first calculating moles of gas, then using m = n × M.
摩尔–气体体积关系常与反应质量计算联用。先计算气体物质的量，再通过 m = n × M 求气体质量。
4. Percentage Yield & Atom Economy | 产率与原子经济
The percentage yield tells you how much product was actually obtained compared to the theoretical maximum. It is rarely 100% because of incomplete reactions, side reactions, or loss during purification.
产率表示实际获得的产物量与理论最大产量之比。受反应不完全、副反应或纯化损失影响，产率很少达到 100%。
Percentage yield = (actual yield / theoretical yield) × 100%
产率 = (实际产量 / 理论产量) × 100%
Atom economy measures how much of the total mass of reactants ends up in the desired product. It helps to evaluate the efficiency and greenness of a reaction pathway.
原子经济衡量的是反应物总质量中有多少进入了目标产物。它有助于评估反应路径的效率和绿色性。
Atom economy = (Mr of desired product / sum of Mr of all reactants) × 100%
原子经济 = (目标产物相对式量 / 所有反应物相对式量之和) × 100%
Important: A high atom economy reduces waste and makes industrial processes more sustainable. Reactions with only one product (e.g. addition polymerisation) have 100% atom economy.
注意：高原子经济可减少废物，使工业流程更可持续。只有一种产物的反应（如加成聚合）原子经济达 100%。
5. Rates of Reaction | 反应速率
The rate of a chemical reaction describes how quickly reactants are used up or products are formed. It can be measured by following a change in mass, volume of gas, colour, or turbidity over time.
化学反应速率描述反应物被消耗或产物生成的快慢。可通过跟踪质量、气体体积、颜色或浊度随时间的变化来测定。
Rate = quantity of reactant used or product formed / time
速率 = 消耗的反应物量或生成的产物量 / 时间
For a reaction producing a gas, rate can be measured as rate = volume of gas / time (cm3
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