📚 IGCSE Chemistry: What Else You Need to Know | IGCSE化学：你还需知道的知识点
Beyond the basic syllabus, IGCSE Chemistry often tests a set of “extra” concepts that students overlook — from subtle laboratory techniques to precise definitions. This guide highlights the essential additional knowledge you need to secure top marks.
除了基础考纲，IGCSE化学常常考查一些学生容易忽略的“额外”概念——从微妙的实验技巧到严谨的定义。本指南将重点介绍你取得高分所需的关键补充知识点。
1. Accuracy in Definitions | 定义的准确性
Many students lose marks by using vague definitions. For example, “oxidation is gaining oxygen” only applies in one context. You must know the electron transfer definitions: oxidation is loss of electrons, reduction is gain of electrons (OIL RIG).
许多学生因定义含糊而失分。例如，“氧化是得到氧”仅适用于一种情境。你必须掌握电子转移定义：氧化是失电子，还原是得电子（OIL RIG）。
Similarly, an acid is a proton donor, not just something that “tastes sour”. A base is a proton acceptor. Be precise with terms like “empirical formula” (simplest whole-number ratio of atoms) versus “molecular formula” (actual number of atoms in a molecule).
同样，酸是质子供体，而不只是“尝起来酸的东西”。碱是质子受体。要精确区分“实验式”（原子的最简整数比）和“分子式”（分子中实际原子数）。
2. Observing Colour Changes in Titrations | 滴定中的颜色变化观察
Titration is a core practical, but many students forget the endpoint colour details. With methyl orange, the colour changes from red (acidic) to yellow (alkaline), with orange in between. With phenolphthalein, it changes from colourless (acid) to pink (alkaline).
滴定是核心实验，但许多学生忘记终点颜色细节。用甲基橙时，颜色从红色（酸性）变为黄色（碱性），中间为橙色。用酚酞时，从无色（酸）变为粉红色（碱）。
When reading the burette, always read from the bottom of the meniscus, and remember that the burette reading is to two decimal places, e.g. 25.00 cm³. Also, rinse the pipette with the solution it will contain, not with distilled water.
读取滴定管时，必须读取弯月面底部，并记住滴定管读数精确到两位小数，如25.00 cm³。另外，移液管要用待装溶液润洗，而不是用蒸馏水。
3. The Mole and Gas Volumes | 摩尔与气体体积
You know that one mole of any gas occupies 24 dm³ at room temperature and pressure (r.t.p.). But did you know that this is 24,000 cm³? Always convert units carefully: 1 dm³ = 1000 cm³.
你知道在室温常压下，任何气体1摩尔体积为24 dm³。但你是否知道这是24,000 cm³？务必仔细换算单位：1 dm³ = 1000 cm³。
For reactions involving gases, use the ratio of volumes directly. For example, in N₂(g) + 3H₂(g) → 2NH₃(g), 20 cm³ of nitrogen reacts with 60 cm³ of hydrogen to give 40 cm³ of ammonia (if all volumes measured at the same conditions).
对于涉及气体的反应，可直接使用体积比。例如，N₂(g) + 3H₂(g) → 2NH₃(g) 中，20 cm³氮气与60 cm³氢气反应生成40 cm³氨气（所有体积在相同条件下测量）。
4. Ionic Equations and Spectator Ions | 离子方程式与旁观离子
In precipitation and neutralisation reactions, you need to write ionic equations. For example, AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq). The full ionic equation is Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq).
在沉淀和中和反应中，你需要写离子方程式。例如，AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)。完整离子方程式是 Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)。
Cancel the spectator ions (Na⁺ and NO₃⁻) to get the net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Always include state symbols: (s), (l), (g), (aq).
消去旁观离子（Na⁺ 和 NO₃⁻）得到净离子方程式：Ag⁺(aq) + Cl⁻(aq) → AgCl(s)。始终包含状态符号：(s)、(l)、(g)、(aq)。
5. Electrolysis of Aqueous Solutions | 水溶液电解的特殊规则
In electrolysis of molten compounds, the ions are simple. But in aqueous solutions, water also provides H⁺ and OH⁻ ions. At the cathode, H⁺ is discharged instead of Na⁺ or K⁺ if the metal is more reactive than hydrogen. At the anode, OH⁻ is discharged instead of Cl⁻ if the solution is dilute, producing oxygen.
熔融化合物电解时，离子很简单。但在水溶液中，水也提供H⁺ 和 OH⁻ 离子。在阴极，如果金属比氢更活泼，则H⁺ 优先放电，而不是Na⁺ 或 K⁺。在阳极，如果是稀溶液，OH⁻ 优先于Cl⁻ 放电，产生氧气。
Remember the mnemonic: “reactivity series” – metals above hydrogen stay in solution, hydrogen is produced. For concentrated chloride solutions, chlorine gas is produced at the anode. Also, the mass of substance liberated is proportional to the charge passed (Q = It).
记住口诀：“金属活动性顺序”——氢之前的金属留在溶液中，产生氢气。对于浓氯化物溶液，阳极产生氯气。此外，析出物质的质量与通过的电量成正比（Q = It）。
6. Energy Level Diagrams and Bond Energies | 能级图与键能
An exothermic reaction has products lower than reactants; the enthalpy change ΔH is negative. An endothermic reaction has products higher than reactants; ΔH is positive. You must label the activation energy (Eₐ) on the diagram.
放热反应中产物能量低于反应物，焓变ΔH为负。吸热反应中产物能量高于反应物，ΔH为正。你必须在图上标注活化能（Eₐ）。
Bond breaking is endothermic (energy taken in), bond making is exothermic (energy given out). ΔH = energy to break bonds – energy to make bonds. For example, in H₂ + Cl₂ → 2HCl, bond energies: H–H 436 kJ/mol, Cl–Cl 243 kJ/mol, H–Cl 432 kJ/mol. ΔH = (436 + 243) – 2×432 = 679 – 864 = –185 kJ/mol.
断键是吸热的（吸收能量），成键是放热的（释放能量）。ΔH = 断键总能量 – 成键总能量。例如，H₂ + Cl₂ → 2HCl，键能：H–H 436 kJ/mol，Cl–Cl 243 kJ/mol，H–Cl 432 kJ/mol。ΔH = (436 + 243) – 2×432 = 679 – 864 = –185 kJ/mol。
7. Rate of Reaction: Surface Area and Catalysts | 反应速率：表面积与催化剂
Increasing surface area (e.g. using powder instead of lumps) increases the frequency of collisions between reacting particles, so the rate increases. A catalyst works by providing an alternative pathway with lower activation energy.
增大表面积（例如用粉末代替块状）会增加反应粒子之间的碰撞频率，从而提高反应速率。催化剂通过提供活化能更低的其他途径来加快反应。
In rate experiments, measure either the volume of gas produced over time or the loss of mass. Remember to draw a tangent at a given time to find the instantaneous rate. Also, a catalyst is not used up in the reaction and does not change the position of equilibrium.
在速率实验中，测量随时间产生的气体体积或质量损失。记住在给定时间画切线以求瞬时速率。此外，催化剂不消耗，也不改变平衡位置。
8. Reversible Reactions and Le Chatelier’s Principle | 可逆反应与勒夏特列原理
For a reversible reaction at equilibrium, if you increase the temperature, the equilibrium shifts in the endothermic direction. If you increase pressure, it shifts to the side with fewer moles of gas. The Haber process uses this: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = –92 kJ/mol.
对于处于平衡的可逆反应，升高温度，平衡向吸热方向移动；增大压强，平衡向气体分子数减少的方向移动。哈伯工艺利用此原理：N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，ΔH = –92 kJ/mol。
A note on catalysts: they speed up both forward and reverse reactions equally, so they do not shift equilibrium. However, they help reach equilibrium faster, which improves productivity. The industrial conditions are a compromise: 450°C, 200 atm, iron catalyst.
关于催化剂：它们同等加快正逆反应速率，因此不移动平衡。但它们帮助更快达到平衡，从而提高产率。工业条件是折中结果：450°C、200 atm、铁催化剂。
9. Organic Chemistry: Naming and Functional Groups | 有机化学：命名与官能团
You must know the prefixes: meth- (1 C), eth- (2 C), prop- (3 C), but- (4 C). The suffixes indicate the functional group: -ane (alkane), -ene (alkene), -ol (alcohol), -oic acid (carboxylic acid). For example, propene is CH₃–CH=CH₂, ethanol is CH₃CH₂OH.
你必须知道前缀：甲（1个C）、乙（2个C）、丙（3个C）、丁（4个C）。后缀表示官能团：-ane（烷烃）、-ene（烯烃）、-ol（醇）、-oic acid（羧酸）。例如，丙烯是 CH₃–CH=CH₂，乙醇是 CH₃CH₂OH。
In IGCSE, you also need to know the combustion of alkanes: CH₄ + 2O₂ → CO₂ + 2H₂O. And the addition reaction of alkenes with bromine water: orange to colourless, which is a test for unsaturation.
在IGCSE中，你还需要知道烷烃燃烧：CH₄ + 2O₂ → CO₂ + 2H₂O。以及烯烃与溴水的加成反应：橙色变无色，这是不饱和性的测试。
10. Separation Techniques: Fractional Distillation | 分离技术：分馏
Fractional distillation is used to separate miscible liquids with different boiling points. It uses a fractionating column, which provides a temperature gradient; the liquid with the lowest boiling point distills first. This is how crude oil is separated into fractions.
分馏用于分离沸点不同的互溶液体。它使用分馏柱，提供温度梯度；沸点最低的液体先蒸出。这就是原油分离成馏分的方法。
For separating a solid from a liquid, use filtration if the solid is insoluble, or evaporation/crystallisation if it is soluble. Remember: a separating funnel separates immiscible liquids (e.g. oil and water).
若分离固体和液体：不溶性固体用过滤；可溶性固体用蒸发或结晶。注意：分液漏斗分离不互溶液体（如油和水）。
11. Testing for Gases and Ions | 气体与离子的检验
You are expected to recall these tests: hydrogen (squeaky pop test with a lit splint), oxygen (relights a glowing splint), carbon dioxide (turns limewater milky), chlorine (bleaches damp litmus paper).
你要记住这些检验：氢气（点燃木条有尖锐爆鸣声）、氧气（使带火星木条复燃）、二氧化碳（使石灰水变浑浊）、氯气（使湿润石蕊试纸褪色）。
For cations: copper(II) gives a blue precipitate with NaOH, iron(II) gives green, iron(III) gives brown. For anions: carbonate releases CO₂ when acid added; sulfate gives a white precipitate with BaCl₂/HCl; halides give precipitates with AgNO₃ – chloride white, bromide cream, iodide yellow.
阳离子检验：铜(II)与NaOH生成蓝色沉淀，亚铁离子生成绿色，铁(III)生成红棕色。阴离子检验：碳酸盐加酸释放CO₂；硫酸盐与BaCl₂/HCl生成白色沉淀；卤化物与AgNO₃生成沉淀——氯化物白色，溴化物奶油色，碘化物黄色。
12. Moles in Solution: Concentration Calculations | 溶液中的摩尔：浓度计算
Concentration (mol/dm³) = number of moles ÷ volume (dm³). When performing dilutions, use C₁V₁ = C₂V₂. For example, if 25.0 cm³ of 0.10 mol/dm³ HCl is neutralised by 20.0 cm³ of NaOH, the NaOH concentration is:
浓度（mol/dm³）= 物质的量 ÷ 体积（dm³）。进行稀释时，使用 C₁V₁ = C₂V₂。例如，25.0 cm³ 的 0.10 mol/dm³ HCl 被 20.0 cm³ NaOH 中和，NaOH 浓度为：
Moles HCl = 0.0250 dm³ × 0.10 mol/dm³ = 0.00250 mol
Reaction: HCl + NaOH → NaCl + H₂O, so moles NaOH = 0.00250 mol
[NaOH] = 0.00250 mol ÷ 0.0200 dm³ = 0.125 mol/dm³
Always convert cm³ to dm³ by dividing by 1000. In titrations, use the average titre, excluding any anomalous results. Write down all workings clearly to gain method marks.
始终将 cm³ 除以1000换算为 dm³。在滴定中，使用平均滴定读数，排除异常值。清楚写出所有计算步骤以获得方法分。
Published by TutorHao | Chemistry Revision Series | aleveler.com
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