📚 Year 11 WJEC Chemistry: Unit Test Mock Exam Analysis | Year 11 WJEC 化学：单元测试模拟卷解析
Mock exams are essential for identifying gaps in knowledge before the real assessment. This analysis covers typical WJEC Year 11 unit test questions, breaking down common pitfalls and showing you how to maximise marks using precise scientific terminology and clear working.
模拟考试是在真正评估前查漏补缺的关键。本文解析 WJEC 十一年级单元测试常见题型，拆解易错点，展示如何运用准确科学术语与清晰步骤拿到高分。
1. Atomic Structure and Isotopes | 原子结构与同位素
WJEC questions often ask you to state the number of protons, neutrons and electrons for an atom or ion using data from the Periodic Table. Recall that the atomic number (bottom number) equals protons, and mass number (top number) equals protons plus neutrons.
WJEC 题目常要求利用周期表数据说明某原子或离子的质子、中子和电子数。记住原子序数（下方数字）等于质子数，质量数（上方数字）等于质子数加中子数。
n lost. Protons remain 11, neutrons = 23 - 11 = 12, but electrons become 10 instead of 11.
离子如 ²³₁₁Na⁺，学生常忘记正电荷表示失去一个电子。质子仍为 11，中子 = 23 - 11 = 12，但电子变为 10 而非 11。
sum, then divide by 100. For example, for chlorine: Ar = (35 × 75 + 37 × 25) / 100 = 35.5.
计算相对原子质量时，必须清晰展示步骤：每种同位素质量乘以其丰度百分比，求和后除以 100。例如氯：相对原子质量 = (35 × 75 + 37 × 25) / 100 = 35.5。
2. Chemical Bonding and Structure | 化学键与结构
A common gap is linking bonding type to bulk properties. Ionic compounds like NaCl have giant ionic lattices and conduct electricity only when molten or dissolved because ions are free to move.
常见知识漏洞是无法将键型与物质性质关联。离子化合物如 NaCl 具有巨型离子晶格，仅在熔融或溶解时导电，因为离子能自由移动。
Simple molecular substances such as water or methane have low melting and boiling points because of weak intermolecular forces, not weak covalent bonds. Examiners penalise mixing up 'intermolecular forces' with 'covalent bonds'.
水、甲烷等简单分子物质熔沸点低，是分子间作用力弱，而非共价键弱。考官若见“分子间力”与“共价键”混淆会扣分。
When drawing dot-and-cross diagrams for covalent compounds, ensure only outer shell electrons are shown and that bonding pairs are clearly placed between the two atoms. For ionic compounds, show square brackets and charges, with correct electron transfer.
绘制共价化合物的点叉图时，确保只画最外层电子，且共用电子对清楚放在两原子间。离子化合物则要标出方括号和电荷，并正确展示电子转移。
3. Quantitative Chemistry: Moles and Concentration | 定量化学：摩尔与浓度
with precision. Always convert mass to grams and show units. In a titration calculation, do not forget the mole ratio from the balanced equation.
摩尔公式 n = m / Mᵣ 必须精确运用。质量务必换算成克并标注单位。在滴定计算中，不要忽略从配平方程式得出的摩尔比。
³). If volume is given in cm³, divide by 1000. Many candidates lose marks by failing to convert cm³ to dm³.
浓度公式：浓度 (mol dm⁻³) = 摩尔数 / 体积 (dm³)。若体积单位是 cm³，需除以 1000。许多考生因未将 cm³ 转化为 dm³ 而失分。
When calculating masses in reactions, a structured approach helps: write the balanced equation, note molar masses, find moles of the known substance, use the ratio to find moles of the target substance, then convert to mass.
计算反应中的质量时，结构化方法有效：写出配平方程式，标注摩尔质量，求出已知物质的摩尔数，用摩尔比求目标物质的摩尔数，再换算成质量。
4. Acids, Bases and Salts | 酸、碱和盐
H₂O. Also, link pH to hydrogen ion concentration: lower pH means higher H⁺ concentration.
WJEC 考生必须能用离子方程式 H⁺ + OH⁻ → H₂O 描述中和反应。同时能将 pH 与氢离子浓度关联：pH 越低，H⁺ 浓度越高。
For salt preparation methods, choosing the correct method is crucial. Use titration for soluble group 1 or ammonium salts, and use excess insoluble base plus filtration and crystallisation for soluble salts like copper sulfate.
制盐方法的选择至关重要。制备可溶性第Ⅰ族盐或铵盐用滴定法，制备硫酸铜等可溶性盐则用过量不溶碱加过滤和结晶法。
When naming salts, hydrochloric acid produces chlorides, sulfuric acid produces sulfates, and nitric acid produces nitrates. Students often confuse 'chloride' and 'chlorate', or miss the Roman numeral in transition metal salts such as iron(III) chloride.
命名盐时，盐酸产生 chloride，硫酸产生 sulfate，硝酸产生 nitrate。学生常混淆 chloride 和 chlorate，或漏写过渡金属盐中的罗马数字，如 iron(III) chloride。
5. Electrolysis | 电解
at the anode (positive electrode). For lead bromide: Pb²⁺ + 2e⁻ → Pb and 2Br⁻ → Br₂ + 2e⁻.
在熔融电解中，金属阳离子在阴极（负极）被还原，非金属阴离子在阳极（正极）被氧化。如溴化铅：Pb²⁺ + 2e⁻ → Pb 与 2Br⁻ → Br₂ + 2e⁻。
For aqueous solutions, the products depend on the relative reactivity. At the cathode, if the metal is more reactive than hydrogen, hydrogen gas is produced; otherwise the metal forms. At the anode, halide ions discharge preferentially unless they are dilute; otherwise oxygen is produced.
在溶液电解中，产物取决于相对活泼性。阴极上若金属比氢活泼，则产生氢气；否则金属析出。阳极上卤素离子优先放电，除非浓度极稀；否则产生氧气。
A common error is forgetting that the anode' ydroxide ions: 4OH⁻ → 2H₂O + O₂ + 4e⁻. Always balance half-equations for atoms and charge.
常见错误是遗忘阳极产生的氧气来自氢氧根离子氧化：4OH⁻ → 2H₂O + O₂ + 4e⁻。务必配平半方程式中的原子与电荷。
6. Redox Reactions and Reactivity Series | 氧化还原反应与金属活动性顺序
Oxidation is loss of electrons, reduction is gain of electrons – metal will displace a less reactive metal from its compound, e.g. Fe + CuSO₄ → FeSO₄ + Cu.
氧化是失电子，还原是得电子——OILRIG。在置换反应中，较活泼金属能把较不活泼金属从其化合物中置换出来，如 Fe + CuSO₄ → FeSO₄ + Cu。
⁺ + Cu. Identify which species is oxidised (Fe to Fe²⁺) and which is reduced (Cu²⁺ to Cu).
写氧化还原离子方程式时，删除旁观离子。上述反应的离子方程式为 Fe + Cu²⁺ → Fe²⁺ + Cu。指出哪种物质被氧化（Fe 变为 Fe²⁺），哪种被还原（Cu²⁺ 变为 Cu）。
To predict reactions with water or dilute acid, refer to the reactivity series. Potassium, sodium and lithium react vigorously with water; magnesium requires steam; zinc and iron only react with dilute acid; copper and silver do not react with acid.
预测与水或稀酸的反应需参考活动性顺序。钾、钠、锂与水剧烈反应；镁需与水蒸气反应；锌和铁仅与稀酸反应；铜和银不与酸反应。
7. The Periodic Table and Group Trends | 元素周期表与族趋势
For Group 1 (alkali metals), reactivity increases down the group because the outer electron is further from the nucleus and more easily lost. Melting points decrease down the group.
对于第Ⅰ族（碱金属），反应活性向下增强，因为最外层电子离核更远，更易失去。熔点向下降低。
Group 7 (halogens) show the opposite trend: reactivity decreases down the group as it becomes harder to gain an electron. Fluorine is the most reactive, iodine the least among the common halogens.
第Ⅶ族（卤素）趋势相反：向下反应活性减弱，因为得电子更难。常见卤素中氟最活泼，碘最不活泼。
de solution liberates bromine, turning the solution orange-brown: Cl₂ + 2KBr → 2KCl + Br₂.
卤素与卤化物溶液间的置换反应可展示该趋势。氯水加入溴化钾溶液中释出溴，使溶液变为橙棕色：Cl₂ + 2KBr → 2KCl + Br₂。
8. Rate of Reaction | 反应速率
Collision theory states that for a reaction to occur, particles must collide with sufficient energy (activation energy) and correct orientation. Factors affecting rate include temperature, concentration, surface area and catalysts.
碰撞理论指出，发生反应的粒子必须以足够能量（活化能）和正确取向碰撞。影响速率的因素包括温度、浓度、表面积和催化剂。
When interpreting rate graphs, remember that a steeper slope means a faster rate. The total amount of product formed is usually the same if the same mass of limiting reactant is used, regardless of rate differences.
解读速率图时记住，斜率越陡速率越快。若使用相同质量的限量反应物，无论速率快慢，生成物总量最终相同。
For a required practical, to investigate the effect of concentration on rate, you might measure the volume of gas evolved over time or the time for a cross to disappear. Always identify the independent variable, dependent variable and controls.
在必做实验中探究浓度对速率的影响，可测量气体体积随时间的变化或十字消失所需时间。务必确定自变量、因变量和控制变量。
9. Energetics: Exothermic and Endothermic Reactions | 能量学：放热与吸热反应
Exothermic reactions release heat to the surroundings, resulting in a temperature rise, while endothermic reactions absorb heat, causing a temperature drop. Bond making is exothermic, bond breaking is endothermic.
放热反应向环境放热，导致温度升高；吸热反应吸热，导致温度降低。成键放热，断键吸热。
al energy of bonds broken − total energy of bonds formed. A negative value indicates an exothermic reaction; a positive value indicates an endothermic reaction.
键能计算是常见考点。总能量变化 = 断裂键总能量 − 形成键总能量。负值表示放热反应；正值表示吸热反应。
t form 2 C=O bonds and 4 O−H bonds. Always draw full structural displays to avoid mistakes.
常见错误是数错键。例如甲烷燃烧 CH₄ + 2O₂ → CO₂ + 2H₂O 中，断裂 4 个 C−H 键和 2 个 O=O 键，形成 2 个 C=O 键和 4 个 O−H 键。始终画出完整结构式以避免出错。
10. Organic Chemistry Introduction: Alkanes and Alkenes | 有机化学入门：烷烃和烯烃
₂O. Alkenes are unsaturated, containing a C=C double bond, with the general formula CₙH₂ₙ.
烷烃是通式为 CₙH₂ₙ₊₂ 的饱和烃。它们完全燃烧生成 CO₂ 和 H₂O。烯烃是不饱和烃，含 C=C 双键，通式为 CₙH₂ₙ。
The bromine water test distinguishes between alkanes and alkenes: alkenes decolourise bromine water from orange to colourless via an addition reaction, while alkanes do not react (no colour change).
溴水测试区分烷烃和烯烃：烯烃通过加成反应使溴水从橙色褪至无色，而烷烃不反应（无颜色变化）。
When naming organic compounds in WJEC, identify the longest carbon chain, number to give the double bond or functional group the lowest number, and use prefixes like meth-, eth-, prop-, but-.
在 WJEC 命名有机化合物时，确定最长碳链，编号使双键或官能团位次最小，并运用 meth-、eth-、prop-、but- 等前缀。
Draw displayed structural formulas carefully, showing every bond. For but-1-ene, the double bond is between carbon 1 and carbon 2. Ensure no hydrogen atoms are missing, as marks are allocated for precision.
仔细绘制展开结构式，展示每个键。如 but-1-ene 的双键位于第 1 和第 2 碳之间。确保不缺氢原子，因为这涉及评分精确性。
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