📚 GCSE WJEC Science: Common Mistake Questions Explained | GCSE WJEC 科学：易错题精讲
As students prepare for GCSE WJEC Science, many find that seemingly straightforward questions can lead to lost marks due to common pitfalls. This article dissects typical error-prone questions across Physics, Chemistry and Biology, offering clear explanations to help you avoid these mistakes.
在准备 GCSE WJEC 科学考试时，许多学生发现一些看似简单的题目常常因为常见陷阱而失分。本文剖析了物理、化学和生物中典型的易错题，提供清晰的讲解，帮助你避免这些错误。
1. Unit Conversion in Speed Calculations | 速度计算中的单位换算
One of the most frequent errors involves converting km/h to m/s. Students often take the number of km/h and treat it as m/s directly. For example, a car travels 72 km in 1 hour; some will write the speed as 72 m/s, forgetting that 1 km = 1000 m and 1 hour = 3600 s.
最常见的错误之一是 km/h 与 m/s 的单位换算。很多学生直接把 km/h 的数字当作 m/s 使用。例如，一辆汽车 1 小时行驶 72 km，有人会直接写下速度 72 m/s，却忘了 1 km = 1000 m，1 h = 3600 s。
Correct method: speed (m/s) = speed (km/h) × (1000 m / 3600 s) = speed (km/h) ÷ 3.6. So 72 km/h ÷ 3.6 = 20 m/s. Always check the units and use the factor 3.6 when converting between these common measures of speed.
正确方法：速度 (m/s) = 速度 (km/h) × (1000 m / 3600 s) = 速度 (km/h) ÷ 3.6。因此 72 km/h ÷ 3.6 = 20 m/s。在转换这两种常见速度单位时，务必检查单位并使用换算因子 3.6。
2. Confusing Atom and Ion Symbols | 原子符号与离子符号的混淆
A common slip in Chemistry is writing ion symbols without their charge or adding a charge where none exists. For instance, a sodium ion must be written as Na⁺, not Na, because it has lost one electron. Conversely, a neutral sodium atom is Na, never Na⁺. The same applies to calcium: Ca for the atom, Ca²⁺ for the ion.
化学中常见的疏忽是离子符号不写电荷，或者给中性原子添加电荷。例如，钠离子必须写成 Na⁺，而不是 Na，因为它失去了一个电子。相反，中性钠原子是 Na，绝不能写成 Na⁺。钙也是如此：原子为 Ca，离子为 Ca²⁺。
In equations, missing charges can break the balancing of ionic compounds or prevent you from scoring marks. Remember: the charge must appear as a superscript—Na⁺, Cl⁻, O²⁻, Al³⁺. When drawing ions in dot-and-cross diagrams, the charge must be clearly indicated.
在方程式中，遗漏电荷会破坏离子化合物的配平或导致丢分。记住：电荷必须以上标形式出现——Na⁺、Cl⁻、O²⁻、Al³⁺。在绘制离子点叉图时，也必须清晰标出电荷。
3. Interpreting Enzyme Rate Graphs | 解读酶反应速率图
Many candidates incorrectly state that the rate of an enzyme-catalysed reaction keeps increasing as temperature rises. A typical graph shows the rate rising to an optimum point and then falling steeply. The mistake is failing to mention denaturation: above the optimum temperature (often around 37°C for human enzymes), the enzyme’s active site changes shape and loses its function.
许多考生错误地声称，随着温度升高，酶催化反应的速率会持续上升。典型的图线显示速率上升到最适点，然后急剧下降。常见的错误是没提到变性作用：当温度超过最适温度（人体酶通常在 37°C 左右）后，酶的活性部位形状发生改变，失去功能。
When describing such a graph, always structure your answer as: ‘As temperature increases, particles gain kinetic energy, so the rate rises up to the optimum. Beyond this temperature, the enzyme denatures, so the active site is no longer complementary to the substrate and the rate decreases.’ This full explanation secures full marks. The same principle applies to pH graphs, where extremes of acidity or alkalinity cause denaturation.
描述这类图线时，始终按以下结构回答：“随着温度升高，粒子获得动能，因此速率上升至最适温度。超过此温度后，酶变性，活性部位不再与底物互补，速率下降。”完整的解释才能拿满分。同样的原则也适用于 pH 图线，过酸或过碱都会导致变性。
4. Calculating Total Resistance in Parallel | 并联电路总电阻的计算
A classic trap is adding resistances in parallel as if they were in series. If two 4 Ω resistors are connected in parallel, some students will write Rtotal = 4 + 4 = 8 Ω. The correct rule for two resistors in parallel is 1/Rtotal = 1/R₁ + 1/R₂. Therefore, 1/Rtotal = 1/4 + 1/4 = 1/2, giving Rtotal = 2 Ω.
典型的陷阱是把并联电阻当成串联来计算。如果把两个 4 Ω 电阻并联，一些学生会写下 R总 = 4 + 4 = 8 Ω。正确的并联电阻公式是 1/R总 = 1/R₁ + 1/R₂。因此 1/R总 = 1/4 + 1/4 = 1/2，得出 R总 = 2 Ω。
Remember this essential check: the total resistance in parallel is always smaller than the smallest individual resistance. If your answer is larger than any single resistor, you have made a mistake. For more than two resistors, simply add more terms: 1/Rtotal = 1/R₁ + 1/R₂ + 1/R₃ + …
记住一个关键的检验方法：并联总电阻总是小于最小的单个电阻值。如果算出的答案比任何一个分电阻都大，那就是出错了。当超过两个电阻并联时，只需在倒数求和中增加更多的项：1/R总 = 1/R₁ + 1/R₂ + 1/R₃ + …
5. Relative Atomic Mass from Isotopic Abundances | 根据同位素丰度求相对原子质量
When calculating relative atomic mass (Aᵣ) from isotopic data, many students simply take the average of the mass numbers, ignoring percentage abundances. For chlorine, which has 75% ³⁵Cl and 25% ³⁷Cl, an incorrect calculation is (35 + 37) ÷ 2 = 36. The correct approach uses a weighted average: Aᵣ = (35 × 75 + 37 × 25) ÷ 100 = 35.5.
根据同位素数据计算相对原子质量（Aᵣ）时，许多学生直接取质量数的平均值，而忽略了百分比丰度。氯有 75% 的 ³⁵Cl 和 25% 的 ³⁷Cl，错误的计算是 (35 + 37) ÷ 2 = 36。正确方法应使用加权平均：Aᵣ = (35 × 75 + 37 × 25) ÷ 100 = 35.5。
In the WJEC exam, you must show the full expression, even if the answer is a decimal. Pay attention to the units: relative atomic mass has no unit. Set out your working step by step to avoid missing any abundance. If you are given a table, always check that the percentages add up to 100% or use the given ratios carefully.
在 WJEC 考试中，即使答案是小数，也必须写出完整的计算式。注意单位：相对原子质量没有单位。一步一步列出计算过程，避免遗漏任何丰度。如果给出的是表格，始终检查百分比之和是否为 100%，或谨慎使用给出的比值。
6. Predicting Monohybrid Cross Outcomes | 单基因杂交结果的预测
Using Punnett squares is straightforward, yet students frequently mix up genotype and phenotype ratios. In a cross between two heterozygous organisms (Bb × Bb), the genotype ratio is 1 BB : 2 Bb : 1 bb. Some will incorrectly give this as a phenotype ratio. The phenotype ratio depends on dominance: if B is dominant, the phenotype ratio is 3 dominant : 1 recessive.
庞纳特方格法本身很简单，但学生经常混淆基因型比例和表现型比例。在两个杂合子（Bb × Bb）的杂交中，基因型比例为 1 BB : 2 Bb : 1 bb。有人误把该比例当作表现型比例。表现型比例取决于显隐性：若 B 为显性，表现型比例则是 3 显性 : 1 隐性。
When a question asks for ‘the probability of an offspring being heterozygous,’ identify the specific genotype Bb out of the four boxes. In a Bb × Bb cross, the chance is 2/4 = 1/2 or 50%. Always label your gametes and offspring genotypes clearly. Use the terms ‘homozygous dominant,’ ‘heterozygous’ and ‘homozygous recessive’ precisely to earn communication marks.
当题目问“后代为杂合子的概率”时，要从四个方格中确定具体的 Bb 基因型。在 Bb × Bb 杂交中，概率为 2/4 = 1/2 或 50%。始终要清晰标注配子和后代的基因型。准确使用“纯合显性”“杂合子”和“纯合隐性”等术语，才能拿到表达分。
7. Balancing Equations with Polyatomic Ions | 含原子团离子的方程式配平
When polyatomic ions (such as CO₃²⁻, SO₄²⁻, NO₃⁻) appear in an equation, many candidates break them up prematurely and then struggle with atom counts. A typical example is Na₂CO₃ + HCl → NaCl + H₂O + CO₂. Attempting to balance each atom individually often leads to confusion. The trick is to treat the carbonate group CO₃ as a single unit where possible.
当方程式里出现原子团离子（如 CO₃²⁻、SO₄²⁻、NO₃⁻）时，许多考生过早把它们拆开，然后在原子数中出错。典型例子是 Na₂CO₃ + HCl → NaCl + H₂O + CO₂。如果试图逐个原子配平，常常造成混乱。诀窍是尽可能将碳酸根 CO₃ 视为一个整体。
Balanced equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. Notice how the CO₃ unit stays intact in Na₂CO₃ and CO₂. In exams, check that each side has the same number of each atom and that charges balance. If a polyatomic ion does not break apart, do not split it prematurely—this simplifies your work enormously.
配平后的方程式：Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂。注意 CO₃ 整体在 Na₂CO₃ 和 CO₂ 中保持不变。考试中要检查两边各原子数目相同，且电荷平衡。如果原子团没有分解，就不要过早把它拆开——这会极大简化你的步骤。
8. Energy Efficiency Calculations | 能量效率的计算
Efficiency questions lose marks for two main reasons: forgetting to multiply by 100 to give a percentage, and using inconsistent units. The formula Efficiency = (useful output energy / total input energy) × 100% must be applied correctly. If a filament lamp takes in 100 J of electrical energy and gives 10 J of light, the efficiency is (10/100) × 100% = 10%. Writing 0.1 alone scores no percentage mark.
效率题的丢分原因主要有两个：忘记乘以 100 得出百分比，以及单位不一致。公式 效率 =（有用输出能量 / 总输入能量）× 100% 必须正确使用。如果一个白炽灯输入 100 J 电能，输出 10 J 光能，效率为 (10/100) × 100% = 10%。只写 0.1 是拿不到百分比分数的。
Always double-check that both energy values are in the same unit—joules or kilojoules. Some exam questions deliberately give input in kJ and output in J to catch you out. Convert everything to joules first, substitute into the formula, and finish with the % sign. Remember, efficiency has no unit, but you must state it as a percentage in GCSE exams.
始终要再次确认两个能量值的单位是否相同——焦耳 (J) 或千焦 (kJ)。有些考题会故意把输入以 kJ 给出，输出以 J 给出，诱你犯错。先将所有单位换算成焦耳，代入公式，最后加上 % 符号。记住，效率没有单位，但在 GCSE 考试中必须用百分比表示。
9. Predicting Products of Electrolysis | 电解产物的预测
Electrolysis predictions are a frequent source of error. Candidates often ignore the state of the substance (molten or aqueous) and the nature of the electrodes. In molten sodium chloride (NaCl), there are only Na⁺ and Cl⁻ ions, so the cathode produces sodium metal and the anode produces chlorine gas. Students sometimes mistakenly write ‘oxygen’ here because they confuse it with aqueous electrolysis.
电解产物的预测是常出错的地方。考生经常忽略物质的状态（熔融还是水溶液）以及电极的性质。在熔融氯化钠 (NaCl) 中，只有 Na⁺ 和 Cl⁻ 离子，因此阴极产生钠金属，阳极产生氯气。学生有时会误写成“氧气”，因为他们将其与电解水溶液的情况混淆了。
For aqueous solutions, you must know the reactivity series and anion discharge order. At the cathode, if a metal is more reactive than hydrogen (e.g. Na⁺, K⁺, Ca²⁺), hydrogen gas is produced. If less reactive (e.g. Cu²⁺, Ag⁺), the metal is formed. At the anode, in concentrated halide solutions, the halide (Cl⁻, Br⁻, I⁻) may be discharged; otherwise, oxygen from OH⁻ is released. Carbon or platinum electrodes do not react. Always apply given concentration clues.
对于水溶液，必须掌握金属活动性顺序和阴离子放电顺序。在阴极，如果金属比氢活泼（如 Na⁺、K⁺、Ca²⁺），则生成氢气；如果不如氢活泼（如 Cu²⁺、Ag⁺），则析出金属。在阳极，浓卤化物溶液中，卤素离子 (Cl⁻, Br⁻, I⁻) 可能放电，否则由 OH⁻ 放电产生氧气。碳或铂电极不参与反应。一定要根据题目给出的浓度信息进行分析。
10. Accurately Reading Values from Graphs | 准确读取图表数据
Data-extraction questions test a fundamental scientific skill, yet many marks are dropped through careless reading. For a line graph showing the change in temperature over time, a student might read the temperature at 30 s as ‘approximately 35°C’ when the fine grid shows it is 34°C. Always use a ruler (or your eyes carefully) to draw vertical and horizontal lines to the axes.
数据提取题考验基本科学技能，但很多分数因粗心读数而丢失。对于一条显示温度随时间变化的线图，学生可能把 30 秒时的温度读作“大约 35°C”，而细格子显示的是 34°C。应始终用直尺（或仔细目测）作垂直线和水平线到坐标轴上。
Before reading, determine the value of each small division on both axes. For example, if the y-axis runs from 0 to 50 over 10 large
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