📚 GCSE WJEC Science: Worked Examples Explained | GCSE WJEC 科学：典型例题详解
This selection of worked examples covers typical GCSE WJEC Science questions in Biology, Chemistry and Physics. Each example shows you how to break down a problem, apply key formulas and explain scientific concepts clearly. Mastering these patterns will help you gain confidence for your exams.
本系列典型例题覆盖 GCSE WJEC 科学考试中生物、化学和物理的常见题型。每道例题都会展示如何拆解问题、运用核心公式并清晰地解释科学概念。熟练掌握这些解题模式将增强你的应试信心。
1. Osmosis in Potato Strips | 马铃薯条渗透作用
A student investigated osmosis by placing potato strips in different sugar solutions. The initial mass of each strip was 5.0 g. After one hour the strips were blotted dry and their final masses recorded.
一名学生将马铃薯条放入不同浓度的蔗糖溶液中探究渗透作用。每条马铃薯的初始质量均为 5.0 g。一小时后将薯条吸干并记录最终质量。
- 0.0 mol/dm³ sugar: final mass 5.8 g, change +0.8 g
- 0.2 mol/dm³ sugar: final mass 5.4 g, change +0.4 g
- 0.4 mol/dm³ sugar: final mass 5.0 g, change 0.0 g
- 0.6 mol/dm³ sugar: final mass 4.6 g, change −0.4 g
- 0.8 mol/dm³ sugar: final mass 4.3 g, change −0.7 g
- 1.0 mol/dm³ sugar: final mass 4.0 g, change −1.0 g
- 0.0 mol/dm³ 蔗糖溶液: 最终质量 5.8 g，变化 +0.8 g
- 0.2 mol/dm³ 蔗糖溶液: 最终质量 5.4 g，变化 +0.4 g
- 0.4 mol/dm³ 蔗糖溶液: 最终质量 5.0 g，变化 0.0 g
- 0.6 mol/dm³ 蔗糖溶液: 最终质量 4.6 g，变化 −0.4 g
- 0.8 mol/dm³ 蔗糖溶液: 最终质量 4.3 g，变化 −0.7 g
- 1.0 mol/dm³ 蔗糖溶液: 最终质量 4.0 g，变化 −1.0 g
The percentage change in mass is calculated using:
质量变化百分比的计算公式如下：
percentage change = (final mass − initial mass) ÷ initial mass × 100%
For the 0.0 mol/dm³ solution: (5.8 − 5.0) ÷ 5.0 × 100% = +16.0%. This positive change indicates water entered the potato by osmosis because the external solution had a higher water potential than the potato cells.
以 0.0 mol/dm³ 溶液为例：(5.8 − 5.0) ÷ 5.0 × 100% = +16.0%。该正值表明水分通过渗透作用进入马铃薯，因为外部溶液的水势高于马铃薯细胞。
At 0.4 mol/dm³ the mass remained unchanged. This is the isotonic point where water potential inside and outside the cells is equal, so there is no net movement of water.
在 0.4 mol/dm³ 时质量未变。这是等渗点，细胞内外水势相等，因此无净水移动。
2. Investigating Enzyme Activity | 酶活性探究
An enzyme called amylase breaks down starch. A student tested the rate of reaction at different temperatures by measuring the time taken for starch to disappear.
一种名为淀粉酶的酶能分解淀粉。一名学生通过测量淀粉消失所需的时间来测试不同温度下的反应速率。
- 0 °C: time > 300 s
- 10 °C: 180 s
- 20 °C: 90 s
- 30 °C: 45 s
- 40 °C: 30 s
- 50 °C: 80 s
- 60 °C: > 300 s
- 0 °C: 时间 > 300 s
- 10 °C: 180 s
- 20 °C: 90 s
- 30 °C: 45 s
- 40 °C: 30 s
- 50 °C: 80 s
- 60 °C: > 300 s
The shortest time (fastest rate) occurred at 40 °C, so this is the optimum temperature for amylase. As temperature rises towards the optimum, particles move faster and collide more frequently, increasing enzyme activity.
最短用时（最快速率）出现在 40 °C，因此这是淀粉酶的最适温度。当温度升至最适点时，粒子运动更快、碰撞更频繁，从而增强酶活性。
Above 40 °C the time increased sharply. High temperatures disrupt the bonds that maintain the enzyme’s active site shape, causing denaturation. The substrate can no longer fit, and the rate falls to nearly zero.
高于 40 °C 时，用时急剧增加。高温破坏了维持酶活性位点形状的键，导致变性。底物无法契合，速率几乎降至零。
3. Balancing Equations and Moles | 配平方程式与摩尔计算
Methane burns in oxygen to produce carbon dioxide and water. The balanced equation is:
甲烷在氧气中燃烧生成二氧化碳和水，其配平方程式为：
CH₄ + 2O₂ → CO₂ + 2H₂O
Calculate the mass of oxygen needed to completely burn 16 g of methane. (Ar: H = 1, C = 12, O = 16)
计算完全燃烧 16 g 甲烷所需氧气的质量。（原子量：H = 1，C = 12，O = 16）
Step 1: Molar mass of CH₄ = 12 + (4 × 1) = 16 g/mol. Moles of CH₄ = mass ÷ Mᵣ = 16 ÷ 16 = 1.0 mol.
步骤一：CH₄ 的摩尔质量 = 12 + (4 × 1) = 16 g/mol。CH₄ 的摩尔数 = 质量 ÷ 摩尔质量 = 16 ÷ 16 = 1.0 mol。
Step 2: From the equation, 1 mole of CH₄ reacts with 2 moles of O₂. So O₂ required = 2.0 mol.
步骤二：由方程式可知，1 mol CH₄ 与 2 mol O₂ 反应。因此需要 O₂ 2.0 mol。
Step 3: Molar mass of O₂ = 2 × 16 = 32 g/mol. Mass of O₂ = moles × molar mass = 2.0 × 32 = 64 g.
步骤三：O₂ 的摩尔质量 = 2 × 16 = 32 g/mol。O₂ 的质量 = 摩尔数 × 摩尔质量 = 2.0 × 32 = 64 g。
4. Electrolysis of Aqueous Sodium Chloride | 氯化钠水溶液电解
When electricity is passed through brine (concentrated NaCl solution), three products are obtained: hydrogen gas at the cathode, chlorine gas at the anode, and sodium hydroxide left in solution.
当电流通过浓盐水（NaCl 溶液）时，可获得三种产物：阴极产生氢气，阳极产生氯气，溶液中留下氢氧化钠。
At the cathode, H⁺ ions from water are discharged in preference to Na⁺ ions because H⁺ is less reactive. The half-equation is:
阴极处，水中的 H⁺ 离子比 Na⁺ 离子优先放电，因为 H⁺ 活泼性较低。半反应式为：
2H⁺ + 2e⁻ → H₂
At the anode, Cl⁻ ions are discharged instead of OH⁻ ions because the solution is concentrated chloride. The half-equation is:
阳极处，由于溶液为浓氯化物，Cl⁻ 离子优先于 OH⁻ 放电。半反应式为：
2Cl⁻ → Cl₂ + 2e⁻
The Na⁺ and OH⁻ ions remain behind, forming sodium hydroxide (NaOH) in solution.
Na⁺ 和 OH⁻ 离子留在溶液中，形成氢氧化钠 (NaOH)。
5. Ohm’s Law and Resistance in a Circuit | 电路中的欧姆定律与电阻
A student measured the potential difference across a resistor and the current flowing through it. The results were:
一名学生测量了电阻器两端的电势差和流过它的电流。结果如下：
- V = 0.0 V, I = 0.00 A
- V = 2.0 V, I = 0.50 A
- V = 4.0 V, I = 1.00 A
- V = 6.0 V,
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