Typical Worked Examples for WJEC A-Level Science | WJEC A-Level 科学典型例题详解

📚 Typical Worked Examples for WJEC A-Level Science | WJEC A-Level 科学典型例题详解

WJEC A-Level Science papers require both conceptual understanding and the ability to apply knowledge to unfamiliar situations. This article walks you through detailed worked examples across physics, chemistry and biology, showing how marks are awarded and where common mistakes occur. Each example is paired with a full Chinese translation so you can follow every step.

WJEC A-Level 科学考试既考概念理解，也考在新情境中应用知识的能力。本文带你深入物理、化学和生物的典型例题详解，展示评分要点和常见错误。每道例题都配有完整的中文翻译，方便你跟紧每一步。

1. Kinematics: Uniform Acceleration | 运动学：匀加速直线运动

A car accelerates uniformly from rest at 2.0 m s⁻² for 5.0 s. Calculate its final velocity and the displacement during this time.

一辆汽车从静止开始以 2.0 m s⁻² 匀加速运动 5.0 s。计算末速度和这段时间内的位移。

Step 1: Use v = u + at. u = 0, a = 2.0 m s⁻², t = 5.0 s → v = 0 + (2.0)(5.0) = 10 m s⁻¹.

步骤 1：使用 v = u + at。u = 0, a = 2.0 m s⁻², t = 5.0 s → v = 0 + (2.0)(5.0) = 10 m s⁻¹。

Step 2: Displacement s = ut + ½at² = 0×5.0 + ½ × 2.0 × (5.0)² = 0 + ½ × 2.0 × 25 = 25 m. (Check with average velocity: s = (u+v)/2 × t = 25 m.)

步骤 2：位移 s = ut + ½at² = 0×5.0 + ½ × 2.0 × (5.0)² = 0 + ½ × 2.0 × 25 = 25 m。（用平均速度验证：s = (u+v)/2 × t = 25 m。）

Key point: Always list the known quantities (u, v, a, t, s) and choose the equation that links them. Here, using s = (u+v)t/2 was quicker once v was known.

要点：务必列出已知量 (u, v, a, t, s) 并选择关联它们的方程。本题中知道 v 后用 s = (u+v)t/2 更快。

2. Newton’s Second Law on an Inclined Plane | 斜面上的牛顿第二定律

A 5.0 kg block slides down a smooth slope inclined at 30° to the horizontal. Ignoring friction, calculate its acceleration.

一个 5.0 kg 的木块沿光滑斜面（与水平夹角 30°）下滑。忽略摩擦，计算它的加速度。

Step 1: Resolve weight mg into components: parallel to slope = mg sin 30°, perpendicular = mg cos 30°. Only the parallel component causes acceleration.

步骤 1：将重力 mg 分解：平行于斜面的分量为 mg sin 30°，垂直于斜面的分量为 mg cos 30°。只有平行分量产生加速度。

Step 2: Apply F = ma along the slope: mg sin 30° = ma → a = g sin 30°. Since g = 9.81 m s⁻², a = 9.81 × 0.5 = 4.91 m s⁻².

步骤 2：沿斜面应用 F = ma：mg sin 30° = ma → a = g sin 30°。g = 9.81 m s⁻²，a = 9.81 × 0.5 = 4.91 m s⁻²。

Key point: With no friction, mass cancels out; acceleration depends only on g and the angle.

要点：无摩擦时，质量约掉；加速度仅取决于 g 和角度。

3. Circuit Analysis: Series and Parallel Resistors | 电路分析：串并联电阻

Three resistors: R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 12 Ω. R₂ and R₃ are connected in parallel, and this combination is in series with R₁. The supply is 12 V. Find the current through R₁ and the p.d. across R₂.

三个电阻：R₁ = 4 Ω，R₂ = 6 Ω，R₃ = 12 Ω。R₂ 与 R₃ 并联，该并联组合再与 R₁ 串联。电源电压 12 V。求流过 R₁ 的电流和 R₂ 两端的电压。

Step 1: Calculate equivalent resistance of R₂ ∥ R₃: 1/Rₚ = 1/6 + 1/12 = 3/12 = 1/4 → Rₚ = 4 Ω.

步骤 1：计算 R₂ 与 R₃ 并联的等效电阻：1/Rₚ = 1/6 + 1/12 = 3/12 = 1/4 → Rₚ = 4 Ω。

Step 2: Total resistance Rₜ = R₁ + Rₚ = 4 + 4 = 8 Ω. Total current I = V/Rₜ = 12/8 = 1.5 A. This current flows through R₁, so I(R₁) = 1.5 A.

步骤 2：总电阻 Rₜ = R₁ + Rₚ = 4 + 4 = 8 Ω。总电流 I = V/Rₜ = 12/8 = 1.5 A。此电流流过 R₁，所以 R₁ 的电流为 1.5 A。

Step 3: The p.d. across the parallel block Vₚ = I × Rₚ = 1.5 × 4 = 6 V. This is the same voltage across R₂, so V(R₂) = 6 V. Verify: current through R₂ = Vₚ/6 = 1 A, through R₃ = Vₚ/12 = 0.5 A, total = 1.5 A, correct.

步骤 3：并联块两端的电压 Vₚ = I × Rₚ = 1.5 × 4 = 6 V。这也是 R₂ 两端的电压，所以 V(R₂) = 6 V。验证：流过 R₂ 的电流 = 6/6 = 1 A，流过 R₃ 的电流 = 6/12 = 0.5 A，总和 1.5 A，正确。

4. Titration Calculation: Acid-Base Neutralisation | 滴定计算：酸碱中和

25.0 cm³ of 0.100 mol dm⁻³ NaOH solution is titrated against H₂SO₄ of unknown concentration. The equivalence point is reached after adding 18.5 cm³ of the acid. Determine the concentration of the H₂SO₄.

用 25.0 cm³ 0.100 mol dm⁻³ NaOH 溶液滴定未知浓度的 H₂SO₄。加入 18.5 cm³ 酸后到达等当点。计算 H₂SO₄ 的浓度。

Step 1: Write the balanced equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Mole ratio NaOH : H₂SO₄ = 2 : 1.

步骤 1：写出配平方程式：2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O。摩尔比 NaOH : H₂SO₄ = 2 : 1。

Step 2: Moles of NaOH = c × V = 0.100 × (25.0/1000) = 0.00250 mol.

步骤 2：NaOH 的物质的量 = c × V = 0.100 × (25.0/1000) = 0.00250 mol。

Step 3: From mole ratio, moles of H₂SO₄ = 0.00250 / 2 = 0.00125 mol.

步骤 3：根据摩尔比，H₂SO₄ 的物质的量 = 0.00250 / 2 = 0.00125 mol。

Step 4: Concentration of H₂SO₄ = moles/volume (dm³) = 0.00125 / (18.5/1000) = 0.00125 / 0.0185 = 0.0676 mol dm⁻³ (3 s.f.).

步骤 4：H₂SO₄ 浓度 = 物质的量 / 体积 (dm³) = 0.00125 / (18.5/1000) = 0.00125 / 0.0185 = 0.0676 mol dm⁻³（3 位有效数字）。

5. Chemical Equilibrium: Calculating Kc | 化学平衡：计算 Kc

For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), a mixture initially containing 0.500 mol of H₂ and 0.500 mol of I₂ in a vessel of volume 2.0 dm³ reaches equilibrium. The equilibrium mixture contains 0.800 mol of HI. Calculate Kc.

反应 H₂(g) + I₂(g) ⇌ 2HI(g)，在 2.0 dm³ 容器中初始含有 0.500 mol H₂ 和 0.500 mol I₂，达到平衡。平衡混合物中含有 0.800 mol HI。计算 Kc。

Step 1: Set up an ICE table (Initial, Change, Equilibrium). Moles: H₂ and I₂ both start at 0.500. HI starts at 0. Change: let x = moles of H₂ reacted. Then H₂ and I₂ each decrease by x, HI increases by 2x. At equilibrium: HI = 0 + 2x = 0.800 → x = 0.400 mol. So H₂ at eqm = 0.500 – 0.400 = 0.100 mol; I₂ also 0.100 mol.

步骤 1：建立 ICE 表格（初始值、变化量、平衡值）。物质的量：H₂ 和 I₂ 初始均为 0.500 mol，HI 为 0。设反应掉的 H₂ 为 x mol，则 H₂ 和 I₂ 各减少 x，HI 增加 2x。平衡时：HI = 0 + 2x = 0.800 → x = 0.400 mol。所以平衡时 H₂ = 0.500 – 0.400 = 0.100 mol；I₂ 也是 0.100 mol。

Step 2: Convert to concentrations (volume 2.0 dm³): [H₂] = 0.100/2.0 = 0.0500 mol dm⁻³, [I₂] = 0.0500 mol dm⁻³, [HI] = 0.800/2.0 = 0.400 mol dm⁻³.

步骤 2：换算浓度（体积 2.0 dm³）：[H₂] = 0.100/2.0 = 0.0500 mol dm⁻³，[I₂] = 0.0500 mol dm⁻³，[HI] = 0.400 mol dm⁻³。

Step 3: Kc = [HI]² / ([H₂][I₂]) = (0.400)² / (0.0500 × 0.0500) = 0.160 / 0.00250 = 64.0 (no units, as Δn = 0).

步骤 3：Kc = [HI]² / ([H₂][I₂]) = (0.400)² / (0.0500 × 0.0500) = 0.160 / 0.00250 = 64.0（无单位，因 Δn = 0）。

6. Organic Reaction Mechanism: Electrophilic Addition of Br₂ to Ethene | 有机反应机理：Br₂ 与乙烯的亲电加成

Describe the mechanism for the reaction between ethene (C₂H₄) and bromine (Br₂), explaining the formation of 1,2-dibromoethane.

描述乙烯 (C₂H₄) 与溴 (Br₂) 反应的机理，解释 1,2-二溴乙烷的生成。

Step 1: The double bond in ethene is an electron-rich region. As bromine approaches, the π-bond polarises the Br–Br bond, causing heterolytic fission. Br₂ → Br⁺ + Br⁻. The Br⁺ acts as an electrophile.

步骤 1：乙烯的双键是电子富集区。当溴靠近时，π 键使 Br–Br 键极化，导致异裂：Br₂ → Br⁺ + Br⁻。Br⁺ 作为亲电试剂。

Step 2: The Br⁺ accepts a pair of π-electrons from ethene, forming a C–Br bond and leaving a carbocation (bromonium ion intermediate, or a cyclic bromonium ion). The intermediate is unstable and carries a positive charge on the three-membered ring.

步骤 2：Br⁺ 接受来自乙烯的一对 π 电子，形成一个 C–Br 键，同时留下一个碳正离子（溴鎓离子中间体，或环状溴鎓离子）。中间体不稳定，三元环带一个正电荷。

Step 3: The bromide ion (Br⁻) attacks the carbocation rapidly from the opposite side (anti addition), forming the second C–Br bond. Product: 1,2-dibromoethane, CH₂Br–CH₂Br.

步骤 3：溴离子 Br⁻ 从背面快速进攻碳正离子（反式加成），形成第二个 C–Br 键。产物：1,2-二溴乙烷，CH₂Br–CH₂Br。

Key point: The mechanism involves an electrophilic attack and subsequent nucleophilic attack, with anti addition leading to stereospecificity.

要点：该机理涉及亲电进攻和随后的亲核进攻，反式加成导致立体专一性。

7. Energetics: Hess’s Law Calculation | 能量学：Hess 定律计算

Given the following standard enthalpy changes of formation: ΔHf°[CO₂(g)] = -394 kJ mol⁻¹, ΔHf°[H₂O(l)] = -286 kJ mol⁻¹, ΔHf°[C₂H₅OH(l)] = -278 kJ mol⁻¹. Calculate the standard enthalpy of combustion of ethanol, C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l).

已知下列标准生成焓变：ΔHf°[CO₂(g)] = -394 kJ mol⁻¹, ΔHf°[H₂O(l)] = -286 kJ mol⁻¹, ΔHf°[C₂H₅OH(l)] = -278 kJ mol⁻¹。计算乙醇的标准燃烧焓变：C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)。

Step 1: Apply Hess’s Law: ΔH°comb = Σ ΔHf°(products) – Σ ΔHf°(reactants). Remember elements in standard state (like O₂) have ΔHf° = 0.

步骤 1：应用 Hess 定律：ΔH°comb = Σ ΔHf°(产物) – Σ ΔHf°(反应物)。注意标准状态下的单质（如 O₂）的 ΔHf° = 0。

Step 2: Σ ΔHf°(products) = 2 × (-394) + 3 × (-286) = -788 – 858 = -1646 kJ mol⁻¹.

步骤 2：Σ ΔHf°(产物) = 2 × (-394) + 3 × (-286) = -788 – 858 = -1646 kJ mol⁻¹。

Step 3: Σ ΔHf°(reactants) = 1 × (-278) + 3 × (0) = -278 kJ mol⁻¹.

步骤 3：Σ ΔHf°(反应物) = 1 × (-278) + 3 × 0 = -278 kJ mol⁻¹。

Step 4: ΔH°comb = -1646 – (-278) = -1368 kJ mol⁻¹.

步骤 4：ΔH°comb = -1646 – (-278) = -1368 kJ mol⁻¹。

8. Genetics: Monohybrid Cross and Punnett Square | 遗传学：单因子杂交与旁氏表

In pea plants, tall (T) is dominant over dwarf (t). Two heterozygous tall plants (Tt) are crossed. Predict the genotypic and phenotypic ratios of the offspring.

在豌豆中，高茎 (T) 对矮茎 (t) 为显性。两株杂合高茎 (Tt) 植株杂交。预测子代的基因型比例和表现型比例。

Step 1: Determine gametes: Each parent (Tt) produces gametes T and t with equal probability.

步骤 1：确定配子：每个亲本 (Tt) 产生 T 和 t 配子，概率相等。

Step 2: Construct a Punnett square:

步骤 2：构建旁氏表：

	T	t
T	TT	Tt
t	Tt	tt

Step 3: Genotypic ratio: 1 TT : 2 Tt : 1 tt. Phenotypic ratio: Since both TT and Tt are tall, the ratio tall : dwarf = 3 : 1.

步骤 3：基因型比例：1 TT : 2 Tt : 1 tt。表现型比例：由于 TT 和 Tt 都是高茎，高茎 : 矮茎 = 3 : 1。

9. Enzyme Kinetics: Michaelis-Menten Calculation | 酶动力学：Michaelis-Menten 计算

An enzyme-catalysed reaction has Vmax = 100 μmol min⁻¹ and Km = 0.05 mM. Calculate the initial rate (v₀) when the substrate concentration [S] is 0.20 mM.

某酶催化反应的 Vmax = 100 μmol min⁻¹，Km = 0.05 mM。当底物浓度 [S] 为 0.20 mM 时，计算初始速率 v₀。

Step 1: Use the Michaelis-Menten equation: v₀ = (Vmax [S]) / (Km + [S]).

步骤 1：使用米氏方程：v₀ = (Vmax [S]) / (Km + [S])。

Step 2: Substitute values: v₀ = (100 × 0.20) / (0.05 + 0.20) = 20 / 0.25 = 80 μmol min⁻¹.

步骤 2：代入数值：v₀ = (100 × 0.20) / (0.05 + 0.20) = 20 / 0.25 = 80 μmol min⁻¹。

Key point: When [S] is well above Km, v₀ approaches Vmax. Here [S]/Km = 4, giving 80% Vmax.

要点：当 [S] 远大于 Km 时，v₀ 趋近于 Vmax。此处 [S]/Km = 4，达到 Vmax 的 80%。

10. Ecology: Mark-Release-Recapture | 生态学：标记重捕法

In a lake, researchers capture 120 perch, mark them with a harmless tag and release them. Two days later, a second sample of 95 perch is caught, of which 19 are marked. Estimate the total population N.

在一个湖泊中，研究者捕获了 120 条鲈鱼，用无害标签标记后放回。两天后，第二次捕获了 95 条鲈鱼，其中 19 条带有标记。估算总种群数量 N。

Step 1: Use the Lincoln Index: N = (M × C) / R, where M = number initially marked (120), C = size of second sample (95), R = number of marked recaptures (19).

步骤 1：使用林肯指数：N = (M × C) / R，其中 M = 初次标记数 (120)，C = 第二次样本大小 (95)，R = 重捕标记数 (19)。

Step 2: N = (120 × 95) / 19 = 11400 / 19 = 600. Estimated population ≈ 600 perch.

步骤 2：N = (120 × 95) / 19 = 11400 / 19 = 600。估算种群约 600 条鲈鱼。

Assumptions: no migration, no births/deaths, marks are not lost, and marked fish mix randomly. These must be stated for full marks.

假设：无迁入迁出，无出生死亡，标记不脱落，标记鱼随机混合。这些需陈述才能得满分。

11. Wave Optics: Young’s Double-Slit Experiment | 波动光学：杨氏双缝干涉

In a double-slit experiment, light of wavelength 600 nm illuminates two slits separated by 0.25 mm. Calculate the fringe spacing on a screen 2.0 m away.

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Typical Worked Examples for WJEC A-Level Science | WJEC A-Level 科学典型例题详解