📚 OCR Science: Typical Example Questions Explained | OCR 科学：典型例题详解
This article walks you through ten carefully selected, exam-style questions from OCR GCSE Science. Each example includes a step-by-step bilingual solution, helping you switch smoothly between English and Chinese while mastering key concepts in Biology, Chemistry and Physics.
本文精选十道 OCR 科学考试风格的典型例题，逐一进行中英双语详解。每道题都配有分步解答，帮助你在巩固生物、化学和物理核心知识的同时，提升中英双语切换表达的能力。
1. Microscopy Calculation | 显微镜计算
A student observes a plant cell under a light microscope. The image of the cell measures 7.5 mm across, and the actual cell diameter is 0.25 mm. Calculate the magnification used.
一名学生在光学显微镜下观察植物细胞，图像中细胞的直径为 7.5 mm，实际细胞直径为 0.25 mm。求使用的放大倍数。
Write down the formula: magnification = image size ÷ actual size. Ensure both measurements are in the same units.
列出公式：放大倍数 = 图像大小 ÷ 实际大小。务必确保两个量使用相同单位。
Substitute the values: magnification = 7.5 mm ÷ 0.25 mm = 30. Magnification has no unit because it is a ratio.
代入数值：放大倍数 = 7.5 mm ÷ 0.25 mm = 30。放大倍数是比值，因此没有单位。
Examiner’s tip: Always check whether the actual size needs to be converted, e.g. from micrometres to millimetres (1 mm = 1000 µm).
考官提示：务必检查实际尺寸是否需要单位换算，如从微米换算为毫米（1 mm = 1000 µm）。
2. Enzyme Activity and Denaturation | 酶活性与变性
An experiment measures the rate of an enzyme‑controlled reaction at different temperatures. The rate increases from 10 °C up to 40 °C, but above 50 °C the rate falls rapidly to zero. Explain these observations.
一项实验测量了不同温度下某种酶促反应的速率。在 10 °C 到 40 °C 之间反应速率升高，但超过 50 °C 后速率迅速下降至零。请解释上述现象。
At lower temperatures, increasing thermal energy makes enzyme and substrate molecules move faster, so they collide more often and the rate increases.
在较低温度下，热能增加使酶和底物分子运动加快，碰撞频率提高，因此反应速率上升。
Above the optimum temperature (around 40 °C), the weak bonds holding the enzyme’s tertiary structure break. The active site changes shape and the substrate can no longer fit. The enzyme is denatured and the reaction stops.
超过最适温度（约 40 °C）后，维持酶三级结构的弱键断裂。活性位点形状改变，底物不再契合。酶已变性，反应终止。
Key term: Denaturation is permanent – cooling the enzyme will not restore its activity.
关键词：变性是不可逆的——冷却酶也无法恢复其活性。
3. Ionic Bonding and Properties | 离子键与性质
Explain why solid sodium chloride does not conduct electricity, but molten sodium chloride does. Use ideas about structure and bonding.
解释为何固态氯化钠不导电，而熔融态氯化钠可以导电。请从结构与成键的角度回答。
In solid NaCl, Na⁺ and Cl⁻ ions are held in a fixed lattice by strong electrostatic forces. The ions cannot move, so no electrical current can flow.
固态 NaCl 中，Na⁺ 和 Cl⁻ 离子被强静电作用固定在晶格中。离子无法自由移动，因此电流不能通过。
When melted, the lattice breaks down and the ions become free to move. These mobile charged particles can carry the electric current towards the electrodes.
熔化后晶格被破坏，离子可以自由移动。这些可移动的带电粒子能将电流传导至电极。
Many students confuse electrons with ions – in ionic compounds, electricity is conducted by mobile ions, not by mobile electrons.
许多学生混淆电子与离子——在离子化合物中，导电靠的是自由移动的离子，而非自由电子。
4. Mole Calculation in a Reaction | 化学反应中的摩尔计算
Magnesium burns in oxygen: 2Mg + O₂ → 2MgO. Calculate the mass of magnesium oxide formed when 4.8 g of magnesium reacts completely. (Aᵣ: Mg = 24, O = 16)
镁在氧气中燃烧：2Mg + O₂ → 2MgO。当 4.8 g 镁完全反应时，计算生成的氧化镁质量。（相对原子质量：Mg = 24，O = 16）
Step 1: Moles of Mg = mass ÷ molar mass = 4.8 g ÷ 24 g mol⁻¹ = 0.20 mol.
步骤1：Mg 的物质的量 = 质量 ÷ 摩尔质量 = 4.8 g ÷ 24 g mol⁻¹ = 0.20 mol。
Step 2: From the equation, 2 mol Mg produce 2 mol MgO, so mole ratio is 1:1. Thus moles of MgO = 0.20 mol.
步骤2：由方程式，2 mol Mg 生成 2 mol MgO，物质的量之比为 1:1。所以 MgO 的物质的量为 0.20 mol。
Step 3: Mᵣ of MgO = 24 + 16 = 40. Mass of MgO = moles × Mᵣ = 0.20 × 40 = 8.0 g.
步骤3：MgO 的相对分子质量 = 24 + 16 = 40。MgO 质量 = 物质的量 × Mᵣ = 0.20 × 40 = 8.0 g。
5. Electrolysis Half‑Equations | 电解半方程式
Aqueous copper(II) chloride is electrolysed using inert carbon electrodes. Write the half‑equations for the reactions at the cathode and the anode, and name the products.
用惰性碳电极电解氯化铜水溶液。写出阴极和阳极的半方程式，并指明产物。
At the cathode (negative electrode): Cu²⁺ ions gain electrons. Cu²⁺ + 2e⁻ → Cu. The product is copper metal, a reddish‑brown solid.
阴极（负极）：Cu²⁺ 离子得电子。Cu²⁺ + 2e⁻ → Cu。产物为铜金属，红棕色固体。
At the anode (positive electrode): Cl⁻ ions lose electrons. 2Cl⁻ → Cl₂ + 2e⁻. The product is chlorine gas, which turns damp blue litmus paper red then white.
阳极（正极）：Cl⁻ 离子失电子。2Cl⁻ → Cl₂ + 2e⁻。产物为氯气，可使湿润的蓝色石蕊试纸先变红后变白。
Remember: In aqueous solutions, if halide ions are present, they are discharged at the anode in preference to OH⁻ ions.
注意：在水溶液中，卤素离子存在时，会比 OH⁻ 优先在阳极放电。
6. Acceleration and Distance | 加速度与路程
A cyclist accelerates from rest to 24 m/s in 8.0 seconds. Calculate the acceleration and the total distance travelled during this time.
一名自行车手从静止加速到 24 m/s，用时 8.0 s。计算加速度和这段时间内的总路程。
Acceleration a = (final velocity – initial velocity) ÷ time = (24 m/s – 0) ÷ 8.0 s = 3.0 m/s².
加速度 a =（末速度 – 初速度）÷ 时间 = (24 m/s – 0) ÷ 8.0 s = 3.0 m/s²。
Distance s can be found using s = ut + ½at². With u = 0, s = ½ × 3.0 × (8.0)² = 1.5 × 64 = 96 m.
路程 s 使用 s = ut + ½at²。u = 0，因此 s = ½ × 3.0 × (8.0)² = 1.5 × 64 = 96 m。
Alternatively, average velocity = (0 + 24)/2 = 12 m/s; distance = average velocity × time = 12 × 8.0 = 96 m. Both methods are valid.
也可用平均速度 = (0 + 24)/2 = 12 m/s；路程 = 平均速度 × 时间 = 12 × 8.0 = 96 m。两种方法均正确。
7. Ohm’s Law and Resistance | 欧姆定律与电阻
A resistor is connected to a 9.0 V battery and a current of 0.45 A flows through it. Calculate the resistance and state what happens to the current if the voltage is doubled while the temperature remains constant.
一个电阻接在 9.0 V 电池上，通过的电流为 0.45 A。计算电阻值，并说明保持温度不变、电压加倍时电流如何变化。
Using Ohm’s law: R = V ÷ I = 9.0 V ÷ 0.45 A = 20 Ω.
应用欧姆定律：R = V ÷ I = 9.0 V ÷ 0.45 A = 20 Ω。
For an ohmic conductor at constant temperature, current is directly proportional to potential difference. Doubling the voltage to 18 V would double the current to 0.90 A.
对于恒定温度下的欧姆导体，电流与电压成正比。电压加倍至 18 V，电流随之加倍至 0.90 A。
Common mistake: Do not assume V = I × R can be used without checking if the component is ohmic – filament lamps and diodes are non‑ohmic.
常见错误：未检查元件是否为欧姆导体就直接使用 V = I × R——灯丝和二极管均非欧姆元件。
8. Radioactive Decay and Half‑Life | 放射性衰变与半衰期
A radioactive isotope has a half‑life of 3 hours. The initial count rate is 1200 counts per minute. What will the count rate be after 12 hours?
一种放射性同位素的半衰期为 3 小时，初始计数率为每分钟 1200 次。求 12 小时后的计数率。
Number of half‑lives elapsed = total time ÷ half‑life = 12 h ÷ 3 h = 4 half‑lives.
已过去的半衰期数 = 总时间 ÷ 半衰期 = 12 小时 ÷ 3 小时 = 4 个半衰期。
After each half‑life, the count rate halves. Sequence: 1200 → 600 → 300 → 150 → 75 counts per minute.
每经过一个半衰期，计数率减半。变化序列：1200 → 600 → 300 → 150 → 75 次/分钟。
Conclusion: After 12 hours the count rate drops to 75 counts per minute. Background count rate may need to be subtracted for accurate measurement.
结论：12 小时后计数率降为 75 次/分钟。精确测量时可能需要扣除本底计数率。
9. Limiting Factors of Photosynthesis | 光合作用的限制因素
A graph shows the rate of photosynthesis increasing with light intensity until it levels off. Explain what limits the rate at low and high light intensities.
图像显示，光合作用速率随光照强度增加而升高，随后趋于平稳。解释在低光照和高光照条件下，分别是什么因素限制了速率。
At low light intensity, light is the limiting factor. Although CO₂ concentration and temperature may be adequate, the plant cannot photosynthesise faster without more energy from light.
在低光照强度下，光照是限制因素。即使 CO₂ 浓度和温度可能充裕，没有更多的光能，光合速率也无法提高。
At high light intensity, the rate plateaus because another factor becomes limiting – typically CO₂ concentration or temperature, depending on which is in shortest supply.
在高光照下，速率趋于平稳是因为另一个因素成为限制因素——通常是 CO₂ 浓度或温度，取决于哪一个最稀缺。
On the plateau, adding more light does not increase the rate; the enzyme Rubisco may be working at its maximum capacity.
在平稳区增加光照已不能提高速率；酶 Rubisco 可能已满负荷工作。
10. Acid–Base Titration Calculation | 酸碱滴定计算
25.0 cm³ of 0.200 mol dm⁻³ sodium hydroxide (NaOH) neutralises 20.0 cm³ of hydrochloric acid (HCl). Calculate the concentration of the acid.
25.0 cm³ 的 0.200 mol dm⁻³ 氢氧化钠溶液（NaOH）可中和 20.0 cm³ 的盐酸（HCl）。计算盐酸的浓度。
Balanced equation: NaOH + HCl → NaCl + H₂O. The mole ratio is 1:1.
配平方程式：NaOH + HCl → NaCl + H₂O。物质的量之比为 1:1。
Moles of NaOH = concentration × volume = 0.200 mol dm⁻³ × (25.0/1000) dm³ = 0.00500 mol.
NaOH 的物质的量 = 浓度 × 体积 = 0.200 mol dm⁻³ × (25.0/1000) dm³ = 0.00500 mol。
From the 1:1 ratio, moles of HCl = 0.00500 mol. Concentration of HCl = moles ÷ volume = 0.00500 mol ÷ (20.0/1000) dm³ = 0.250 mol dm⁻³.
由 1:1 的比例得，HCl 的物质的量 = 0.00500 mol。HCl 浓度 = 物质的量 ÷ 体积 = 0.00500 mol ÷ (20.0/1000) dm³ = 0.250 mol dm⁻³。
Remember to convert cm³ to dm³ by dividing by 1000. Using consistent units is essential for correct answers.
务必记得将 cm³ 除以 1000 换算为 dm³。单位统一是得出正确答案的关键。
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