📚 Year 11 CIE Science: Unit Test Mock Paper Walkthrough | 11年级 CIE 科学：单元测试模拟卷解析
This article walks you through a typical Year 11 CIE Coordinated Science unit test mock paper. Each question is broken down with a step-by-step solution in both English and Chinese to help you understand key concepts, avoid common mistakes, and build confidence for your final examination. The mock paper covers topics from Biology (B1–B4), Chemistry (C1–C4), and Physics (P1–P4) as specified in the CIE syllabus.
本文带你逐步解析一份典型的11年级 CIE 综合科学单元测试模拟卷。每道题目都配有中英双语的详细解答，帮助你理解核心概念、避开常见错误，并为最终考试建立信心。模拟卷涵盖 CIE 考纲中生物（B1–B4）、化学（C1–C4）和物理（P1–P4）的重点内容。
1. Question 1: Diffusion and Osmosis | 第1题：扩散与渗透
A student placed a piece of potato in a beaker of pure water and another piece in a concentrated sugar solution. After 30 minutes, the potato in pure water felt firmer, while the one in sugar solution became soft. Explain these observations using the terms water potential, osmosis, and turgid.
一名学生将一块土豆放入纯水中，另一块放入浓糖水中。30分钟后，纯水中的土豆变得更硬实，而糖水中的土豆变软。请用水势、渗透和硬挺解释这些现象。
English Solution: Pure water has a higher water potential than the potato cell cytoplasm, so water moves into the cells by osmosis. The cells become turgid, making the potato feel firm. The concentrated sugar solution has a lower water potential than the cytoplasm, so water leaves the cells. The cells become flaccid, and the potato turns soft.
中文解析：纯水的水势比土豆细胞质的水势高，因此水通过渗透进入细胞。细胞吸水后变得硬挺，土豆摸起来较硬。浓糖水的水势低于细胞质的水势，因此水从细胞中流失，细胞变得松弛，土豆因此变软。
2. Question 2: Enzyme Activity and pH | 第2题：酶活性与pH
An investigation recorded the time taken for amylase to break down starch at different pH values. The shortest time was observed at pH 7, while at pH 2 and pH 12 no breakdown occurred. Explain these results in terms of enzyme structure and active site.
一项实验记录了淀粉酶在不同pH下分解淀粉所需的时间。pH 7时反应时间最短，而pH 2和pH 12时淀粉完全不被分解。请从酶的结构和活性位点角度解释这些结果。
English Solution: Amylase has an optimum pH of 7. At this pH, the shape of the active site is complementary to the starch substrate, so the enzyme–substrate complex forms easily. At extreme pH values (2 and 12), the enzyme denatures: the bonds holding the tertiary structure break, changing the shape of the active site permanently. The substrate can no longer fit, and no reaction occurs.
中文解析：淀粉酶的最适pH为7。在该pH下，活性位点的形状与淀粉底物互补，酶-底物复合物易于形成。在极端pH值（2和12）下，酶发生变性：维持三级结构的键断裂，活性位点形状永久改变。底物无法再结合，因此不发生反应。
3. Question 3: Ionic Bonding and Lattice Structure | 第3题：离子键与晶格结构
Sodium chloride (NaCl) has a high melting point and conducts electricity when molten but not when solid. Explain these properties using ideas of bonding and structure.
氯化钠 (NaCl) 熔点高，熔融态能导电但固态不能。请用化学键和结构的知识解释这些性质。
English Solution: Sodium chloride is an ionic compound with a giant ionic lattice. Strong electrostatic forces of attraction between oppositely charged Na⁺ and Cl⁻ ions require a lot of energy to overcome, so the melting point is high. In the solid state, the ions are fixed in position and cannot move, so it does not conduct electricity. When molten, the ions are free to move and carry charge, enabling electrical conductivity.
中文解析：氯化钠是一种离子化合物，具有巨型离子晶格结构。带相反电荷的 Na⁺ 和 Cl⁻ 离子之间的强静电吸引力需要大量能量才能克服，因此熔点很高。固态时，离子位置固定不能移动，故不导电。熔融状态下离子可自由移动并携带电荷，因此能够导电。
4. Question 4: Reactivity Series and Displacement | 第4题：金属活动性顺序与置换反应
A student adds magnesium ribbon to copper(II) sulfate solution. A pink-brown solid appears and the blue solution fades. Write the word equation and explain why this reaction occurs with reference to the reactivity series.
一名学生将镁条放入硫酸铜(II)溶液中。溶液中出现粉棕色固体，蓝色逐渐褪去。写出文字方程式，并根据金属活动性顺序解释反应发生的原因。
English Solution: er. Magnesium is more reactive than copper, so it displaces copper from the copper sulfate solution. The pink-brown solid is copper metal, and the blue colour fades because Cu²⁺ ions are removed from the solution and replaced by colourless Mg²⁺ ions.
中文解析：文字方程式：镁 + 硫酸铜(II) → 硫酸镁 + 铜。镁比铜更活泼，因此镁能从硫酸铜溶液中将铜置换出来。粉棕色固体是金属铜，蓝色褪去是因为溶液中的 Cu²⁺ 离子被移除，被无色的 Mg²⁺ 离子取代。
5. Question 5: Moles and Concentration | 第5题：摩尔与浓度
Calculate the concentration in mol/dm³ when 0.40 moles of sodium hydroxide are dissolved in water to make 250 cm³ of solution.
将0.40摩尔氢氧化钠溶于水，配置成250 cm³的溶液，计算其浓度（单位 mol/dm³）。
English Solution: olume (dm³). Volume = 250 cm³ = 0.250 dm³. So concentration = 0.40 / 0.250 = 1.60 mol/dm³.
中文解析：浓度 (mol/dm³) = 物质的量 (mol) / 体积 (dm³)。体积 = 250 cm³ = 0.250 dm³。因此浓度 = 0.40 / 0.250 = 1.60 mol/dm³。
6. Question 6: Forces and Acceleration | 第6题：力与加速度
A car of mass 1200 kg accelerates from rest to 20 m/s in 8.0 seconds. Calculate the resultant force acting on the car, assuming constant acceleration.
一辆质量为1200 kg 的汽车从静止加速到20 m/s，用时8.0秒。假设加速度恒定，计算作用在汽车上的合力。
English Solution: First find acceleration: a = (v – u) / t = (20 – 0) / 8.0 = 2.5 m/s². Then use Newton's second law: F = m × a = 1200 kg × 2.5 m/s² = 3000 N.
中文解析：先求加速度：a = (v – u) / t = (20 – 0) / 8.0 = 2.5 m/s²。再用牛顿第二定律：F = m × a = 1200 kg × 2.5 m/s² = 3000 N。
7. Question 7: Energy Transfers and Efficiency | 第7题：能量转移与效率
An electric motor lifts a 50 N weight through a vertical height of 3.0 m. The motor uses 250 J of electrical energy. Calculate the useful work done and the efficiency of the motor.
一台电动机将一个50 N 的重物竖直提升3.0 m。电动机消耗了250 J 的电能。计算有用功和电动机的效率。
English Solution: nce = 50 N × 3.0 m = 150 J. Efficiency = (useful energy output / total energy input) × 100% = (150 / 250) × 100% = 60%.
中文解析：有用功 = 力 × 距离 = 50 N × 3.0 m = 150 J。效率 = (有用输出能量 / 总输入能量) × 100% = (150 / 250) × 100% = 60%.
8. Question 8: Wave Properties | 第8题：波的性质
A water wave has a frequency of 5.0 Hz and a wavelength of 0.40 m. Calculate the wave speed. State what happens to the frequency and wavelength if the wave enters deeper water and speeds up.
一个水波频率为5.0 Hz，波长为0.40 m。计算波速。如果该波进入更深水域而加速，说明频率和波长会如何变化。
English Solution: ns unchanged. Since v = f λ, if v increases and f stays the same, the wavelength λ must increase.
中文解析：波速 v = f × λ = 5.0 Hz × 0.40 m = 2.0 m/s。波进入更深水域时，速度增加。频率由波源决定，保持不变。根据 v = f λ，若 v 增大而 f 不变，则波长 λ 必然增大。
9. Question 9: Genetic Cross and Monohybrid Inheritance | 第9题：遗传杂交与单因子遗传
In pea plants, the allele for tall stems (T) is dominant over the allele for short stems (t). Two heterozygous tall plants (Tt) are crossed. Use a Punnett square to determine the expected ratio of phenotypes in the offspring.
在豌豆中，高茎等位基因 (T) 对矮茎等位基因 (t) 为显性。将两株杂合高茎植株 (Tt) 杂交。用旁氏表确定子代预期表现型比例。
English Solution: Parental gametes: T and t from each parent.
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
中文解析：亲本配子：每个亲本产生 T 和 t。
| T | t | |
| T | TT | Tt |
| t | Tt | tt |
10. Question 10: Electrolysis of Molten Lead(II) Bromide | 第10题：熔融溴化铅(II) 的电解
Describe what is observed at the anode and cathode during the electrolysis of molten lead(II) bromide. Include half-equations for both electrode reactions.
描述电解熔融溴化铅(II)时阳极和阴极分别观察到什么现象。写出两个电极反应的半方程式。
English Solution: node (positive electrode): brown bromine gas bubbles off. Half-equation: 2Br⁻ → Br₂ + 2e⁻.
中文解析：阴极（负极）：生成灰色金属铅。半方程式：Pb²⁺ + 2e⁻ → Pb。阳极（正极）：产生红棕色溴蒸气气泡。半方程式：2Br⁻ → Br₂ + 2e⁻。
11. Question 11: Momentum and Collisions | 第11题：动量与碰撞
A 0.50 kg trolley moving at 4.0 m/s collides with a stationary 1.0 kg trolley. They stick together. Calculate the velocity of the combined trolleys after the collision. State the principle used.
一辆0.50 kg 的小车以4.0 m/s 的速度撞上一辆静止的1.0 kg 小车，两者粘在一起。计算碰撞后组合小车的速度，并说明所使用的原理。
English Solution: ed mass = 1.50 kg, velocity = v. So 1.50 × v = 2.0, v = 2.0 / 1.50 = 1.33 m/s (to 3 s.f.).
中文解析：动量守恒定律：碰撞前总动量 = 碰撞后总动量。碰撞前：(0.50 × 4.0) + (1.0 × 0) = 2.0 kg m/s。碰撞后总质量 = 1.50 kg，速度为 v。因此 1.50 × v = 2.0，v = 2.0 / 1.50 ≈ 1.33 m/s（保留三位有效数字）。
Published by TutorHao | Coordinated Science Revision Series | aleveler.com
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