Year 11 Cambridge Science: Mock Unit Test Walkthrough | 11年级剑桥科学：单元测试模拟卷解析

📚 Year 11 Cambridge Science: Mock Unit Test Walkthrough | 11年级剑桥科学：单元测试模拟卷解析

Preparing for your Cambridge IGCSE Science exam means practising with realistic mock tests – not just to check facts, but to sharpen your application skills. This walkthrough takes you through a full unit test covering Biology, Chemistry and Physics. Each question is broken down step by step, so you can see exactly how marks are earned and deepen your understanding of core concepts like enzymes, electrolysis, forces and titration.

准备剑桥 IGCSE 科学考试，不只是核对知识点，更要通过模拟卷提升应用能力。本文带你逐题解析一份涵盖生物、化学和物理的完整单元测试。每道题都分步详解，让你清楚得分要点，同时深入理解酶、电解、力和滴定等核心概念。

1. Biology Multiple Choice: Effect of pH on Amylase Activity | 生物选择题：pH 对淀粉酶活性的影响

Question: A student investigates the breakdown of starch by amylase at pH 4, pH 7 and pH 9. Iodine solution is used to indicate the presence of starch. After 10 minutes, the iodine remains blue‑black at pH 4 and pH 9, but turns yellow‑brown at pH 7. Which conclusion is correct?

问题：某学生研究 pH 4、pH 7 和 pH 9 下淀粉酶对淀粉的分解。用碘液指示淀粉。10 分钟后，pH 4 和 pH 9 的碘液保持蓝黑色，而 pH 7 变为黄褐色。哪项结论正确？

A	Amylase works best at pH 4.
B	Amylase is denatured at pH 9 only.
C	The optimum pH for amylase is around 7.
D	Starch is broken down faster in acidic conditions.

Blue‑black indicates starch is still present. Yellow‑brown at pH 7 shows starch has been fully broken down, meaning amylase was active. At pH 4 and pH 9, the enzyme’s shape changed, reducing its ability to bind to starch. The optimum pH is therefore near neutral.

蓝黑色表明淀粉仍存在。pH 7 下变黄褐色说明淀粉已被完全分解，淀粉酶具有活性。pH 4 和 pH 9 下酶的形状改变，与淀粉结合能力下降。因此最适 pH 接近中性。

Answer: C | 答案：C

2. Chemistry Multiple Choice: Balancing a Displacement Reaction | 化学选择题：置换反应配平

Question: Magnesium reacts with copper(II) chloride solution according to the equation below. What is the correct set of coefficients?

__ Mg + __ CuCl₂ → __ MgCl₂ + __ Cu

A	1, 2, 1, 2
B	1, 1, 1, 1
C	2, 1, 2, 1
D	1, 1, 2, 1

Count atoms on each side: one Mg, one Cu and two Cl atoms are already balanced with Mg + CuCl₂ → MgCl₂ + Cu. All coefficients equal 1. No further adjustment is needed because the charges also balance: Mg loses two electrons, Cu²⁺ gains them.

点数原子：左右各有一个 Mg、一个 Cu 和两个 Cl，已经配平。所有系数均为 1。电荷也平衡：Mg 失去两个电子，Cu²⁺ 得到它们。

Answer: B | 答案：B

3. Physics Multiple Choice: Interpreting a Potential Divider | 物理选择题：分压器分析

Question: A light‑dependent resistor (LDR) and a fixed resistor form a potential divider. When light intensity increases, the voltmeter reading across the fixed resistor rises. What happens to the resistance of the LDR and the current in the circuit?

问题：光敏电阻 (LDR) 与定值电阻组成分压器。光照增强时，定值电阻两端电压表读数上升。LDR 的电阻和电路中的电流如何变化？

Option	LDR resistance	Current
A	decreases	increases
B	decreases	decreases
C	increases	increases
D	increases	decreases

Higher light intensity lowers the LDR resistance. Total resistance of the series circuit drops, so current increases (Ohm’s law). The fixed resistor then gets a larger share of the supply voltage, matching the rise on the voltmeter.

光照增强降低 LDR 电阻。串联电路总电阻下降，电流增大（欧姆定律）。定值电阻分得更多电源电压，与电压表上升一致。

Answer: A | 答案：A

4. Biology Structured: Structure of the Heart and Double Circulation | 生物结构化题：心脏结构与双循环

(i) Describe the path taken by a red blood cell from the right ventricle through the lungs and back to the left atrium. Include the names of key blood vessels and valves. [3 marks]

(i) 描述红细胞从右心室出发，经肺回到左心房的路径。写出关键血管和瓣膜的英文名称。 [3 分]

The oxygen‑poor blood leaves the right ventricle through the pulmonary valve into the pulmonary artery. It passes through capillary beds in the lungs, where gas exchange occurs. Oxygenated blood returns via the pulmonary veins into the left atrium.

缺氧血经肺动脉瓣离开右心室，进入肺动脉。通过肺部毛细血管网进行气体交换。充氧血经肺静脉返回左心房。

(ii) Explain why the wall of the left ventricle is thicker than the wall of the right ventricle. [2 marks]

(ii) 解释为什么左心室壁比右心室壁更厚。 [2 分]

The left ventricle pumps blood to the whole body (systemic circuit), requiring much higher pressure to overcome resistance in the extensive network of arteries. The right ventricle only pumps blood to the nearby lungs, a shorter circuit with lower resistance, so a thinner wall suffices.

左心室将血液泵至全身（体循环），需要更高的压力以克服广大动脉网的阻力。右心室仅将血泵至邻近的肺，路程短、阻力小，较薄的室壁就足够了。

5. Chemistry Structured: Rate of Reaction and Collision Theory | 化学结构化题：反应速率与碰撞理论

Zinc granules react with dilute hydrochloric acid to produce hydrogen gas. Describe and explain the effect on the rate of reaction when:

锌粒与稀盐酸反应生成氢气。描述并解释以下情况对反应速率的影响：

(a) the acid is heated before adding zinc. [2 marks]

(a) 加入锌粒前先加热酸。 [2 分]

Heating increases the kinetic energy of acid particles. They move faster and collide more frequently with zinc atoms. More of these collisions have energy greater than the activation energy, so the rate of successful collisions rises, speeding up the reaction.

加热使酸粒子的动能增大。它们运动更快，与锌原子碰撞更频繁。更多碰撞的能量超过活化能，因此有效碰撞频率增加，反应加快。

(b) powdered zinc is used instead of granules, keeping mass constant. [2 marks]

(b) 使用锌粉代替锌粒，质量不变。 [2 分]

Powdered zinc has a much larger total surface area in contact with the acid. This provides more sites for collisions between reactant particles per second, increasing the frequency of successful collisions without changing the activation energy.

锌粉与酸的总接触表面积大大增加。每秒有更多反应物粒子碰撞位点，有效碰撞频率上升，而活化能不变。

6. Physics Structured: Interpreting a Velocity‑Time Graph | 物理结构化题：速度‑时间图像分析

A car moves along a straight road. The velocity‑time graph shows three sections: a straight line sloping upward from (0,0) to (10, 20 m/s), a horizontal line at 20 m/s from 10 s to 30 s, and a straight line sloping downward from (30,20) to (40,0).

一辆汽车沿直线行驶。速度‑时间图有三段：从 (0,0) 到 (10, 20 m/s) 的上升直线；10 s 到 30 s 保持在 20 m/s 的水平线；从 (30,20) 到 (40,0) 的下降直线。

(a) Calculate the acceleration during the first 10 s. [1 mark]

(a) 计算前 10 秒的加速度。 [1 分]

Acceleration = change in velocity ÷ time = (20 m/s – 0 m/s) ÷ 10 s = 2.0 m/s².

加速度 = 速度变化量 ÷ 时间 = (20 m/s – 0) ÷ 10 s = 2.0 m/s²。

(b) Determine the total distance travelled from 0 s to 40 s. [2 marks]

(b) 求 0 s 到 40 s 行驶的总距离。 [2 分]

Distance = area under the graph. First triangle: ½ × 10 × 20 = 100 m. Rectangle: 20 s × 20 m/s = 400 m. Final triangle: ½ × 10 × 20 = 100 m. Total = 100 + 400 + 100 = 600 m.

距离 = 图像下方面积。第一个三角形：½ × 10 × 20 = 100 m。矩形：20 × 20 = 400 m。最后一个三角形：½ × 10 × 20 = 100 m。总和 600 m。

7. Chemistry Extended: Acid‑Base Titration Calculation | 化学拓展题：酸碱滴定计算

25.0 cm³ of dilute sulfuric acid (H₂SO₄) is neutralised by 20.0 cm³ of 0.40 mol/dm³ sodium hydroxide (NaOH) solution. The equation is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Calculate the concentration of sulfuric acid in mol/dm³. [3 marks]

用 25.0 cm³ 稀硫酸与 20.0 cm³ 0.40 mol/dm³ 的氢氧化钠溶液恰好中和。计算硫酸的浓度（mol/dm³）。 [3 分]

Step 1: Moles of NaOH = concentration × volume (in dm³) = 0.40 × (20.0/1000) = 0.0080 mol. Step 2: From the equation, 2 mol NaOH react with 1 mol H₂SO₄, so moles of H₂SO₄ = 0.0080 ÷ 2 = 0.0040 mol. Step 3: Concentration of H₂SO₄ = moles / volume in dm³ = 0.0040 ÷ (25.0/1000) = 0.16 mol/dm³.

步骤 1：NaOH 物质的量 = 0.40 × (20.0/1000) = 0.0080 mol。步骤 2：根据方程式，2 mol NaOH 与 1 mol H₂SO₄ 反应，H₂SO₄ 物质的量 = 0.0080 ÷ 2 = 0.0040 mol。步骤 3：硫酸浓度 = 0.0040 ÷ (25.0/1000) = 0.16 mol/dm³。

8. Physics Extended: Energy Efficiency and Sankey Diagram | 物理拓展题：能量效率与桑基图

An electric motor lifts a 50 N weight through 3.0 m in 5.0 s. The motor receives 60 J of electrical energy. (a) Calculate the useful work done on the weight. [1 mark] (b) Calculate the efficiency of the motor. [2 marks] (c) Sketch a labelled Sankey diagram for the energy transfer. [2 marks]

一台电动机在 5.0 秒内将 50 N 的重物提升 3.0 m。电动机共接收 60 J 电能。(a) 计算对重物做的有用功。 (b) 计算电动机效率。 (c) 画出标有数值的桑基图。

(a) Work done = force × distance moved in direction of force = 50 N × 3.0 m = 150 J. Note: 150 J of useful work is larger than the 60 J input – a red flag, but the numbers are illustrative. Let me adjust to make sense: if the motor receives 200 J, not 60 J, to allow efficiency below 1. I’ll correct the question: motor receives 200 J of electrical energy. Then useful work = 150 J. Efficiency = useful output / total input = 150 J / 200 J = 0.75 or 75%. In the Sankey diagram, the arrow representing 200 J splits: a broad arrow labelled ‘useful work (150 J)’ points to the lifted weight, and a thinner arrow ‘thermal energy (50 J)’ branches downward.

修改后的数据：电动机接收 200 J 电能。(a) 有用功 = 力 × 沿力方向移动的距离 = 50 N × 3.0 m = 150 J。(b) 效率 = 有用输出 / 总输入 = 150 J / 200 J = 0.75 即 75%。(c) 桑基图：代表 200 J 的箭头分成较宽的“有用功 150 J”指向重物，较窄的“热能 50 J”分支向下。

9. Biology Practical: Investigating Osmosis in Potato Tissue | 生物实验题：土豆组织渗透作用探究

A student places potato cylinders of equal mass into different sucrose solutions (0.0, 0.2, 0.4, 0.6, 0.8 mol/dm³). After 30 minutes, the cylinders are blotted and reweighed. The percentage change in mass is plotted against sucrose concentration. At 0.4 mol/dm³ the mass change is zero.

某学生将同等质量的土豆条放入不同浓度蔗糖溶液中（0.0、0.2、0.4、0.6、0.8 mol/dm³）。30 分钟后吸干水分重新称重，以质量变化百分比对蔗糖浓度作图。在 0.4 mol/dm³ 时质量变化为零。

(a) Explain why the mass increases in 0.0 mol/dm³ solution. [2 marks]

(a) 解释为什么在 0.0 mol/dm³ 溶液中质量增加。 [2 分]

Distilled water has a higher water potential than the potato cells. Water moves by osmosis from the external solution into the cells, causing the cells to swell and the tissue mass to increase.

蒸馏水的水势高于土豆细胞。水通过渗透作用从外部溶液进入细胞，细胞膨胀，组织质量增加。

(b) State what the zero mass change indicates and estimate the water potential of the potato tissue in terms of sucrose concentration. [2 marks]

(b) 说明质量变化为零表示什么，并用蔗糖浓度估算土豆组织的水势。 [2 分]

Zero change means there is no net water movement between the potato tissue and the solution; the solution is isotonic to the cell sap. The sucrose concentration that gives zero change (0.4 mol/dm³) can be used as a measure of the water potential of the tissue, approximately equivalent to the water potential of a 0.4 mol/dm³ sucrose solution.

质量不变说明土豆组织与溶液间无净水移动，溶液与细胞液等渗。产生零变化时的蔗糖浓度（0.4 mol/dm³）可用来衡量组织的水势，约等于 0.4 mol/dm³ 蔗糖溶液的水势。

10. Chemistry Practical: Electrolysis of Aqueous Sodium Chloride | 化学实验题：氯化钠水溶液的电解

In the electrolysis of concentrated aqueous sodium chloride using inert electrodes:

(a) Name the products formed at the anode and cathode. [2 marks]

用惰性电极电解浓氯化钠水溶液：(a) 写出阳极和阴极产物。

At the cathode (–), hydrogen gas (H₂) is produced because H⁺ ions are discharged more readily than Na⁺. At the anode (+), chlorine gas (Cl₂) is produced because the concentration of Cl⁻ is high, so Cl⁻ is discharged in preference to OH⁻.

阴极 (–) 产生氢气 (H₂)，因为 H⁺ 比 Na⁺ 更易放电。阳极 (+) 产生氯气 (Cl₂)，因为 Cl⁻ 浓度高，优先于 OH⁻ 放电。

(b) Write ionic half‑equations for both electrode reactions. [2 marks]

(b) 写出两个电极反应的离子半方程式。 [2 分]

Cathode: 2H⁺ + 2e⁻ → H₂. Anode: 2Cl⁻ → Cl₂ + 2e⁻.

阴极：2H⁺ + 2e⁻ → H₂；阳极：2Cl⁻ → Cl₂ + 2e⁻。

As H⁺ ions are discharged at the cathode, the equilibrium H₂O ⇌ H⁺ + OH⁻ shifts to replace H⁺, leaving an excess of OH⁻ ions in the solution. The remaining solution becomes sodium hydroxide, which is strongly alkaline.

阴极上 H⁺ 放电，水的电离平衡 H₂O ⇌ H⁺ + OH⁻ 向右移动补充 H⁺，溶液中留下了过量的 OH⁻。剩余溶液为氢氧化钠，呈强碱性。

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Year 11 Cambridge Science: Mock Unit Test Walkthrough | 11年级剑桥科学：单元测试模拟卷解析