📚 IGCSE CIE Science Unit Test: Practice Questions with Answers | IGCSE CIE 科学单元测试卷：练习题与答案
This unit test compilation provides targeted practice for IGCSE CIE Science students, covering key concepts across Biology, Chemistry, and Physics. Each question is designed to reflect the style and depth of real exam papers, helping you reinforce understanding and identify areas for improvement. Use these questions as a self-assessment tool before your end-of-topic tests.
这份单元测试卷汇编为 IGCSE CIE 科学的学生提供了有针对性的练习，涵盖生物、化学和物理的关键概念。每道题都旨在反映真实试卷的风格和深度，帮助你巩固理解并找出薄弱环节。在单元结束测验前，用这些题目进行自我评估。
1. Biology: Cell Organelles | 生物：细胞器
Which organelle is responsible for aerobic respiration?
A. Nucleus
B. Mitochondria
C. Ribosome
D. Chloroplast
哪个细胞器负责有氧呼吸？
A. 细胞核
B. 线粒体
C. 核糖体
D. 叶绿体
Answer: B. Mitochondria are the site of aerobic respiration, producing ATP. Other organelles have different functions: the nucleus stores DNA, ribosomes synthesise proteins, and chloroplasts carry out photosynthesis.
答案：B。线粒体是有氧呼吸的场所，产生 ATP。其他细胞器有不同的功能：细胞核储存 DNA，核糖体合成蛋白质，叶绿体进行光合作用。
2. Biology: Enzymes | 生物：酶
An investigation was carried out into the effect of pH on the activity of the enzyme amylase. The rate of starch breakdown was highest at pH 7. Suggest why the rate of reaction decreased when the pH was changed to 9.
进行了一项研究 pH 对淀粉酶活性影响的实验。淀粉分解的速率在 pH 7 时最高。当 pH 变为 9 时，反应速率下降，请解释原因。
Answer: At pH 9, the enzyme’s active site changes shape due to denaturation. The shape of the active site is no longer complementary to the starch substrate, so the enzyme-substrate complex cannot form effectively, and the rate of reaction decreases.
答案：在 pH 9 时，酶因变性导致活性位点的形状改变。活性位点的形状不再与淀粉底物互补，因此酶-底物复合物无法有效形成，反应速率降低。
3. Biology: Nutrition and Digestion | 生物：营养与消化
Which nutrient is broken down by the enzyme protease?
A. Starch
B. Protein
C. Fat
D. Glucose
蛋白酶分解哪种营养物质？
A. 淀粉
B. 蛋白质
C. 脂肪
D. 葡萄糖
Answer: B. Protease acts on proteins, breaking them down into amino acids. Starch is digested by amylase, fats by lipase, and glucose is a simple sugar that does not require protease.
答案：B。蛋白酶作用于蛋白质，将其分解为氨基酸。淀粉由淀粉酶消化，脂肪由脂肪酶消化，葡萄糖是一种单糖，不需要蛋白酶。
4. Biology: Transport in Plants | 生物：植物运输
Name the plant tissue responsible for transporting water and dissolved minerals from the roots to the leaves. Briefly explain how this tissue is adapted for its function.
说出负责将水分和溶解的矿物质从根部运输到叶片的植物组织。简要解释该组织如何适应其功能。
Answer: Xylem. Xylem vessels are made of dead cells arranged end to end to form continuous hollow tubes, with no cytoplasm to obstruct flow. Their walls are strengthened with lignin, which provides structural support and prevents collapse under tension during transpiration.
答案：木质部。木质部导管由死细胞首尾相连形成连续的空心管道，没有细胞质阻碍流动。其细胞壁由木质素加厚，提供结构支撑，并防止在蒸腾作用产生的张力下坍塌。
5. Chemistry: Atomic Structure | 化学：原子结构
Sodium has an atomic number of 11. State its electronic configuration.
钠的原子序数为 11。写出它的电子排布。
Answer: 2,8,1. The first electron shell holds a maximum of 2 electrons, the second shell holds up to 8, and the remaining 1 electron occupies the third shell. This arrangement gives sodium one electron in its outer shell, making it a Group 1 metal.
答案：2,8,1。第一电子层最多容纳 2 个电子，第二层最多容纳 8 个电子，剩下的 1 个电子位于第三层。这种排布使钠的最外层有 1 个电子，成为第 1 族金属。
6. Chemistry: Chemical Bonding | 化学：化学键
Magnesium (Mg) reacts with oxygen (O) to form the ionic compound magnesium oxide. Describe the changes in electron arrangement that occur during this reaction.
镁（Mg）与氧（O）反应生成离子化合物氧化镁。描述该反应过程中电子排布的变化。
Answer: A magnesium atom loses its two outer electrons to form a Mg²⁺ ion, achieving the electronic configuration of neon (2,8). An oxygen atom gains these two electrons to form an O²⁻ ion, achieving the configuration of neon (2,8). The oppositely charged ions are held together by strong electrostatic forces, forming an ionic bond.
答案：一个镁原子失去其两个最外层电子，形成 Mg²⁺ 离子，达到氖的电子排布（2,8）。一个氧原子得到这两个电子，形成 O²⁻ 离子，同样达到氖的排布。带相反电荷的离子通过强大的静电引力结合在一起，形成离子键。
7. Chemistry: Stoichiometry | 化学：化学计量学
Carbon burns in oxygen according to the equation: C + O₂ → CO₂. Calculate the mass of carbon dioxide produced when 12 g of carbon is completely burned. (Relative atomic masses: C = 12, O = 16)
碳在氧气中燃烧的方程式为：C + O₂ → CO₂。计算 12 克碳完全燃烧时产生的二氧化碳的质量。（相对原子质量：C = 12，O = 16）
Answer: 44 g. First, find the relative molecular mass of CO₂: 12 + (16 × 2) = 44. Number of moles of carbon = mass / Ar = 12 / 12 = 1 mol. From the equation, 1 mol of C produces 1 mol of CO₂, so mass of CO₂ = moles × Mr = 1 × 44 = 44 g.
答案：44 克。首先，求 CO₂ 的相对分子质量：12 + (16 × 2) = 44。碳的物质的量 = 质量 / Ar = 12 / 12 = 1 mol。根据方程式，1 mol 碳生成 1 mol CO₂，因此 CO₂ 的质量 = 物质的量 × Mr = 1 × 44 = 44 g。
8. Chemistry: Acids and Bases | 化学：酸与碱
A farmer finds that the soil in a field has a pH of 5.5, which is too acidic for healthy crop growth. Explain why adding lime (calcium oxide) can neutralise the soil. Include a word equation in your answer.
一位农民发现田地土壤的 pH 为 5.5，酸性太强，不利于作物健康生长。解释为何添加石灰（氧化钙）可以中和土壤。答案中请包括文字方程式。
Answer: Lime is a metal oxide, which acts as a base. It reacts with the acids present in the soil (H⁺ ions) in a neutralisation reaction, producing a salt and water. This removes excess hydrogen ions and raises the pH. Word equation: calcium oxide + acid → calcium salt + water. Ionically: CaO + 2H⁺ → Ca²⁺ + H₂O.
答案：石灰是一种金属氧化物，作为碱起作用。它与土壤中的酸（H⁺ 离子）发生中和反应，生成盐和水。这去除了多余的氢离子，提高了 pH 值。文字方程式：氧化钙 + 酸 → 钙盐 + 水。离子方程式：CaO + 2H⁺ → Ca²⁺ + H₂O。
9. Physics: Motion | 物理：运动
A car accelerates uniformly from 10 m/s to 30 m/s in a time of 5 seconds. Calculate the acceleration of the car.
一辆汽车在 5 秒内从 10 m/s 均匀加速到 30 m/s。计算汽车的加速度。
Answer: 4 m/s². Acceleration = change in velocity / time taken. a = (30 – 10) / 5 = 20 / 5 = 4 m/s². The positive sign indicates the car is speeding up in the direction of motion.
答案：4 m/s²。加速度 = 速度变化量 / 所用时间。a = (30 – 10) / 5 = 20 / 5 = 4 m/s²。正值表示汽车在运动方向上加速。
10. Physics: Energy | 物理：能量
A car of mass 1000 kg is travelling at a speed of 20 m/s. Calculate its kinetic energy. (Use the formula: KE = ½ m v²)
一辆质量为 1000 kg 的汽车以 20 m/s 的速度行驶。计算它的动能。（使用公式：KE = ½ m v²）
Answer: 200 000 J (or 200 kJ). KE = ½ × 1000 × (20)² = 0.5 × 1000 × 400 = 200 000 J. Kinetic energy depends on both mass and the square of speed; doubling the speed quadruples the kinetic energy.
答案：200 000 J（或 200 kJ）。KE = ½ × 1000 × (20)² = 0.5 × 1000 × 400 = 200 000 J。动能取决于质量和速度的平方；速度加倍，动能变为原来的四倍。
11. Physics: Waves | 物理：波
A water wave in a ripple tank has a wavelength of 2 m and a frequency of 0.5 Hz. Calculate the speed of the wave. (v = f × λ)
一个水波在波纹槽中的波长为 2 m，频率为 0.5 Hz。计算波速。（v = f × λ）
Answer: 1 m/s. Wave speed = frequency × wavelength = 0.5 Hz × 2 m = 1 m/s. This means the wave travels 1 metre every second.
答案：1 m/s。波速 = 频率 × 波长 =
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