📚 Typical Example Questions Explained for IB WJEC Science | IB WJEC 科学：典型例题详解
IB Science courses, combined with the rigorous assessment style of WJEC, require students to master both conceptual understanding and problem-solving skills. This article presents a collection of typical example questions from Biology, Chemistry, and Physics, along with step-by-step explanations. Each problem reflects common assessment tasks you may encounter, helping you build confidence and accuracy for your exams.
IB 科学课程结合了 WJEC 严格的评估风格，要求学生在理解概念的同时掌握解题技巧。本文精选了生物、化学和物理学科的典型例题，并配以逐步详解。每道题目都反映了考试中常见的评估任务，有助于你提升信心和准确性，为考试做好准备。
1. Mole Calculation: Mass to Moles | 摩尔计算：从质量到物质的量
Example: Calculate the number of moles in 25.0 g of hydrated copper(II) sulfate, CuSO₄·5H₂O. (Ar: Cu = 63.5, S = 32, O = 16, H = 1)
例题：计算25.0克五水合硫酸铜(CuSO₄·5H₂O)中所含的物质的量（摩尔）。(相对原子质量: Cu=63.5, S=32, O=16, H=1)
Step 1: Determine the molar mass. Add the atomic masses: Cu (63.5) + S (32) + 4×O (4×16 = 64) + 5×[2×H (2) + O (16)] = 63.5 + 32 + 64 + 5×18 = 159.5 + 90 = 249.5 g mol⁻¹.
步骤1：计算摩尔质量。将原子质量相加：Cu (63.5) + S (32) + 4×O (4×16 = 64) + 5×[2×H (2) + O (16)] = 63.5 + 32 + 64 + 5×18 = 159.5 + 90 = 249.5 g mol⁻¹。
Step 2: Use the formula n = m / M. n = 25.0 g / 249.5 g mol⁻¹ ≈ 0.1002 mol. Always include units and round appropriately.
步骤2：使用公式 n = m / M。n = 25.0 g / 249.5 g mol⁻¹ ≈ 0.1002 mol。永远要包含单位并适当取整。
2. Limiting Reactant and Theoretical Yield | 限制反应物与理论产率
Example: 10.0 g of hydrogen gas (H₂) react with 10.0 g of oxygen gas (O₂) to form water. Identify the limiting reactant and calculate the maximum mass of water produced. (H=1, O=16)
例题：10.0克氢气(H₂)与10.0克氧气(O₂)反应生成水。判断哪一种为限制反应物，并计算生成水的最大质量。(H=1, O=16)
Step 1: Write the balanced equation.
2H₂ + O₂ → 2H₂O
步骤1：写出配平方程式。
2H₂ + O₂ → 2H₂O
Step 2: Calculate moles of each reactant. Moles H₂ = 10.0 g / 2 g mol⁻¹ = 5.0 mol. Moles O₂ = 10.0 g / 32 g mol⁻¹ = 0.3125 mol.
步骤2：计算各反应物的物质的量。H₂的物质的量 = 10.0 g / 2 g mol⁻¹ = 5.0 mol。O₂的物质的量 = 10.0 g / 32 g mol⁻¹ = 0.3125 mol。
Step 3: Using the 2:1 ratio, O₂ is limiting (0.3125 mol O₂ would require 0.625 mol H₂, which is available). Moles H₂O produced = 2 × 0.3125 = 0.625 mol. Mass of H₂O = 0.625 mol × 18 g mol⁻¹ = 11.25 g.
步骤3：根据2:1的计量比，O₂为限制反应物（0.3125 mol O₂仅需0.625 mol H₂，而H₂充足）。生成H₂O的物质的量 = 2 × 0.3125 = 0.625 mol。H₂O的质量 = 0.625 mol × 18 g mol⁻¹ = 11.25 g。
3. Newton’s Second Law of Motion | 牛顿第二运动定律
Example: A constant net force of 15 N is applied to a 3.0 kg box initially at rest on a frictionless surface. Calculate the acceleration and the distance the box travels after 5.0 seconds.
例题：一个3.0千克的盒子静止在无摩擦表面上，施加15牛恒定的净力。计算加速度和5.0秒后盒子滑行的距离。
Step 1: Use F = m × a to find acceleration.
a = F / m = 15 N / 3.0 kg = 5.0 m s⁻²
步骤1：使用 F = m × a 求加速度。
a = F / m = 15 N / 3.0 kg = 5.0 m s⁻²
Step 2: Since initial velocity u = 0, use s = u t + ½ a t². s = 0 + ½ × 5.0 m s⁻² × (5.0 s)² = ½ × 5.0 × 25 = 62.5 m.
步骤2：由于初速度 u = 0，使用 s = u t + ½ a t²。s = 0 + ½ × 5.0 m s⁻² × (5.0 s)² = ½ × 5.0 × 25 = 62.5 m。
Always check that the force is the resultant force and that units are consistent (N, kg, m, s).
永远要检查力是否为合力，并确保单位一致（牛、千克、米、秒）。
4. Kinetic Energy and Work-Energy Theorem | 动能与动能定理
Example: A car of mass 1200 kg is travelling at 20 m/s. The brakes apply a constant friction force of 3000 N. Using energy considerations, calculate the braking distance.
例题：一辆质量为1200 kg的汽车以20 m/s的速度行驶。刹车时施加恒定的3000 N摩擦力。用能量方法计算刹车距离。
Step 1: Calculate initial kinetic energy. KE = ½ m v² = ½ × 1200 kg × (20 m/s)² = 600 × 400 = 240,000 J.
步骤1：计算初始动能。KE = ½ m v² = ½ × 1200 kg × (20 m/s)² = 600 × 400 = 240,000 J。
Step 2: Work done by brakes = force × distance = loss in KE. So 3000 N × d = 240,000 J. Therefore d = 240,000 / 3000 = 80 m.
步骤2：刹车做功 = 力 × 距离 = 动能损失量。因此 3000 N × d = 240,000 J。得出 d = 240,000 / 3000 = 80 m。
The work-energy theorem provides a quick alternative to suvat equations when acceleration is not directly needed.
当不需要直接求加速度时，动能定理提供了一种比运动学公式更快捷的方法。
5. Series and Parallel Circuits | 串联与并联电路
Example: A 4 Ω resistor and a 6 Ω resistor are connected to a 12 V battery. First, they are connected in series; then they are reconnected in parallel. For each arrangement, calculate the equivalent resistance, the total current from the battery, and the potential difference across each resistor.
例题：一个4 Ω和一个6 Ω的电阻连接到12 V电池上。先串联，然后改为并联。对每种连接方式，计算等效电阻、电池输出的总电流以及每个电阻两端的电压。
Series: Equivalent resistance R = 4 + 6 = 10 Ω. Total current I = V / R = 12 V / 10 Ω = 1.2 A. Potential difference across 4 Ω = 1.2 × 4 = 4.8 V; across 6 Ω = 1.2 × 6 = 7.2 V.
串联：等效电阻 R = 4 + 6 = 10 Ω。总电流 I = V / R = 12 V / 10 Ω = 1.2 A。4 Ω电阻两端电压 = 1.2 × 4 = 4.8 V；6 Ω电阻两端电压 = 1.2 × 6 = 7.2 V。
Parallel: 1/R = 1/4 + 1/6 = 3/12 + 2/12 = 5/12, so R = 12/5 = 2.4 Ω. Total current I = 12 V / 2.4 Ω = 5 A. In parallel, each resistor has the full 12 V.
并联：1/R = 1/4 + 1/6 = 3/12 + 2/12 = 5/12，因此 R = 12/5 = 2.4 Ω。总电流 I = 12 V / 2.4 Ω = 5 A。并联时，每个电阻均承受12 V电压。
6. Radioactive Decay and Half-Life | 放射性衰变与半衰期
Example: The isotope iodine-131 has a half-life of 8 days. A sample initially contains 80 g of I-131. Calculate the mass remaining after 24 days and the time required for the mass to decay to 5 g.
例题：碘-131的半衰期为8天。某样本最初含有80克I-131。计算24天后剩余的质量以及衰变至5克所需的时间。
Step 1: After 24 days, number of half-lives = 24 / 8 = 3. Mass remaining = initial mass × (½)ⁿ = 80 g × (½)³ = 80 × 1/8 = 10 g.
步骤1：24天后，经过的半衰期个数 = 24 / 8 = 3。剩余质量 = 初始质量 × (½)ⁿ = 80 g × (½)³ = 80 × 1/8 = 10 g。
Step 2: For 5 g, set up 5 = 80 (½)^(t/8). Then (½)^(t/8) = 5/80 = 1/16 = (½)⁴. By equating exponents, t/8 = 4, so t = 32 days.
步骤2：对于5克，列出方程 5 = 80 (½)^(t/8)。则 (½)^(t/8) = 5/80 = 1/16 = (½)⁴。比较指数得 t/8 = 4，故 t = 32 天。
The exponential decay equation N = N₀ (½)^(t/t₁/₂) is a powerful tool for half-life questions.
指数衰变公式 N = N₀ (½)^(t/t₁/₂) 是解决半衰期问题的有力工具。
7. Monohybrid Inheritance and Punnett Squares | 单基因遗传与庞纳特方格
Example: In garden peas, the allele for tall stems (T) is dominant over the allele for dwarf stems (t). A heterozygous tall plant is crossed with a dwarf plant. Predict the genotypic and phenotypic ratios of the offspring.
例题：在豌豆中，高茎等位基因(T)对矮茎等位基因(t)为显性。将一个杂合高茎植株与一个矮茎植株杂交。预测后代的基因型比和表现型比。
Step 1: Determine parental genotypes: Tt (tall) and tt (dwarf). Gametes: T or t from the tall parent, and all t from the dwarf parent.
步骤1：确定亲本基因型：Tt (高茎)和 tt (矮茎)。配子：高茎亲本产生 T 或 t，矮茎亲本全为 t。
Step 2: Set up a Punnett square: rows for t, t; columns for T, t. Offspring genotypes: Tt, Tt, tt, tt. Genotypic ratio is 1 Tt : 1 tt (or 1:1). Since Tt is tall and tt is dwarf, the phenotypic ratio is 1 tall : 1 dwarf.
步骤2：绘制庞纳特方格：行对应 t, t；列对应 T, t。后代基因型：Tt, Tt, tt, tt。基因型比为 1 Tt : 1 tt (即1:1)。由于 Tt 为高茎，tt 为矮茎，表现型比为 1高 : 1矮。
8. Enzyme Kinetics: Effect of Temperature on Reaction Rate | 酶动力学：温度对反应速率的影响
Example: An investigation into the activity of catalase recorded initial rates of oxygen production at different temperatures. Data obtained: 10°C – 0.5, 20°C – 1.2, 30°C – 2.5, 40°C – 4.0, 50°C – 3.0, 60°C – 0.8 (arbitrary units). Explain the observed pattern, referring to collision theory and enzyme structure.
例题：一项关于过氧化氢酶活性的研究记录了不同温度下氧气的初始生成速率。数据如下：10°C – 0.5, 20°C – 1.2, 30°C – 2.5, 40°C – 4.0, 50°C – 3.0, 60°C – 0.8（任意单位）。请结合碰撞理论和酶的结构解释观察到的规律。
As temperature rises from 10°C to 40°C, the rate increases because enzyme and substrate molecules have greater kinetic energy, leading to more frequent successful collisions. At 40°C the enzyme reaches its optimum temperature, where the active site shape is ideal for substrate binding.
温度从10°C升至40°C时，速率增加，因为酶和底物分子的动能增大，导致更频繁的有效碰撞。在40°C时酶达到最适温度，此时活性位点的形状最适合与底物结合。
Above 40°C, the tertiary structure of the enzyme is disrupted as hydrogen bonds and hydrophobic interactions break. The active site loses its specific shape, and the enzyme becomes denatured. Consequently, the reaction rate falls sharply at 50°C and 60°C.
40°C以上，随着氢键和疏水相互作用的断裂，酶的三级结构遭到破坏。活性位点失去其特定形状，酶失活（变性）。因此，反应速率在50°C和60°C时急剧下降。
9. Osmosis and Water Potential | 渗透与水势
Example: A human red blood cell has an internal water potential (ψ) of -300 kPa. It is placed in a sodium chloride solution with ψ = -500 kPa. Predict the net direction of water movement and the resulting effect on the cell. Explain your reasoning.
例题：一个人体红细胞内部水势(ψ)为-300 kPa。将其置于水势为-500 kPa的氯化钠溶液中。预测水净移动的方向及其对细胞的影响。解释你的推理。
Step 1: Water moves from a region of higher water potential to a region of lower water potential. Since -300 kPa > -500 kPa, water potential is higher inside the cell than outside.
步骤1：水从高水势区域向低水势区域移动。由于 -300 kPa > -500 kPa，细胞内水势高于细胞外溶液。
Step 2: Therefore, there is a net movement of water out of the cell to the external solution. As the cell loses water, the cell membrane pulls away from the cell wall (in plant cells this would be plasmolysis; in animal cells it causes crenation). The red blood cell will shrink and develop a spiky, crenated appearance.
步骤2：因此，水有从细胞内向外部溶液净流出的趋势。细胞失水后，细胞膜皱缩（植物细胞中为质壁分离；动物细胞中称为皱缩）。红细胞将缩小并呈现刺状皱缩形态。
10. Equilibrium Constant Kc and Le Chatelier’s Principle | 化学平衡常数 Kc 与勒夏特列原理
Example: For the reversible reaction N₂O₄(g) ⇌ 2NO₂(g), at a certain temperature the equilibrium concentrations are [N₂O₄] = 0.20 mol dm⁻³ and [NO₂] = 0.10 mol dm⁻³. Calculate the equilibrium constant Kc and predict the effect of decreasing the container volume on the equilibrium position.
例题：对于可逆反应 N₂O₄(g) ⇌ 2NO₂(g)，在某一温度下平衡浓度为 [N₂O₄] = 0.20 mol dm⁻³，[NO₂] = 0.10 mol dm⁻³。计算平衡常数 Kc，并预测减小容器体积对平衡位置的影响。
Step 1: Write the expression for Kc.
Kc = [NO₂]² / [N₂O₄]
步骤1：写出 Kc 表达式。
Kc = [NO₂]² / [N₂O₄]
Step 2: Substitute values: Kc = (0.10)² / 0.20 = 0.01 / 0.20 = 0.05 mol dm⁻³.
步骤2：代入数值：Kc = (0.10)² / 0.20 = 0.01 / 0.20 = 0.05 mol dm⁻³。
Step 3: Decreasing volume increases the pressure. According to Le Chatelier’s principle, the equilibrium shifts to the side with fewer moles of gas to reduce the pressure. Left side has 1 mole of gas, right side has 2 moles. So the equilibrium shifts to the left, forming more N₂O₄ and reducing the concentration of NO₂.
步骤3：减小体积会增大压强。根据勒夏特列原理，平衡向气体物质的量较少的方向移动以减弱压强的增加。左侧有1摩尔气体，右侧有2摩尔气体。因此平衡向左移动，生成更多 N₂O₄，降低 NO₂ 的浓度。
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