📚 Year 12 WJEC Science Unit Test: Mock Paper Walkthrough | WJEC 12年级科学单元测试模拟卷解析
This article provides a detailed walkthrough of a typical Year 12 WJEC Science mock paper, covering key topics from physics, chemistry and biology. Each section presents a mock question followed by a step‑by‑step bilingual explanation to help you identify common pitfalls and reinforce core concepts.
本文详细解析一份典型的 WJEC 12 年级科学单元测试模拟卷，涵盖物理、化学和生物的关键主题。每个小节提供一道模拟题，并附上逐步双语讲解，帮助你发现常见错误并巩固核心概念。
1. Physics: Kinematics – Uniform Acceleration | 物理：运动学 – 匀加速
A car accelerates uniformly from rest and reaches 20 m/s in 5.0 s. Calculate the acceleration and the distance travelled during this time.
一辆汽车从静止开始匀加速，5.0 秒后速度达到 20 m/s。计算加速度和在这段时间内行驶的距离。
To find the acceleration we use a = (v − u)/t, where u = 0, v = 20 m/s and t = 5.0 s. This gives a = 4.0 m/s². The distance can be found from s = ut + ½at², so s = 0 + ½ × 4.0 × (5.0)² = 50 m. Alternatively, using average velocity is equally valid: s = (u+v)/2 × t = 10 × 5 = 50 m.
求加速度时我们使用 a = (v − u)/t，其中 u = 0，v = 20 m/s，t = 5.0 s，得出 a = 4.0 m/s²。距离可以用 s = ut + ½at² 求得，即 s = 0 + ½ × 4.0 × (5.0)² = 50 m。也可用平均速度法：s = (u+v)/2 × t = 10 × 5 = 50 m，两种方法完全等价。
a = (v − u) / t = (20 − 0)/5.0 = 4.0 m/s²
s = ut + ½at² = 0 + ½ × 4.0 × 25 = 50 m
Many students forget to square the time when using the equation; always check your units and ensure t is in seconds. This question also tests the ability to choose the right SUVAT equation based on the given variables.
许多同学在代入公式时忘记将时间平方；一定要检查单位并确保 t 以秒为单位。这道题还考查了根据已知变量选择合适的匀加速运动公式的能力。
2. Physics: Electrical Circuits – Ohm’s Law | 物理：电路 – 欧姆定律
A resistor of 120 Ω is connected to a 9.0 V battery. Calculate the current through the resistor and the power dissipated.
一个 120 Ω 的电阻连接到 9.0 V 的电池上。计算通过电阻的电流以及电阻消耗的功率。
Using Ohm’s law, I = V/R = 9.0/120 = 0.075 A (or 75 mA). The power dissipated can be calculated by P = IV = 0.075 × 9.0 = 0.675 W, or equivalently P = V²/R = 9.0²/120 = 0.675 W. Both methods give the same result.
根据欧姆定律，I = V/R = 9.0/120 = 0.075 A（即 75 mA）。消耗的功率可以用 P = IV = 0.075 × 9.0 = 0.675 W 计算，也可用 P = V²/R = 9.0²/120 = 0.675 W，两者结果一致。
I = V / R = 9.0 V / 120 Ω = 0.075 A
P = I × V = 0.075 A × 9.0 V = 0.675 W
It is vital to use volts, ohms and amps consistently. A common mistake is to leave the current in milliamps without converting to amps before calculating power.
关键是要统一使用伏特、欧姆和安培。一个常见错误是在计算功率前没有将毫安转换为安培。
3. Chemistry: Balancing Equations | 化学：配平方程式
Balance the chemical equation for the combustion of propane: C₃H₈ + O₂ → CO₂ + H₂O.
配平丙烷燃烧的化学方程式：C₃H₈ + O₂ → CO₂ + H₂O。
Start by counting the carbon atoms: 3 carbons on the left require 3 CO₂ on the right. Then balance hydrogen: 8 hydrogens need 4 H₂O. Finally, balance oxygen: the right side now has 3×2 = 6 O from CO₂ plus 4×1 = 4 O from H₂O, making 10 oxygen atoms. Therefore 5 O₂ are needed on the left. The balanced equation is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.
先数碳原子：左边有 3 个碳，右边需要 3 个 CO₂。然后配平氢：8 个氢需要 4 个 H₂O。最后配平氧：右边现在从 CO₂ 中共有 6 个氧，从 H₂O 中有 4 个氧，合计 10 个氧原子。因此左边需要 5 个 O₂。配平后的方程式为 C₃H₈ + 5O₂ → 3CO₂ + 4H₂O。
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Always use integer coefficients and re‑check all atoms. Incomplete combustion would produce CO and is not correct here.
始终使用整数系数并重新检查所有原子。不完全燃烧会生成 CO，此处不适用。
4. Chemistry: Mole Calculations | 化学：摩尔计算
Calculate the number of moles in 10.0 g of calcium carbonate, CaCO₃, and the volume of CO₂ produced at STP when this mass reacts with excess acid. (Mᵣ CaCO₃ = 100)
计算 10.0 g 碳酸钙 (CaCO₃) 的物质的量，以及该质量的碳酸钙与过量酸反应后在标况下产生的 CO₂ 体积。（相对分子质量 Mᵣ CaCO₃ = 100）
Moles = mass / Mᵣ = 10.0 / 100 = 0.100 mol. The reaction is CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O, so 1 mole of CaCO₃ yields 1 mole of CO₂. Thus 0.100 mol of CO₂ is produced. At STP, 1 mol of any gas occupies 22.4 dm³, so volume = 0.100 × 22.4 = 2.24 dm³.
物质的量 = 质量 / 相对分子质量 = 10.0 / 100 = 0.100 mol。反应为 CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O，因此 1 摩尔 CaCO₃ 产生 1 摩尔 CO₂。所以生成 0.100 mol CO₂。在标况下，1 mol 任何气体体积为 22.4 dm³，故体积 = 0.100 × 22.4 = 2.24 dm³。
n = m / Mᵣ = 10.0 g / 100 g mol⁻¹ = 0.100 mol
V = n × 22.4 dm³ mol⁻¹ = 2.24 dm³
Ensure you use the molar volume at the correct temperature and pressure; STP is 0 °C and 1 atm. A common error is forgetting to convert the mass into moles before using the stoichiometric ratio.
注意使用正确温度和压力下的气体摩尔体积；标况为 0 °C 和 1 atm。常见错误是在使用化学计量比之前忘记将质量转换为物质的量。
5. Biology: Enzyme Activity and Temperature | 生物：酶活性与温度
Explain why the activity of an enzyme increases up to an optimum temperature but then decreases sharply above that temperature.
解释为什么酶的活性在达到最适温度之前逐步升高，但超过该温度后急剧下降。
As temperature rises, kinetic energy of enzyme and substrate molecules increases, leading to more frequent collisions and a higher rate of enzyme‑substrate complex formation. This raises activity up to the optimum temperature. Beyond this point, the heat disrupts the hydrogen bonds and ionic interactions that maintain the enzyme’s tertiary structure, causing the active site to denature. The substrate can no longer bind, so activity drops rapidly.
随着温度升高，酶和底物分子的动能增加，导致更频繁的碰撞以及酶‑底物复合物形成速率上升。这使得活性升高，直至最适温度。在此温度以上，热量破坏了维持酶三级结构的氢键和离子相互作用，导致活性位点变性。底物无法再结合，因此活性急剧下降。
The lock‑and‑key model illustrates that a precise active site shape is essential. Once denatured, the process is usually irreversible.
锁钥模型说明精确的活性位点形状至关重要。一旦变性，过程通常不可逆。
6. Biology: Genetics – Monohybrid Cross | 生物：遗传学 – 单基因杂交
In pea plants, the allele for tall stems (T) is dominant over dwarf (t). Two heterozygous tall plants are crossed. Determine the expected phenotypic ratio in the offspring.
在豌豆植株中，高茎等位基因 (T) 对矮茎 (t) 呈显性。两株杂合高茎植株杂交。确定后代预期的表现型比例。
Parental genotypes: Tt × Tt. Gametes carry T or t. The Punnett square gives offspring genotypes: TT, Tt, Tt, tt. Phenotypes: TT and Tt are tall, tt is dwarf. The ratio is 3 tall : 1 dwarf.
亲本基因型：Tt × Tt。配子携带 T 或 t。庞纳特方格得出子代基因型：TT、Tt、Tt、tt。表现型：TT 和 Tt 为高茎，tt 为矮茎。比例为 3 高 : 1 矮。
Phenotypic ratio: 3 tall : 1 dwarf
Some students confuse genotype ratio (1:2:1) with phenotype ratio (3:1). Remember that dominant homozygote and heterozygote appear identical phenotypically.
有些同学会混淆基因型比例 (1:2:1) 和表现型比例 (3:1)。记住显性纯合子和杂合子在表型上相同。
7. Chemistry: Rates of Reaction – Collision Theory | 化学：反应速率 – 碰撞理论
Magnesium ribbon reacts with dilute hydrochloric acid. State and explain the effect of increasing the acid concentration on the initial rate, referring to collision theory.
镁条与稀盐酸反应。阐述并解释增加酸浓度对初始速率的影响，并运用碰撞理论加以说明。
Increasing the concentration of HCl means there are more H⁺ ions per unit volume. This leads to more frequent collisions between H⁺ ions and Mg atoms in a given time. According to collision theory, a higher collision frequency increases the chance of successful collisions (those with energy greater than the activation energy). Therefore the initial rate of reaction increases.
增加盐酸浓度意味着单位体积内有更多的 H⁺ 离子。这使得一定时间内 H⁺ 离子与 Mg 原子之间的碰撞更加频繁。根据碰撞理论，更高的碰撞频率增加了成功碰撞（能量超过活化能的碰撞）的机会。因此初始反应速率加快。
Activation energy remains unchanged; the effect is purely due to the number of particles per unit volume. This can be monitored by measuring the volume of H₂ gas produced against time.
活化能保持不变；该影响纯粹由于单位体积粒子数增加所致。这可以通过测量随时间产生的 H₂ 气体体积来监测。
8. Physics: Forces and Newton’s Laws | 物理：力与牛顿定律
A block of mass 6.0 kg is pulled along a smooth horizontal surface by a force of 24 N. Determine the acceleration and the normal reaction force acting on the block.
一个 6.0 kg 的木块在光滑水平面上受到 24 N 的拉力。求木块的加速度和作用于木块的法向反力。
Since the surface is smooth, there is no friction. Using Newton’s second law, F = ma, the acceleration a = F/m = 24/6.0 = 4.0 m/s². The normal reaction force balances the weight, as there is no vertical acceleration. Weight = mg = 6.0 × 9.8 = 58.8 N. Therefore the normal reaction R = 58.8 N vertically upwards.
由于表面光滑，无摩擦。根据牛顿第二定律 F = ma，加速度 a = F/m = 24/6.0 = 4.0 m/s²。法向反力与重力平衡，因为垂直方向无加速度。重力 = mg = 6.0 × 9.8 = 58.8 N。因此法向反力 R = 58.8 N，方向竖直向上。
a = 24 N / 6.0 kg = 4.0 m/s²
R = 6.0 kg × 9.8 m/s² = 58.8 N
A classic mistake is to forget the normal force entirely or to assume it equals the applied force. Always draw a free‑body diagram to identify all forces.
一个典型错误是完全忽略法向力或认为它等于拉力。务必画受力图来识别所有力。
9. Biology: Cell Structure and Organelles | 生物：细胞结构与细胞器
Compare the structure of a typical plant cell and an animal cell, giving two similarities and two differences.
比较典型植物细胞和动物细胞的结构，给出两个相似点和两个不同点。
Similarities: both have a nucleus containing genetic material, and both have mitochondria where aerobic respiration occurs. Differences: plant cells have a rigid cellulose cell wall outside the membrane, while animal cells lack a cell wall; plant cells contain permanent vacuoles and often chloroplasts, which are absent from animal cells.
相似点：两者都有包含遗传物质的细胞核，也都有进行有氧呼吸的线粒体。不同点：植物细胞在细胞膜外有坚硬的纤维素细胞壁，而动物细胞缺乏细胞壁；植物细胞含有永久液泡并常有叶绿体，动物细胞则没有。
Note that both cell types have ribosomes, endoplasmic reticulum and Golgi apparatus. In WJEC questions, you are expected to relate organelle presence to the cell’s function, for example chloroplasts for photosynthesis.
注意两种细胞类型都有核糖体、内质网和高尔基体。在 WJEC 考题中，需要将细胞器的存在与细胞功能联系起来，例如叶绿体用于光合作用。
10. Chemistry: Acids, Bases and pH | 化学：酸、碱与 pH
A solution of hydrochloric acid has a hydrogen ion concentration of 1.0 × 10⁻³ mol dm⁻³. Calculate the pH and state whether the solution is acidic, alkaline or neutral.
某盐酸溶液的氢离子浓度为 1.0 × 10⁻³ mol dm⁻³。计算其 pH 并判断该溶液是酸性、碱性还是中性。
pH is defined as pH = −log₁₀[H⁺]. Substituting [H⁺] = 1.0 × 10⁻³ gives pH = −log₁₀(1.0 × 10⁻³) = 3.0. Since the pH is less than 7, the solution is acidic. A neutral solution has pH 7, and alkaline solutions have pH greater than 7.
pH 定义为 pH = −log₁₀[H⁺]。代入 [H⁺] = 1.0 × 10⁻³，得到 pH = −log₁₀(1.0 × 10⁻³) = 3.0。由于 pH 小于 7，该溶液呈酸性。中性溶液 pH 为 7，碱性溶液 pH 大于 7。
pH = −log₁₀(1.0 × 10⁻³) = 3.0
Remember that each unit change in pH represents a ten‑fold change in [H⁺]. Always use the unrounded value if further calculations are required.
记住 pH 单位每变化 1，[H⁺] 浓度变化 10 倍。如需后续计算，请使用未舍入的值。
11. Data Analysis: Interpreting a Rate Graph | 数据分析：解读速率图
An experiment monitors the volume of gas produced in a reaction. The graph shows a steep initial slope that gradually levels off. Explain the shape of the curve using the concepts of collision theory and limiting reactant.
一项实验监测反应产生的气体体积。图像显示初始陡峭上升，随后逐渐平缓。运用碰撞理论和极限反应物概念解释曲线的形状。
Initially, the reactant concentrations are high, leading to a high collision frequency and a fast rate – hence the steep slope. As the reaction proceeds, reactants are consumed, concentrations fall, and successful collisions become less frequent. The rate decreases, causing the curve to level off. Eventually one reactant is used up (the limiting reactant), and the volume of gas stops increasing.
初始时反应物浓度高，碰撞频率高，速率快，因此曲线陡峭。随着反应进行，反应物被消耗，浓度下降，成功碰撞变得不那么频繁。速率降低，曲线趋于平缓。最终某一反应物耗尽（极限反应物），气体体积不再增加。
Always label the point where the graph plateaus as the end‑point of the reaction. The total volume of gas can be used to quantify the amount of product.
始终将曲线变平处标记为反应终点。总气体体积可用于量化产物量。
12. Physics: Energy and Efficiency | 物理：能量与效率
A motor lifts a 2.0 kg mass through a vertical height of 5.0 m in 4.0 s. The input power is 30 W. Calculate the useful power output and the efficiency of the motor. (g = 9.8 m/s²)
一台电动机在 4.0 秒内将一个 2.0 kg 的重物竖直提升了 5.0 m。输入功率为 30 W。计算有用输出功率和电动机的效率。(g = 9.8 m/s²)
Work done against gravity = mgh = 2.0 × 9.8 × 5.0 = 98 J. Useful power output = work done / time = 98 / 4.0 = 24.5 W. Efficiency = (useful power output / input power) × 100% = (24.5 / 30) × 100% = 81.7% (or 82%).
克服重力所做的功 = mgh = 2.0 × 9.8 × 5.0 = 98 J。有用输出功率 = 做功 / 时间 = 98 / 4.0 = 24.5 W。效率 = (有用输出功率 / 输入功率) × 100% = (24.5 / 30) × 100% = 81.7%（或 82%）。
Efficiency = (24.5 W / 30 W) × 100% ≈ 82%
Energy dissipated as heat and sound accounts for the difference. Always express efficiency as a percentage and check that it does not exceed 100%.
以热和声形式耗散的能量造成了差值。始终以百分比表示效率，并确保不超过 100%。
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