Pre-U CIE Chemistry: Unit Test Mock Paper Analysis | Pre-U CIE 化学：单元测试模拟卷解析

📚 Pre-U CIE Chemistry: Unit Test Mock Paper Analysis | Pre-U CIE 化学：单元测试模拟卷解析

Mock papers are an essential tool for mastering Pre-U CIE Chemistry. This analysis breaks down common question types from a typical unit test, highlighting key concepts, step‑by‑step solutions, and frequent mistakes. By working through these examples, you will strengthen your problem‑solving skills and deepen your understanding of the syllabus.

模拟卷是掌握 Pre‑U CIE 化学的重要工具。本文将对一份典型单元测试中的常见题型进行拆解，重点讲解核心概念、分步解题思路以及常见错误。通过这些例题，你将提升解题能力，加深对考纲内容的理解。

1. Mole Calculations and Stoichiometry | 摩尔计算与化学计量学

250.0 cm³. Calculate the concentration of the solution in mol dm⁻³. (Mᵣ of Na₂CO₃ = 106.0)

题目：某学生将 5.30 g 无水碳酸钠溶于蒸馏水，配成 250.0 cm³ 溶液。计算该溶液的浓度（单位：mol dm⁻³）。(Na₂CO₃ 相对分子质量 Mᵣ = 106.0)

g mol⁻¹ = 0.0500 mol. Always check that the mass is in grams and molar mass is in g mol⁻¹.

第一步：使用公式 n = m / M 求出物质的量。n(Na₂CO₃) = 5.30 g ÷ 106.0 g mol⁻¹ = 0.0500 mol。务必确保质量的单位是克，摩尔质量的单位是 g mol⁻¹。

2: Convert the volume from cm³ to dm³. 250.0 cm³ = 0.2500 dm³. Remember 1 dm³ = 1000 cm³.

第二步：将体积从 cm³ 转换为 dm³。250.0 cm³ = 0.2500 dm³。记住 1 dm³ = 1000 cm³。

l answer should be given to three significant figures, matching the precision of the data.

第三步：使用 c = n / V 计算浓度。c = 0.0500 mol ÷ 0.2500 dm³ = 0.200 mol dm⁻³。最终答案应保留三位有效数字，与数据的精密度一致。

Common mistake: Forgetting to convert cm³ to dm³ leads to a concentration 1000 times too small. Always write the conversion explicitly.

常见错误：忘记将 cm³ 转换为 dm³，导致计算出的浓度值偏小 1000 倍。务必明确写出单位换算过程。

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2. Electron Configuration and Ionisation Energy | 电子排布与电离能

Question: Write the full electron configuration of a Cu⁺ ion (atomic number 29) and explain why it does not follow the simple aufbau prediction for the neutral atom.

题目：写出 Cu⁺ 离子的完整电子排布（原子序数 29），并解释为何中性原子的排布不符合简单的构造原理预测。

The ground‑state configuration of a copper atom is [Ar] 3d¹⁰ 4s¹, not [Ar] 3d⁹ 4s². The completely filled 3d subshell gives extra stability.

铜原子的基态电子排布是 [Ar] 3d¹⁰ 4s¹，而不是 [Ar] 3d⁹ 4s²。全满的 3d 亚层提供了额外的稳定性。

When forming a Cu⁺ ion, the electron is removed from the highest energy level, which is the 4s orbital. Therefore the Cu⁺ ion has the configuration [Ar] 3d¹⁰.

形成 Cu⁺ 离子时，电子从最高能级即 4s 轨道中失去。因此 Cu⁺ 离子的电子排布为 [Ar] 3d¹⁰。

A common error is to remove an electron from the 3d subshell first, yielding [Ar] 3d⁹ 4s¹. Remember: 4s electrons are lost before 3d electrons in transition metal ions.

常见错误是先移除 3d 亚层的电子，得到 [Ar] 3d⁹ 4s¹。请记住：过渡金属离子形成时，4s 电子先于 3d 电子失去。

The Cu⁺ ion is diamagnetic because all its electrons are paired. This can be contrasted with Cu²⁺, which has the configuration [Ar] 3d⁹ and is paramagnetic.

Cu⁺ 离子是抗磁性的，因为所有电子均已配对。而 Cu²⁺ 的排布为 [Ar] 3d⁹，具有顺磁性，两者可作对比。

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3. Chemical Bonding and Molecular Shapes | 化学键与分子形状

Question: Predict the shape and bond angles of PF₅. Use VSEPR theory to justify your answer.

题目：预测 PF₅ 的形状和键角，并用 VSEPR 理论加以论证。

Phosphorus is the central atom with five bonding pairs of electrons and no lone pairs. According to VSEPR, five electron pairs arrange themselves in a trigonal bipyramidal geometry to minimise repulsion.

磷是中心原子，有五对成键电子且无孤电子对。根据 VSEPR 理论，五对电子对采用三角双锥构型，以使排斥力最小。

In this shape, three fluorine atoms lie in a plane with 120° bond angles (equatorial positions), and two fluorine atoms are perpendicular to that plane with 90° angles (axial positions).

在此构型中，三个氟原子处于平面内，彼此夹角为 120°（赤道位），另外两个氟原子垂直于该平面，与赤道原子呈 90° 夹角（轴位）。

A common misconception is to confuse PF₅ with a square pyramidal shape. Square pyramidal arises only when a lone pair occupies one of the octahedral positions, which is not the case here.

常见误区是将 PF₅ 误认为是四方锥形。只有当一个孤电子对占据八面体位置之一时，才会呈现四方锥形，此处并非如此。

The molecule is non‑polar overall because the bond dipoles cancel, despite the P–F bonds being polar. This can be tested by considering the symmetry of the trigonal bipyramid.

尽管 P–F 键是极性的，但由于三角双锥的高度对称性，各键的偶极矩相互抵消，分子整体呈非极性。

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4. Enthalpy Changes and Hess's Law | 焓变与赫斯定律

d enthalpies of combustion to calculate the standard enthalpy of formation of methane, ΔHf⦵(CH₄). ΔHc⦵(C) = –394 kJ mol⁻¹, ΔHc⦵(H₂) = –286 kJ mol⁻¹, ΔHc⦵(CH₄) = –890 kJ mol⁻¹.

题目：利用下列标准燃烧焓数据，计算甲烷的标准生成焓 ΔHf⦵(CH₄ )。已知 ΔHc⦵(C) = –394 kJ mol⁻¹，ΔHc⦵(H₂) = –286 kJ mol⁻¹，ΔHc⦵(CH₄) = –890 kJ mol⁻¹。

The formation reaction is C(s) + 2H₂(g) → CH₄(g). Using Hess' ies of the elements equals the combustion enthalpy of methane plus the formation enthalpy.

生成反应为 C(s) + 2H₂(g) → CH₄(g)。运用赫斯定律构建循环：各元素的燃烧焓之和等于甲烷的燃烧焓加上生成焓。

ΔHf⦵ = Σ ΔHc⦵(reactants) – Σ ΔHc⦵(products) = [ΔHc⦵(C) + 2×ΔHc⦵(H₂)] – [ΔHc⦵(CH₄)]

Substituting the values: ΔHf⦵ = [–394 + 2×(–286)] – (–890) = –394 – 572 + 890 = –76 kJ mol⁻¹.

代入数值：ΔHf⦵ = [–394 + 2×(–286)] – (–890) = –394 – 572 + 890 = –76 kJ mol⁻¹。

A frequent error is to forget the minus sign in the combustion enthalpy of methane, or to incorrectly set up the cycle. Always check that the arrows in your enthalpy cycle point down from elements to combustion products.

常见错误是遗漏甲烷燃烧焓的负号，或循环设定有误。请务必检查焓循环图中箭头是否从元素指向燃烧产物。

The calculated value is close to the literature value of –75 kJ mol⁻¹, any difference is due to small experimental uncertainties in the given data.

计算所得数值与文献值 –75 kJ mol⁻¹ 接近，细微差异来源于所给数据存在的实验误差。

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5. Reaction Rates and Rate Equations | 反应速率与速率方程

the reaction A + 2B → products. Deduce the rate equation and calculate the rate constant.

题目：对于反应 A + 2B → 产物，获得下列数据。推导速率方程并计算速率常数。

Experiment	[A]₀ / mol dm⁻³	[B]₀ / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	2.0×10⁻⁴
2	0.20	0.10	4.0×10⁻⁴
3	0.10	0.20	8.0×10⁻⁴

Compare experiments 1 and 2: [A] doubles while [B] is constant; the rate doubles. Hence the reaction is first order with respect to A.

对比实验 1 和 2：[A] 加倍而 [B] 不变，速率加倍。因此反应对 A 为一级。

Compare experiments 1 and 3: [B] doubles while [A] is constant; the rate quadruples. Hence the reaction is second order with respect to B.

对比实验 1 和 3：[B] 加倍而 [A] 不变，速率变为原来的四倍。因此反应对 B 为二级。

rate = k [A] [B]²

rate / ([A][B]²) = 2.0×10⁻⁴ / (0.10 × (0.10)²) = 2.0×10⁻⁴ / 1.0×10⁻³ = 0.20 dm⁶ mol⁻² s⁻¹.

代入实验 1： rate / ([A][B]²) = 2.0×10⁻⁴ / (0.10 × (0.10)²) = 2.0×10⁻⁴ / 1.0×10⁻³ = 0.20 dm⁶ mol⁻² s⁻¹。

Always check the units of k; for an overall order of 3, the units are dm⁶ mol⁻² s⁻¹. Misinterpreting the units often leads to lost marks.

务必检查 k 的单位；对总反应级数 3，单位是 dm⁶ mol⁻² s⁻¹。误判单位是常见的失分点。

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6. Chemical Equilibrium and Le Chatelier's Principle | 化学平衡与勒夏特列原理

Question: For the equilibrium H₂(g) + I₂(g) ⇌ 20 mol dm⁻³, [I₂] = 0.20 mol dm⁻³, [HI] = 1.60 mol dm⁻³. Calculate Kc and state its units.

题目：在某一温度下，反应 H₂(g) + I₂(g) ⇌ 2HI(g) 达到平衡时，浓度分别为 [H₂] = 0.20 mol dm⁻³，[I₂] = 0.20 mol dm⁻³，[HI] = 1.60 mol dm⁻³。计算 Kc 并注明单位。

Kc = [HI]² / ([H₂][I₂])

Substituting the values: Kc = (1.60)² / (0.20 × 0.20) = 2.56 / 0.040 = 64.

代入数值：Kc = (1.60)² / (0.20 × 0.20) = 2.56 / 0.040 = 64。

In this case, the number of moles of gaseous products equals the number of moles of gaseous reactants, so Kc has no units. Always simplify the expression before determining the units.

在该反应中，气体产物的化学计量数之和等于反应物的化学计量数之和，因此 Kc 无单位。务必在化简表达式后再确定单位。

₂][I₂]) without squaring the HI term. The stoichiometric coefficient becomes the exponent in the equilibrium expression.

常见失误是将表达式写成 Kc = [HI] / ([H₂][I₂])，遗漏了 HI 项的平方。化学计量数在平衡表达式中充当指数。

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7. Acid‑Base Equilibria and pH Calculations | 酸碱平衡与 pH 计算

solution of ethanoic acid (CH₃COOH). Ka = 1.8 × 10⁻⁵ mol dm⁻³. State any assumptions made.

题目：计算 0.100 mol dm⁻³ 乙酸溶液的 pH。Ka = 1.8 × 10⁻⁵ mol dm⁻³。说明所作假设。

For a weak acid, [H⁺] = √Ka × c provided that the acid is less than 5% dissociated. This is valid because Ka is small and c is relatively large.

对弱酸，[H⁺] = √Ka × c，前提是酸的电离度小于 5%。由于 Ka 较小且浓度相对较大，该假设成立。

[H⁺] = √1.8 × 10⁻⁵ × 0.100 = √1.8 × 10⁻⁶ = 1.34 × 10⁻³ mol dm⁻³

Then pH = –log₁₀(1.34 × 10⁻³) ≈ 2.87. Check the assumption: % dissociation = ([H⁺]/c) × 100 = (1.34 × 10⁻³ / 0.100) × 100 = 1.34%, which is < 5%, so the approximation is justified.

然后 pH = –log₁₀(1.34 × 10⁻³) ≈ 2.87。检验假设：电离度 = (1.34 × 10⁻³ / 0.100) × 100 = 1.34% < 5%，因此近似处理合理。

00 mol dm⁻³. Always check the value of Ka to determine whether the acid is weak or strong.

常见错误是将乙酸当作强酸，直接令 [H⁺] = 0.100 mol dm⁻³。务必根据 Ka 值判断酸是强是弱。

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8. Electrochemistry and Standard Electrode Potentials | 电化学与标准电极电势

below, determine whether zinc metal will displace copper(II) ions from aqueous solution. E⦵(Zn²⁺/Zn) = –0.76 V, E⦵(Cu²⁺/Cu

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