Year 13 CIE Chemistry Unit Test Mock Paper Analysis | 13年级CIE化学单元测试模拟卷解析

📚 Year 13 CIE Chemistry Unit Test Mock Paper Analysis | 13年级CIE化学单元测试模拟卷解析

This article provides a thorough breakdown of a typical Year 13 CIE Chemistry unit test mock paper. Each section deconstructs one key question, explaining the core concepts, model answers, and examiner expectations. By working through these worked examples, you will strengthen your problem-solving skills and deepen your understanding of equilibrium, electrochemistry, transition metals, organic synthesis, polymers, analytical techniques, and thermodynamics.

本文全面解析一份典型的13年级CIE化学单元测试模拟卷。每个部分拆解一道核心题目，讲解关键概念、标准答案和考官期望。通过研读这些例题精解，你将提升解题能力，加深对平衡、电化学、过渡金属、有机合成、聚合物、分析技术及热力学的理解。

1. Overall Structure of the Mock Paper | 模拟卷整体结构

The mock paper is designed to mirror the CIE A2 examination format, featuring a mix of structured questions covering physical, inorganic, and organic chemistry. It contains eight compulsory questions with multiple sub-parts, including calculations, mechanisms, explanations, and data analysis. Time management is critical: allocate roughly 1.5 minutes per mark.

模拟卷依照CIE A2考试格式设计，包含覆盖物理化学、无机化学和有机化学的结构化问题。试卷共有8道必答题，包含计算、机理、解释和数据分析等子问题。时间管理至关重要：大约每分分配1.5分钟。

Question topics range from equilibrium constant calculations, electrode potential predictions, and transition metal complex colours to multi-step organic synthesis, polymer properties, NMR spectroscopy, acid–base titration curves, and Gibbs free energy. A solid grasp of Year 13 learning outcomes is essential.

题目主题涵盖平衡常数计算、电极电势预测、过渡金属配合物的颜色，以及多步有机合成、聚合物性质、核磁共振波谱、酸碱滴定曲线和吉布斯自由能。牢固掌握13年级学习目标是必不可少的。

Examiners frequently test the ability to apply knowledge to unfamiliar contexts, evaluate data, and communicate chemical ideas precisely. Common pitfalls include sign errors in electrochemical cells, incorrect use of curly arrows, and confusion between thermodynamic and kinetic stability.

考官经常考查将知识应用于陌生情境、评估数据和精确表达化学思想的能力。常见错误包括电化学电池中的符号错误、卷曲箭头使用不当以及热力学稳定性与动力学稳定性的混淆。

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2. Question 1: Equilibrium Constant Calculations | 问题1：平衡常数计算

Question: A mixture of 0.80 mol of N₂O₄ and 0.20 mol of NO₂ is allowed to reach equilibrium in a 2.0 dm³ container at 298 K. The equilibrium mixture contains 0.30 mol of NO₂. Calculate the equilibrium constant Kc for the reaction N₂O₄(g) ⇌ 2NO₂(g).

题目：将0.80 mol N₂O₄与0.20 mol NO₂的混合物置于2.0 dm³容器中，在298 K下达到平衡。平衡混合物中含有0.30 mol NO₂。计算反应N₂O₄(g) ⇌ 2NO₂(g)的平衡常数Kc。

Step 1: Determine the change in moles. Initially, n(NO₂) = 0.20 mol, at equilibrium n(NO₂) = 0.30 mol, so 0.10 mol of NO₂ has been formed. According to the stoichiometry, 2 mol NO₂ is produced for every 1 mol N₂O₄ consumed, so the decrease in N₂O₄ is 0.10 ÷ 2 = 0.05 mol. Equilibrium n(N₂O₄) = 0.80 – 0.05 = 0.75 mol.

步骤1：确定物质的量的变化。初始时n(NO₂) = 0.20 mol，平衡时n(NO₂) = 0.30 mol，故生成了0.10 mol NO₂。根据化学计量，每消耗1 mol N₂O₄生成2 mol NO₂，因此N₂O₄减少量为0.10 ÷ 2 = 0.05 mol。平衡时n(N₂O₄) = 0.80 – 0.05 = 0.75 mol。

Step 2: Calculate equilibrium concentrations. Volume = 2.0 dm³. [NO₂] = 0.30 mol / 2.0 dm³ = 0.15 mol·dm⁻³. [N₂O₄] = 0.75 mol / 2.0 dm³ = 0.375 mol·dm⁻³.

步骤2：计算平衡浓度。体积 = 2.0 dm³。[NO₂] = 0.30 mol / 2.0 dm³ = 0.15 mol·dm⁻³。[N₂O₄] = 0.75 mol / 2.0 dm³ = 0.375 mol·dm⁻³。

Step 3: Write the Kc expression and substitute values.

步骤3：写出Kc表达式并代入数值。

Kc = [NO₂]² / [N₂O₄] = (0.15)² / 0.375 = 0.0225 / 0.375 = 0.060 mol·dm⁻³

The Kc value is moderate, indicating the equilibrium lies somewhat to the left. Note that units depend on the stoichiometry; here Kc has units of mol·dm⁻³. Always check that you have used equilibrium concentrations, not initial amounts.

Kc值中等，表明平衡稍向左偏。注意单位取决于化学计量；此处Kc的单位为mol·dm⁻³。务必确保使用的是平衡浓度而非初始量。

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3. Question 2: Redox and Electrode Potentials | 问题2：氧化还原与电极电势

Question: Use the standard electrode potential data to determine whether acidified dichromate(VI) ions can oxidise Fe²⁺ to Fe³⁺ under standard conditions. E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V; E°(Fe³⁺/Fe²⁺) = +0.77 V. Write the overall ionic equation and calculate the cell potential.

题目：根据标准电极电势数据，判断酸性重铬酸根离子能否在标准条件下将Fe²⁺氧化为Fe³⁺。E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V；E°(Fe³⁺/Fe²⁺) = +0.77 V。写出总离子方程式，并计算电池电动势。

First, identify the half-reactions. The more positive E° value undergoes reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. The Fe²⁺/Fe³⁺ system has a less positive potential, so it is forced to undergo oxidation: Fe²⁺ → Fe³⁺ + e⁻. Multiply the oxidation half-equation by 6 to balance electrons.

首先，识别半反应。E°更正的发生还原：Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O。Fe²⁺/Fe³⁺体系的电势较低，将被强制发生氧化：Fe²⁺ → Fe³⁺ + e⁻。将氧化半反应乘以6以平衡电子。

Overall ionic equation: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺.

总离子方程式：Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺。

E°cell = E°(reduction) – E°(oxidation) = +1.33 V – (+0.77 V) = +0.56 V

A positive E°cell confirms the reaction is thermodynamically feasible under standard conditions. The dichromate(VI) orange colour disappears and a green Cr³⁺ solution forms. This test is commonly used to distinguish between Fe²⁺ and Fe³⁺.

正的E°cell证实该反应在标准条件下热力学上可行。重铬酸根离子的橙色消失，生成绿色的Cr³⁺溶液。此反应常用于区分Fe²⁺与Fe³⁺。

Common mistake: students sometimes reverse the formula as E°cell = E°(oxidation) – E°(reduction), yielding a negative value. Always subtract the less positive potential, or use E°cell = E°(right) – E°(left) for the spontaneous cell.

常见错误：学生有时会将公式颠倒为E°cell = E°(氧化) − E°(还原)，得到负值。应始终减去较负的电势，或对自发电池使用E°cell = E°(右) − E°(左)。

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4. Question 3: Transition Metal Complexes | 问题3：过渡金属配合物

Question: Explain why [Cu(H₂O)₆]²⁺ appears pale blue whereas [Cu(NH₃)₄(H₂O)₂]²⁺ is deep blue. Include the role of ligand field splitting and the spectrochemical series.

题目：解释为何[Cu(H₂O)₆]²⁺呈浅蓝色，而[Cu(NH₃)₄(H₂O)₂]²⁺呈深蓝色。需包含配体场分裂及光谱化学序列的作用。

In an octahedral field, the five d orbitals split into two sets: lower energy t₂g and higher energy e_g. In Cu²⁺ (d⁹), the absorption of visible light promotes an electron from t₂g to e_g. The complementary colour to the absorbed wavelength is observed.

在八面体场中，五个d轨道分裂为两组：能量较低的t₂g和较高的e_g。在Cu²⁺ (d⁹)中，吸收可见光使一个电子从t₂g跃迁到e_g。观察到的是被吸收波长的互补色。

H₂O is a weak field ligand, so the splitting energy Δ is small. [Cu(H₂O)₆]²⁺ absorbs in the red/orange region, transmitting mainly pale blue. NH₃ is a stronger field ligand and lies higher in the spectrochemical series, producing a larger Δ. The stronger field shifts the absorption towards higher energy (yellow-orange), giving a deeper blue colour.

H₂O是弱场配体，分裂能Δ较小。[Cu(H₂O)₆]²⁺吸收红/橙光区，主要透过浅蓝色。NH₃是较强场配体，位于光谱化学序列较高位置，产生较大的Δ。较弱的场使吸收移向高能区（橙黄色），呈现更深的蓝色。

The spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < CN⁻. A larger Δ leads to absorption of shorter wavelength light, thus shifting the observed colour according to the colour wheel. This concept is essential for explaining the colours of transition metal complexes.

光谱化学序列：I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < CN⁻。较大的Δ导致吸收较短波长的光，从而根据色轮移动观察色。这一概念对解释过渡金属配合物的颜色至关重要。

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5. Question 4: Organic Synthesis Pathways | 问题4：有机合成路线

Question: Outline a synthesis of 4-nitrophenol from benzene, clearly showing reagents, conditions, and intermediate products. Discuss why direct nitration of phenol is not used.

题目：由苯合成4-硝基苯酚，需清晰给出试剂、条件及中间产物。讨论为何不采用苯酚的直接硝化。

Direct nitration of phenol is problematic because phenol is highly activated towards electrophilic substitution. Nitration with concentrated HNO₃ leads to multiple nitration and oxidation, giving a mixture of 2‑ and 4‑nitrophenols, further nitration to dinitrophenols, and tar formation. Therefore, a route with deactivating protection is preferred.

苯酚的直接硝化存在困难，因为苯酚对亲电取代反应高度活化。浓HNO₃硝化会导致多硝化和氧化，生成2‑和4‑硝基苯酚的混合物，并进一步硝化成二硝基苯酚和焦油。因此，优选使用钝化保护基团的路线。

Step 1: Electrophilic substitution of benzene: Benzene + Cl₂, AlCl₃ → chlorobenzene. Step 2: Nitration of chlorobenzene: Chlorobenzene + conc. HNO₃ / conc. H₂SO₄, 50–60 °C → 4‑chloronitrobenzene (major) + 2‑nitro isomer. The Cl group is ortho/para directing and deactivating. Step 3: Nucleophilic substitution: 4‑chloronitrobenzene + NaOH(aq) under high pressure, 300 °C → 4‑nitrophenoxide, then acidification yields 4‑nitrophenol.

步骤1：苯的亲电取代：苯 + Cl₂, AlCl₃ → 氯苯。步骤2：氯苯的硝化：氯苯 + 浓HNO₃ / 浓H₂SO₄, 50–60 °C → 4‑氯硝基苯（主产物）+ 2‑硝基异构体。Cl基团是邻对位定位基且钝化。步骤3：亲核取代：4‑氯硝基苯 + NaOH(aq) 高温高压，300 °C → 4‑硝基苯酚钠，然后酸化得到4‑硝基苯酚。

Alternatively, a route via diazonium salts can be used. Nitration of benzene gives nitrobenzene, then reduction to phenylamine (aniline). Diazotisation followed by coupling with water yields phenol, but nitration of phenol suffers the same issues. The chlorobenzene route illustrates strategic use of activating/deactivating groups.

或者，可通过重氮盐路线：苯硝化得硝基苯，还原成苯胺，重氮化后与水偶联得苯酚，但硝化仍有同样问题。氯苯路线展示了活化/钝化基团的策略性应用。

Reaction conditions and safety are frequent marks. Benzene is carcinogenic, so fume cupboards are needed. The nucleophilic aromatic substitution step requires extreme conditions because chlorine is normally unreactive; the electron-withdrawing nitro group activates the ring towards attack.

反应条件与安全是常考得分点。苯是致癌物，需使用通风橱。亲核芳香取代步骤需要极端条件，因为氯通常不活泼；吸电子的硝基活化了芳环使其易受进攻。

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6. Question 5: Polymers and Biodegradability | 问题5：聚合物与可生物降解性

Question: Compare the structure and environmental fate of poly(propene) and poly(lactic acid), PLA. Explain why PLA is biodegradable whereas poly(propene) is not.

题目：比较聚丙烯与聚乳酸的结构和环境归宿。解释为何PLA可生物降解而聚丙烯不能。

Poly(propene) is an addition polymer with a hydrocarbon backbone: –[CH₂–CH(CH₃)]ₙ–. Its C–C backbone is chemically inert and resistant to microbial attack. It persists in landfills for hundreds of years and contributes to microplastic pollution.

聚丙烯是一种加成聚合物，具有烃骨架：–[CH₂–CH(CH₃)]ₙ–。其C–C骨架化学惰性，难以被微生物侵蚀。它在垃圾填埋场中存留数百年，造成微塑料污染。

PLA is a condensation polymer derived from lactic acid, containing ester linkages: –[O–CH(CH₃)–CO]ₙ–. The ester groups are susceptible to hydrolysis, which breaks the polymer chain into smaller, water-soluble lactic acid units. Microorganisms can then metabolise these fragments into CO₂ and H₂O.

PLA是由乳酸形成的缩聚物，含有酯键：–[O–CH(CH₃)–CO]ₙ–。酯基易水解，将聚合物链断裂成较小的水溶性乳酸单元。微生物随后可将这些碎片代谢为CO₂和H₂O。

Hydrolysis occurs more rapidly under warm, moist, and slightly acidic or basic conditions, such as in industrial composting facilities. The chiral nature of lactic acid (L‑ and D‑isomers) also affects the degree of crystallinity and degradation rate.

在温暖、潮湿和微酸性或碱性条件下（如工业堆肥设施），水解发生得更快。乳酸的手性（L‑和D‑异构体）也会影响结晶度和降解速率。

Thus, the key difference is the presence of polar, hydrolysable functional groups in the polymer backbone. Condensation polymers like polyesters and polyamides are generally biodegradable, whereas addition polymers of alkenes are not.

因此，关键区别在于聚合物主链中是否存在极性、可水解官能团。聚酯和聚酰胺等缩聚物通常可生物降解，而烯烃的加成聚合物则不能。

Property	Poly(propene)	PLA
Backbone bonds	C–C	Ester (C–O–CO)
Susceptibility to hydrolysis	Very low	High
Biodegradable	No	Yes

Students should avoid describing addition polymers as ‘saturated’ in a way that implies chemical inertness; both contain C–H bonds, but only hydrolysable links enable biodegradation.

学生应避免将加成聚合物描述为“饱和”来暗示其化学惰性；两者都含C–H键，但只有可水解键才能实现生物降解。

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7. Question 6: Analytical Techniques – ¹H NMR Spectroscopy | 问题6：分析技术——¹H核磁共振波谱

Question: A compound C₃H₆O₂ gives the following ¹H NMR data: δ 1.3 (3H, triplet), δ 4.1 (2H, quartet), δ 11.2 (1H, singlet, exchangeable). Deduce the structure and account for the splitting patterns.

题目：某化合物C₃H₆O₂的¹H NMR数据如下：δ 1.3 (3H, 三重峰), δ 4.1 (2H, 四重峰), δ 11.2 (1H, 单峰, 可交换)。推导结构并解释分裂峰型。

The exchangeable proton at δ 11.2 strongly suggests a carboxylic acid –OH. The triplet at δ 1.3 integrating for 3H indicates a CH₃ group adjacent to a CH₂ (n+1 rule: 2 neighbouring protons → triplet). The quartet at δ 4.1 integrating for 2H indicates a CH₂ group adjacent to a CH₃ (3 neighbouring protons → quartet).

δ 11.2处的可交换质子强烈指示为羧酸–OH。δ 1.3的三重峰积分为3H，表明一个CH₃基团邻接一个CH₂（n+1规则：2个相邻质子→三重峰）。δ 4.1的四重峰积分为2H，表明一个CH₂基团邻接一个CH₃（3个相邻质子→四重峰）。

Hence, we have an ethyl group –CH₂CH₃. The remaining atoms are CO₂H. The molecular formula C₃H₆O₂ corresponds to propanoic acid, CH₃CH₂COOH. The CH₂ is attached to the electronegative carbonyl group, causing its downfield shift to δ 4.1. The CH₃ is further away, at δ 1.3.

因此，我们有一个乙基–CH₂CH₃。剩余原子为CO₂H。分子式C₃H₆O₂对应于丙酸CH₃CH₂COOH。CH₂与吸电子的羰基相连，导致其移向低场至δ 4.1。CH₃距离较远，在δ 1.3出现。

The absence of coupling between the –OH and the CH₂ is due to rapid exchange of the acidic proton, which averages the coupling to zero, giving a singlet. Integration ratios are 3:2:1, consistent with the structure.

–OH与CH₂之间不产生耦合，是因为酸性质子快速交换，使耦合平均为零，呈现单峰。积分比例为3:2:1，与结构吻合。

When interpreting NMR spectra, always check the molecular formula first. Here, DBE (double bond equivalents) = 1, consistent with one C=O. Avoid common pitfalls: confusing quartet-triplet with other arrangements such as isopropyl fragments.

解析NMR谱时，务必先核对分子式。此处DBE（不饱和度）为1，与一个C=O相符。避免常见错误：将四重峰-三重峰与异丙基碎片等其他排列混淆。

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8. Question 7: Acid–Base Equilibria and pH Curves | 问题7：酸碱平衡与pH曲线

Question: 25.0 cm³ of 0.10 mol·dm⁻³ CH₃COOH (Ka = 1.8 × 10⁻⁵ mol·dm⁻³) is titrated with 0.10 mol·dm⁻³ NaOH. Sketch the pH curve, calculate the pH at half-equivalence, and select a suitable indicator.

题目：用0.10 mol·dm⁻³ NaOH滴定25.0 cm³ 0.10 mol·dm⁻³ CH₃COOH (Ka = 1.8 × 10⁻⁵ mol·dm⁻³)。绘制pH曲线，计算半中和点的pH，并选择合适的指示剂。

At half-equivalence, half of the weak acid has been neutralised, so [HA] = [A⁻]. From the Henderson–Hasselbalch equation:

在半中和点时，一半的弱酸被中和，因此[HA] = [A⁻]。由Henderson–Hasselbalch方程：

pH = pKa + log₁₀([A⁻]/[HA]) = pKa + log₁₀(1) = pKa

pKa = –log₁₀(1.8 × 10⁻⁵) ≈ 4.74. Therefore, pH at half-equivalence = 4.74.

pKa = –log₁₀(1.8 × 10⁻⁵) ≈ 4.74。因此半中和点pH = 4.74。

The equivalence point occurs when 25.0 cm³ of NaOH has been added. At this point, the solution is CH₃COONa(aq). Acetate ion hydrolyses: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, giving a basic pH (≈8.7). The steep rise occurs between pH ≈7 and 11.

等当点出现在加入25.0 cm³ NaOH时。此时溶液为CH₃COONa(aq)。醋酸根离子水解：CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻，pH约为8.7。突跃范围在pH≈7到11之间。

A suitable indicator must change colour entirely within the steep portion. Phenolphthalein (pH range 8.3–10.0, colourless to pink) is ideal. Methyl orange (3.1–4.4) changes too early and would give a significant titration error. Always draw the curve starting at pH ≈2.9 for the 0.10 M weak acid and level off at high pH.

合适的指示剂必须在突跃范围内完全变色。酚酞（pH范围8.3–10.0，无色到粉红）是最佳选择。甲基橙（3.1–4.4）变色过早，会造成较大滴定误差。绘制曲线时，0.10 M弱酸起点pH≈2.9，并在高pH处趋于平缓。

Buffer region extends around the half-equivalence point, where pH changes slowly. This is a classic weak acid – strong base titration curve, requiring careful labelling of axes and identification of equivalence point volume.

缓冲区域在半中和点附近，pH变化平缓。这是典型的弱酸–强碱滴定曲线，需仔细标注坐标轴并标出等当点体积。

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9. Question 8: Thermodynamics and Entropy | 问题8：热力学与熵

Question: For the decomposition of calcium carbonate, CaCO₃(s) → CaO(s) + CO₂(g), the standard enthalpy change ΔH° = +178 kJ·mol⁻¹ and the standard entropy change ΔS° = +161 J·K⁻¹·mol⁻¹ at 298 K. Calculate ΔG° and determine the temperature at which the reaction becomes feasible.

题目：碳酸钙分解CaCO₃(s) → CaO(s) + CO₂(g)的标准焓变ΔH° = +178 kJ·mol⁻¹，标准熵变ΔS° = +161 J·K⁻¹·mol⁻¹（298 K）。计算ΔG°，并确定反应可自发进行的温度。

Step 1: ΔG° = ΔH° – TΔS°. Ensure unit consistency: convert ΔS° to kJ·K⁻¹·mol⁻¹: +0.161 kJ·K⁻¹·mol⁻¹. At 298 K: ΔG° = +178 – (298 × 0.161) = +178 – 48.0 = +130 kJ·mol⁻¹ (approx.). A positive ΔG° means the reaction is not spontaneous at 298 K.

步骤1：ΔG° = ΔH° – TΔS°。保持单位一致：将ΔS°转换为kJ·K⁻¹·mol⁻¹：+0.161 kJ·K⁻¹·mol⁻¹。298 K时：ΔG° = +178 – (298 × 0.161) = +178 – 48.0 = +130 kJ·mol⁻¹（约）。正ΔG°表示该反应在298 K不自发。

Step 2: Feasibility occurs when ΔG° ≤ 0. Set ΔG° = 0: T = ΔH° / ΔS° = 178 kJ·mol⁻¹ / 0.161 kJ·K⁻¹·mol⁻¹ ≈ 1106 K (833 °C). Above this temperature, TΔS° outweighs ΔH° and ΔG° becomes negative.

步骤2：当ΔG° ≤ 0时反应可行。令ΔG° = 0：T = ΔH° / ΔS° = 178 kJ·mol⁻¹ / 0.161 kJ·K⁻¹·mol⁻¹ ≈ 1106 K (833 °C)。高于此温度，TΔS°超过ΔH°，ΔG°变为负值。

The entropy increase (ΔS° > 0) is primarily due to the production of a gas from a solid, which greatly increases disorder. Many thermal decomposition reactions have a large positive ΔS°, making them feasible at high temperatures despite endothermicity.

熵增（ΔS° > 0）主要由于由固体生成气体，大大增加了混乱度。许多热分解反应具有较大的正ΔS°，尽管吸热，但在高温下可行。

Common error: failing to convert ΔS° to kJ or misplacing the decimal. Also, ‘feasibility’ in thermodynamics refers to spontaneity; kinetics may still prevent the reaction from occurring at an observable rate.

常见错误：未将ΔS°转换为kJ或小数点错位。此外，热力学“可行性”指自发性；动力学因素

Published by TutorHao | Year 13 Chemistry Revision Series | aleveler.com

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