Year 13 Edexcel Chemistry Unit Test Mock Paper Walkthrough | 爱德思化学单元测试模拟卷精讲

📚 Year 13 Edexcel Chemistry Unit Test Mock Paper Walkthrough | 爱德思化学单元测试模拟卷精讲

This walkthrough provides a detailed analysis of a full Year 13 Edexcel Chemistry mock paper, covering key topics such as thermodynamics, kinetics, equilibria, organic synthesis, and spectroscopic analysis. Each question is broken down to highlight common pitfalls, essential revision points, and model answers that meet Edexcel mark scheme expectations.

本文详细讲解一套完整的 Year 13 爱德思化学模拟试卷，覆盖热力学、动力学、平衡、有机合成及波谱分析等核心主题。通过逐题拆解，点出常见失分点、关键复习要点，并提供符合爱德思考官评分标准的满分答案。

1. Lattice Energy and Born–Haber Cycles | 晶格能与玻恩–哈伯循环

Question 1 asked students to construct a Born–Haber cycle for calcium oxide and calculate its lattice energy. Many candidates lost marks by forgetting to double the first ionisation energy of calcium or by incorrectly treating the second electron affinity of oxygen as a single step. The standard enthalpy change of atomisation for oxygen is +249 kJ mol⁻¹, and for calcium it is +178 kJ mol⁻¹. The first and second ionisation energies of calcium are +590 kJ mol⁻¹ and +1145 kJ mol⁻¹ respectively. The first electron affinity of oxygen is –141 kJ mol⁻¹, and the second is +798 kJ mol⁻¹. The enthalpy of formation of CaO(s) is –635 kJ mol⁻¹.

第一题要求学生构建氧化钙的玻恩–哈伯循环并计算其晶格能。许多考生因忘记将钙的第一电离能乘以2，或将氧的第二电子亲和能错误地合并处理而失分。氧的标准原子化焓为 +249 kJ mol⁻¹，钙的为 +178 kJ mol⁻¹。钙的第一、第二电离能分别为 +590 kJ mol⁻¹ 和 +1145 kJ mol⁻¹。氧的第一电子亲和能为 –141 kJ mol⁻¹，第二电子亲和能为 +798 kJ mol⁻¹。CaO(s) 的生成焓为 –635 kJ mol⁻¹。

To calculate lattice energy, the cycle must balance: ΔH_f = ΔH_atomisation(Ca) + ΔH_atomisation(O) + 1st IE(Ca) + 2nd IE(Ca) + 1st EA(O) + 2nd EA(O) + LE. Plugging in numbers: –635 = 178 + 249 + 590 + 1145 – 141 + 798 + LE. This gives LE = –635 – (178+249+590+1145–141+798) = – mitted the double ionisation or treated the second EA incorrectly arrived at values around –2000 to –2500 kJ mol⁻¹, which were not credited.

计算晶格能需保持循环平衡：ΔH_f = ΔH_原子化(Ca) + ΔH_原子化(O) + 第一电离能(Ca) + 第二电离能(Ca) + 第一电子亲和能(O) + 第二电子亲和能(O) + LE。代入数据得：–635 = 178 + 249 + 590 + 1145 – 141 + 798 + LE，解得 LE = –3454 kJ mol⁻¹。漏掉双倍电离或错误处理第二电子亲和能的学生得到的数值在 –2000 至 –2500 kJ mol⁻¹ 之间，无法得分。

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2. Entropy and Gibbs Free Energy | 熵与吉布斯自由能

Question 2 required the calculation of ΔS^⦵ ndard entropies are: NaHCO₃(s) = 102 J K⁻¹ mol⁻¹, Na₂CO₃(s) = 136 J K⁻¹ mol⁻¹, H₂O(g) = 189 J K⁻¹ mol⁻¹, CO₂(g) = 214 J K⁻¹ mol⁻¹. ΔS^⦵ = ΣS^⦵(products) – ΣS^⦵(reactants) = (136 + 189 + 214) – (2 × 102) = 539 – ol⁻¹. Candidates must convert this to kJ K⁻¹ mol⁻¹ (0.335 kJ K⁻¹ mol⁻¹) for use in ΔG = ΔH – TΔS.

第二题要求计算反应的 ΔS^⦵ 并确定反应变得可行的温度。反应 2NaHCO₃(s) → Na₂CO₃(s) + H₂O(g) + CO₂(g) 的标准熵为：NaHCO₃(s) = 102 J K⁻¹ mol⁻¹，Na₂CO₃(s) = 136 J K⁻¹ mol⁻¹，H₂O(g) = 189 J K⁻¹ mol⁻¹，CO₂(g) = 214 J K⁻¹ mol⁻¹。ΔS^⦵ = ΣS^⦵(产物) – ΣS^⦵(反应物) = (136 + 189 + 214) – (2 × 102) = 539 – 204 = +335 J K⁻¹ mol⁻¹。考生必须将其转换为 kJ K⁻¹ mol⁻¹（0.335 kJ K⁻¹ mol⁻¹），以便代入 ΔG = ΔH – TΔS 使用。

Given ΔH^⦵ = +135 kJ mol⁻¹, the reaction becomes feasible when ΔG ≤ 0. Set 135 – eme insists on correct unit conversion and stating that the reaction is feasible above 403 K (130 °C), which explains why solid sodium hydrogencarbonate decomposes on heating.

已知 ΔH^⦵ = +135 kJ mol⁻¹，反应可行时 ΔG ≤ 0。令 135 – T(0.335) ≤ 0 → T ≥ 135 / 0.335 ≈ 403 K。常见错误是未转换 ΔS 单位，直接用 J 计算得出 T ≥ 0.4 K，结果显然荒谬。评分标准要求正确换算单位，并阐明反应在 403 K（130 °C）以上可行，这解释了碳酸氢钠固体受热分解的现象。

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3. Rates and Orders of Reaction | 反应速率与反应级数

uestion 3 presented kinetic data for the reaction 2A + B → C, as shown in the table below.

第三题给出反应 2A + B → C 的动力学数据，如下表所示。

Experiment	[A] / mol dm⁻³	[B] / mol dm⁻³	Initial rate / mol dm⁻³ s⁻¹
1	0.10	0.10	1.20 × 10⁻³
2	0.20	0.10	4.80 × 10⁻³
3	0.20	0.30	1.44 × 10⁻²

riments 2 and 3, tripling [B] while keeping [A] constant multiplies the rate by 3 (1.44 × 10⁻² / 4.80 × 10⁻³ = 3), so the order with respect to B is 1. The rate equation is rate = k[A]²[B]. The rate constant k can be calculated from any experiment, e.g., from experiment 1: k = rate / ([A]²[B]) = 1.20 × 10⁻³ / (0.10² × 0.10) = 1.20 × 10⁻³ / 1.0 × 10⁻³ = 1.2 dm⁶ mol⁻² s⁻¹. Remember to include units.

对比实验 1 和 2，[A] 加倍而 [B] 不变，速率乘以 4，因此对 A 为二级。对比实验 2 和 3，[B] 增至三倍而 [A] 不变，速率乘以 3（1.44 × 10⁻² / 4.80 × 10⁻³ = 3），因此对 B 为一级。速率方程为 rate = k[A]²[B]。可通过任一实验数据求速率常数 k，例如由实验 1： / ([A]²[B]) = 1.20 × 10⁻³ / (0.10² × 0.10) = 1.20 × 10⁻³ / 1.0 × 10⁻³ = 1.2 dm⁶ mol⁻² s⁻¹。务必注明单位。

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4. Equilibrium Constants and Le Chatelier | 平衡常数与勒夏特列原理

Question 4 involved the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = – low temperature favours products, but rate is too slow at low T; a compromise temperature around 450 °C is used with a V₂O₅ catalyst. High pressure favours the side with fewer gas molecules (3 mol → 2 mol), so about 2 atm pressure is enough because too high pressure costs more. The catalyst does not affect equilibrium position but speeds up attainment of equilibrium.

第四题涉及平衡 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = –196 kJ mol⁻¹。第 (a) 小题要求说明最大化 SO₃ 产率的条件。根据勒夏特列原理，正反应放热，低温有利于产物，但低温下速率过慢；实际采用约 450 °C 的折中温度，并使用 V₂O₅ 催化剂。高压有利于气体分子数较少的一侧（3 mol → 2 mol），但约 2 atm 压力即可，过高压力增加成本。催化剂不影响平衡位置，但加快达到平衡的速率。

20, [SO₃] = 0.10 mol dm⁻³. Kc = [SO₃]² / ([SO₂]²[O₂]) = (0.10)² / (0.40² × 0.20) = 0.01 / (0.16 × 0.20) = 0.01 / 0.032 = 0.3125 mol⁻¹ dm³. Many students forgot to square concentrations or misapplied the ratio. Always check the balanced equation and use correct powers.

第 (b) 小题给出了 700 K 时的平衡混合物：在 2.0 dm³ 容器中含 0.80 mol SO₂、0.40 mol O₂ 和 0.20 mol SO₃。浓度为：[SO₂] = 0.40，[O₂] = 0.20，[SO₃] = 0.10 mol dm⁻³。 ²[O₂]) = (0.10)² / (0.40² × 0.20) = 0.01 / (0.16 × 0.20) = 0.01 / 0.032 = 0.3125 mol⁻¹ dm³。许多学生忘记浓度取平方或弄错比例。务必核对配平方程并正确使用幂次。

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5. Acid–Base Equilibria and Buffers | 酸碱平衡与缓冲溶液

mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵ mol dm⁻³), the standard approximation gives [H⁺] = √Ka × c = √1.8 × 10⁻⁵ × 0.15 = √2.7 × 10⁻⁶ = 1.64 × 10⁻³ mol dm⁻³, so pH = – 0 mol. NaOH reacts 1:1 to form ethanoate ions. After addition, moles of acid remaining = 0.20 – mol dm⁻³. Moles of ethanoate = 0.10 mol → [ethanoate] = 0.20 mol dm⁻³. Using the Henderson– f you had miscalculated concentrations, the ratio would be off, leading to a different pH.

第五题考查弱酸和缓冲溶液的 pH 计算。对于 0.15 mol dm⁻³ 的乙酸（Ka = 1.8 × 10⁻⁵ mol dm⁻³），标准近似公式 [H⁺] = √Ka × c = √1.8 × 10⁻⁵ × 0.15 = √2.7 × 10⁻⁶ = 1.64 × 10⁻³ mol dm⁻³，因此 pH = –log(1.64 × 10⁻³) = 2.78。缓冲溶液由 0.10 mol NaOH 加入 0.50 dm³ 的 0.40 mol dm⁻³ 乙酸中制成。初始酸摩尔数 = 0.50 × 0.40 = 0.20 mol。NaOH 按 1:1 反应生成乙酸根离子。加入后剩余酸摩尔数 = 0.20 – 0.10 = 0.10 mol，体积 0.50 dm³ → [酸] = 0.20 mol dm⁻³。乙酸根摩尔数 = 0.10 mol → [乙酸根] = 0.20 mol dm⁻³。使用亨德森–哈塞尔巴尔赫方程：pH = pKa + log([盐]/[酸]) = 4.74 + log(0.20/0.20) = 4.74。浓度计算错误会导致比例偏差，得出不同 pH 值。

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6. Redox Titrations and Electrochemical Cells | 氧化还原滴定与电化学电池

verall: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O. A 25.0 cm³ sample of iron(II) solution required 18.80 cm³ of 0.0200 mol dm⁻³ KMnO₄ for complete oxidation. Moles of MnO₄⁻ = 0.0200 × 18.80/1000 = 3.76 × 10⁻⁴ mol. Moles of Fe²⁺ in 25.0 cm³ = 5 × 3.76 × 10⁻⁴ = 1.88 × 10⁻³ mol. Concentration of Fe²⁺ = 1.88 × 10⁻³ / 0.0250 = 0.0752 mol dm⁻³. Students must show the 1:5 ratio clearly and avoid confusion with units.

第六题是一道关于锰/铁氧化还原滴定的多部分题目。半反应：MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O；Fe²⁺ → Fe³⁺ + e⁻。总反应：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。取 25.0 cm³ 的铁(II) 溶液样品，需要 18.80 cm³ 的 0.0200 mol dm⁻³ KMnO₄ 才能完全氧化。MnO₄⁻ 摩尔数 = 0.0200 × 18.80/1000 = 3.76 × 10⁻⁴ mol。25.0 cm³ 样品中 Fe²⁺ 的摩尔数 = 5 × 3.76 × 10⁻⁴ = 1.88 × 10⁻³ mol。Fe²⁺ 浓度 = 1.88 × 10⁻³ / 0.0250 = 0.0752 mol dm⁻³。考生需清晰展示 1:5 的比例，并避免单位混淆。

⁻/Mn²⁺ cell, given E⁰ MnO₄⁻/Mn²⁺ = +1.52 V and E⁰ Fe³⁺/Fe²⁺ = +0.77 V. The cell EMF = 1.52 – he question asked to write the conventional cell diagram, the correct representation is Pt | Fe²⁺, Fe³⁺ || MnO₄⁻, Mn²⁺, H⁺ | n to phase boundaries and the use of platinum electrodes where no solid metal is involved.

第 (b) 小题询问 Fe²⁺/Fe³⁺ 与 MnO₄⁻/Mn²⁺ 构成的电池电动势，已知 E⁰ MnO₄⁻/Mn²⁺ = +1.52 V，E⁰ Fe³⁺/Fe²⁺ = +0.77 V。电池 EMF = 1.52 – 0.77 = 0.75 V。由于 EMF 为正，反应可行。若题目要求书写传统电池图示，正确表示为 Pt | Fe²⁺, Fe³⁺ || MnO₄⁻, Mn²⁺, H⁺ | Pt，注意相界面以及无固体金属时使用铂电极。

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7. Carbonyl Compounds and Organometallic Synthesis | 羰基化合物与有机金属合成

Question 7 asked to identify the organic product when propanone reacts with HCN under acidic conditions. The mechanism is nucleophilic addition: CN⁻ attacks the electrophilic carbonyl carbon, forming a tetrahedral intermediate, which is protonated to give 2-hydroxy-2-methylpropanenitrile (CH₃)₂C(OH)CN. The mark scheme required use of curly arrows showing attack by CN⁻ on carbon, movement of a pair of electrons to oxygen, and protonation by H⁺/H₂O. A common error was drawing cyanide as the nucleophile but omitting the role of acid in the final step.

第七题要求写出丙酮与 HCN 在酸性条件下反应的有机产物。机理为亲核加成：CN⁻ 进攻亲电的羰基碳，形成四面体中间体，再质子化生成 2-羟基-2-甲基丙腈 (CH₃)₂C(OH)CN。评分标准需用弯箭头显示 CN⁻ 对碳的进攻、一对电子转移至氧，以及 H⁺/H₂O 的质子化作用。常见错误是画出了氰根作为亲核试剂，但忽略了酸在最后一步的作用。

The question also covered the reaction with LiAlH₄ in dry ether, which reduces propanone to propan-2-ol. The mechanism involves hydride (H⁻) transfer from AlH₄⁻, followed by hydrolysis to release the alcohol. Candidates must remember that NaBH₄ is also capable of this reduction but LiAlH₄ is stronger and can reduce carboxylic acids as well. Care must be taken to show the alcohol as the final product, not the alkoxide.

题目还涉及在无水乙醚中用 LiAlH₄ 还原丙酮的反应，生成丙-2-醇。机理为 AlH₄⁻ 提供氢负离子 (H⁻) 转移，随后水解得到醇。考生需记住 NaBH₄ 也能实现此还原，但 LiAlH₄ 更强，还能还原羧酸。必须小心展示最终产物为醇，而非醇盐。

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8. Aromatic Chemistry and Directing Effects | 芳香化学与定位效应

Question 8 required an explanation of why nitrobenzene reacts with bromine more slowly than benzene, and why the bromination occurs at the meta position. The nitro group is an electron-withdrawing group (deactivator) that removes electron density from the pi system, making the ring less susceptible to electrophilic attack. It directs to meta because the ortho and para positions are destabilised by adjacent positive charges in the resonance structures, while the meta position carries a partial negative charge that is slightly more favourable.

第八题要求解释为何硝基苯与溴的反应比苯更慢，以及为何溴代发生在间位。硝基是吸电子基团（钝化基团），从 π 体系中抽走电子云密度，使苯环不易受亲电进攻。它定位到间位是因为邻、对位在共振结构中因相邻正电荷而去稳定化，而间位带有部分负电荷，相对较为有利。

In the mock paper, many students confused deactivation with meta-directing, stating that meta is always preferred for deactivated rings. This is not true; halogen substituents are deactivating yet ortho/para directing. Understanding the interplay between inductive and mesomeric effects is crucial. For the nitro group, the powerful -M effect dominates, leading to meta substitution. The mechanism must show formation of the electrophile Br⁺ (from Br₂/FeBr₃), attack at the meta position, and regeneration of the catalyst.

在此次模拟卷中，许多学生混淆了钝化与间位定位，声称钝化环一定导向间位。这并不准确；卤素取代基虽是钝化基团，却为邻/对位定位。理解诱导效应与共轭效应的相互作用至关重要。对于硝基，强 -M 效应占主导，导致间位取代。机理中必须展示亲电试剂 Br⁺ 的生成（由 Br₂/FeBr₃ 产生）、对间位的进攻以及催化剂的再生。

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9. Organic Synthesis and Retrosynthetic Analysis | 有机合成与逆合成分析

om benzene to 4-nitrobenzoic acid. Students needed to propose a two-step sequence: Friedel– ₄ at low temperature. The key challenge was recognising that oxidation of the side chain to a carboxylic acid should be done before nitration, or that nitration must be performed after the ketone is oxidised? Actually, the correct order is to acylate, then oxidise the acetyl group to – , giving 3-nitrobenzoic acid. However, the question asked for 4-nitrobenzoic acid (para). Therefore, the route must ensure para substitution. This would require nitration first (nitro group is meta-directing, so would not give para). The better route: alkylation to toluene, then nitrate → predominantly ortho/para, separate para-nitrotoluene, then oxidise methyl to COOH with KMnO₄ to yield 4-nitrobenzoic acid. This exemplifies how retrosynthetic planning determines the order of steps based on directing effects.

第九题给出由苯合成 4-硝基苯甲酸的路线。学生需提出一个两步序列：苯的傅克酰基化（用乙酰氯 CH₃COCl 和 AlCl₃）制得苯乙酮，随后在低温下用浓 HNO₃/H₂SO₄ 进行硝化。关键难点在于认识到侧链氧化为羧酸应在硝化之前进行，还是硝化必须在酮氧化之后进行？实际上，正确顺序是：酰基化，然后用 KMnO₄/OH⁻ 将乙酰基氧化为 –COOH，最后硝化 HNO₃/H₂SO₄。但 COOH 基团为间位定位，硝化将发生在相对于 COOH 的间位，得到 3-硝基苯甲酸。然而，题目要求的是 4-硝基苯甲酸（对位）。因此，路线必须确保对位取代。这需要先硝化吗？硝基也是间位定位，不会给出对位。更好路线：烷基化生成甲苯，再硝化 → 主要得邻、对位，分离出对硝基甲苯，然后用 KMnO₄ 将甲基氧化为 COOH，得到 4-硝基苯甲酸。这体现了逆合成分析如何根据定位效应决定步骤顺序。

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10. Spectral Analysis and Structure Determination | 波谱分析与结构推断

NMR data for an unknown organic compound C₄H₈O₂. The IR showed a broad absorption at 2500–3300 cm⁻¹ (O– als: δ 1.2 (3H, triplet), δ 2.4 (2H, quartet), δ 11.5 (1H, singlet). The triplet and quartet indicate an ethyl group (CH₃CH₂– The fragmentation pattern confirmed loss of OH and COOH groups, consistent with propanoic acid.

第十题给出了未知有机物 C₄H₈O₂ 的红外、质谱和 ¹H 核磁共振数据。红外光谱在 2500–3300 cm⁻¹ 有宽吸收（羧酸 O–H 伸缩振动），1710 cm⁻¹ 有强吸收峰（C=O 伸缩）。质谱显示分子离子峰 m/z = 88。¹H NMR 有三个信号：δ 1.2 (3H, 三重峰)，δ 2.4 (2H, 四重峰)，δ 11.5 (1H, 单峰)。三重峰和四重峰指示一个乙基 (CH₃CH₂–) 与羰基相邻。δ 11.5 的宽单峰是羧酸质子的特征。因此，该化合物是丙酸，CH₃CH₂COOH。碎片模式证实了 OH 和 COOH 基团的丢失，与丙酸吻合。

This question highlighted common pitfalls: misidentifying the broad O–H as an alcohol rather than carboxylic acid (alcohol O–H appears around 3200–3600 cm⁻¹ but usually sharper unless hydrogen-bonded extensively). The integration of 1H for the COOH proton is key; sometimes it may be broad and exchangeable with D₂O, which is a classic test. Students should practice interpreting combined spectra and using the molecular formula to confirm their structure.

本题凸显了常见错误：将宽 O–H 吸收误认为醇而非羧酸（醇 O–H 出现在 3200–3600 cm⁻¹ 左右，但通常较尖锐，除非存在广泛氢键）。COOH 质子的积分面积为 1H 是关键；有时它可能很宽，且可与 D₂O 交换，这是经典验证方法。学生应练习综合解析谱图，并利用分子式确认结构。

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11. Common Errors and How to Avoid Them | 常见错误与避坑指南

Throughout the mock paper, several recurring mistakes were evident: (1) forgetting to convert units, especially in entropy and kinetics calculations; (2) misapplying squared terms in Kc and rate equations; (3) drawing incomplete mechanisms, missing lone pairs or curly arrows; (4) confusing nucleophilic addition with electrophilic substitution; (5) poor understanding of directing effects in benzene substitution. To avoid these, use checklists: Units, Significant figures, Balanced equation, All arrows and charges, and always ask whether the mechanism matches the reagent type.

整份模拟卷中，反复出现几个错误：(1) 忘记单位换算，尤其是在熵和动力学计算中；(2) 在 Kc 和速率方程中错误使用平方项；(3) 机理图不完整，遗漏孤对电子或弯箭头；(4) 混淆亲核加成与亲电取代；(5) 对苯环取代定位效应理解不透彻。为避免这些错误，务必使用检查清单：单位、有效数字、配平方程、全部箭头和电荷，并时刻自问机理是否与试剂类型匹配。

Another big issue was time management. Several students spent too long on thermodynamics and did not attempt the organic section. The paper allocates roughly equal marks across physical and organic chemistry; practice under timed conditions is essential. Also, in extended response questions, structure your answer logically, using bullet points or clear steps, to ensure the examiner can follow your reasoning.

另一大问题是时间管理。部分学生在热力学部分耗时过多，未能完成有机板块。试卷在物理化学和有机化学间大致平分分值，计时练习至关重要。此外，在扩展型问题中，答案结构要条理清晰，使用要点或清晰步骤，确保考官能跟上你的推理逻辑。

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12. Summary and Revision Focus | 总结与复习侧重点

This mock paper walkthrough has emphasised the core skills Edexcel Year 13 chemistry students must master: quantitative manipulation of energy cycles, application of entropy and Gibbs free energy, deduction of rate equations, equilibrium constant calculations, buffer solutions, redox stoichiometry, organic mechanisms with curly arrows, aromatic substitution rules, multi-step synthesis planning, and spectroscopic analysis. The key to success is not just knowing the facts but practising them in exam-style contexts. Review the mark schemes for similar past papers; they reveal exactly how marks are awarded for each step.

本次模拟卷精讲强调了爱德思 Year 13 化学学生必须掌握的核心技能：能量循环的定量运算、熵与吉布斯自由能的应用、速率方程的推导、平衡常数计算、缓冲溶液、氧化还原计量关系、带弯箭头的有机机理、芳环取代规则、多步合成规划以及波谱分析。成功的关键不仅是记住知识点，更要在考试情境中反复练习。复习类似历年真题的评分标准，它们能清晰展示每一步的给分点。

We recommend focusing revision on areas where you lost marks in this mock. Use active recall, draw mechanisms repeatedly until they become second nature, and attempt full papers under exam conditions. With targeted practice, you can turn your weaknesses into strengths before the final assessment.

我们建议将复习重点放在本模拟卷中失分的部分。运用主动回忆法，反复绘制机理直到滚瓜烂熟，并在模拟考试环境下完成整套试卷。通过有针对性的训练，你能在最终大考前将薄弱环节转化为优势。

Published by TutorHao | Chemistry Revision Series | aleveler.com

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