A-Level Chemistry Calculation Questions in June 2018 Paper 2 | A-Level 化学：2018年6月试卷2计算题型精讲

📚 A-Level Chemistry Calculation Questions in June 2018 Paper 2 | A-Level 化学：2018年6月试卷2计算题型精讲

Calculation questions form the backbone of A-Level Chemistry examinations, testing not only your ability to recall facts but also your capacity to apply numerical reasoning to chemical situations. The June 2018 Paper 2 of the A-Level Chemistry specification (often CIE 9701/22 or a similar board) contains a rich variety of calculation tasks spanning stoichiometry, energetics, kinetics, and equilibria. Mastering these question types is essential for a top grade. This article breaks down the key calculation types found in that examination, provides step-by-step strategies, and illustrates common pitfalls, helping you transform challenging numbers into clear marks.

计算题是 A-Level 化学考试的核心组成部分，不仅检验你对知识点的记忆，更侧重于将量化推理应用于化学情境的能力。2018年6月A-Level化学试卷2（常见于 CIE 9701/22 或同类考试局）囊括了涵盖化学计量、能量学、动力学与平衡的丰富计算题型。掌握这些题型是取得高分的关键。本文将逐一剖析该试卷出现的核心计算类型，提供分步策略，揭示常见易错点，帮助你将复杂的数字转化为清晰的得分点。

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

Mole concept underpins almost all quantitative chemistry. In June 2018 Paper 2, a typical question asks you to find the mass of a product formed from a given mass of reactant, using balanced equations and molar masses. Start by converting the given mass to moles using Mᵣ. Apply the mole ratio from the equation. Then convert moles of the desired substance back to mass or volume. Always check the stoichiometric coefficients carefully – especially when dealing with reactions involving Group 2 carbonates or thermal decomposition.

摩尔概念几乎是所有量化化学的基石。2018年6月试卷2中，典型的问题要求从给定反应物质量求算生成物的质量，需利用配平方程式和摩尔质量。首先用相对分子质量 Mᵣ 将给定质量转换为物质的量。根据方程式中的摩尔比进行换算，再将目标物质的物质的量转换为质量或体积。务必仔细核对化学计量系数 – 尤其在涉及第2族碳酸盐或热分解反应时。

Example: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂. Calculate mass of Na₂CO₃ from 8.4 g of NaHCO₃.
示例：2NaHCO₃ → Na₂CO₃ + H₂O + CO₂。由8.4 g NaHCO₃ 计算 Na₂CO₃ 的质量。

n(NaHCO₃) = 8.4 / 84.0 = 0.100 mol. Mole ratio 2:1 gives n(Na₂CO₃) = 0.0500 mol. Mass = 0.0500 × 106.0 = 5.30 g.

n(NaHCO₃) = 8.4 / 84.0 = 0.100 mol。摩尔比 2:1，得 n(Na₂CO₃) = 0.0500 mol。质量 = 0.0500 × 106.0 = 5.30 g。

2. Titration and Concentration | 滴定与浓度计算

Titration calculations appear frequently. A common format in Paper 2 involves a back titration or direct acid-base titration. You must relate the volume and concentration of the titrant to the amount of unknown. Use the formula n = c × V (in dm³). Don’t forget to convert cm³ to dm³ by dividing by 1000. Many students lose marks by forgetting the mole ratio between the analyte and titrant, especially in redox titrations (e.g., MnO₄⁻ with Fe²⁺).

滴定计算频繁出现。试卷2常见的题型包括返滴定或直接酸碱滴定。你需要将滴定剂的体积和浓度与未知物的物质的量关联。使用公式 n = c × V（V 以 dm³ 为单位）。不要忘记将 cm³ 转换为 dm³（除以1000）。许多学生遗忘待测物与滴定剂之间的摩尔比而失分，尤其在氧化还原滴定中（如 MnO₄⁻ 与 Fe²⁺）。

Step	Action	步骤
1	n(titrant) = c × V(dm³)	计算滴定剂物质的量
2	Use mole ratio to find n(analyte)	利用摩尔比求待测物的 n
3	Scale to original volume if an aliquot was taken	若取用部分溶液，需按比例放大

For instance, in a typical June 2018 question: 25.0 cm³ of a Na₂CO₃ solution required 22.40 cm³ of 0.100 mol dm⁻³ HCl for neutralisation. n(HCl) = 0.100 × 0.0224 = 0.00224 mol. 2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂, so n(Na₂CO₃) = 0.00112 mol in 25.0 cm³. Concentration = 0.00112 / 0.025 = 0.0448 mol dm⁻³.

例如，2018年6月一道典型题：25.0 cm³ Na₂CO₃ 溶液需 22.40 cm³ 0.100 mol dm⁻³ HCl 中和。n(HCl) = 0.100 × 0.0224 = 0.00224 mol。2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂，故 n(Na₂CO₃) = 0.00112 mol（在25.0 cm³中）。浓度 = 0.00112 / 0.025 = 0.0448 mol dm⁻³。

3. Percentage Yield and Atom Economy | 产率与原子经济性

Many synthesis questions in Paper 2 ask for percentage yield and atom economy. Percentage yield compares the actual mass (or moles) of product to the theoretical maximum. % yield = (actual yield / theoretical yield) × 100. Atom economy assesses the efficiency of a reaction in incorporating atoms from reactants into the desired product: % atom economy = (Mᵣ of desired product / sum of Mᵣ of all reactants) × 100. Both are indicators of sustainability. In the June 2018 paper, you might be given an experimental yield and asked to evaluate a reaction’s greenness.

试卷2的许多合成题会要求计算产率与原子经济性。产率（百分收率）将产品的实际质量（或物质的量）与理论最大值比较。产率(%) = (实际产量/理论产量) × 100。原子经济性衡量反应物中的原子进入目标产物的效率：原子经济性(%) = (目标产物 Mᵣ / 所有反应物 Mᵣ 总和) × 100。两者均是可持续性的指标。在2018年6月试卷中，可能给出实验产量，要求评价反应绿色性。

Example: Synthesis of ethyl ethanoate: CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O. Mr desired = 88, sum of reactants = 60+46=106. Atom economy = 88/106 × 100 = 83.0%. If 5.00 g of ester were obtained from a theoretical 6.50 g, % yield = 5.00/6.50 × 100 = 76.9%.

示例：乙酸乙酯的合成：CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O。目标产物 Mr = 88，反应物总和 = 60+46=106。原子经济性 = 88/106 × 100 = 83.0%。若实际得酯5.00 g，理论产量6.50 g，则产率 = 5.00/6.50 × 100 = 76.9%。

4. Ideal Gas Equation and Molar Volume | 理想气体方程与气体摩尔体积

The ideal gas equation pV = nRT is a staple calculation. Paper 2 frequently asks for the volume of gas produced at a certain temperature and pressure, or the relative molecular mass of a volatile liquid. Remember: p in Pa, V in m³, T in K, R = 8.31 J mol⁻¹ K⁻¹. When using kPa, R = 8.31 × 10³ Pa, but it’s safer to convert p to Pa. Also, molar volume at RTP (20°C, 101 kPa) is 24.0 dm³ mol⁻¹; at STP (0°C, 101 kPa) it’s 22.4 dm³ mol⁻¹. Ensure you identify which conditions are being used.

理想气体方程 pV = nRT 是常考计算。试卷2经常要求计算在某温度压力下生成气体的体积，或求挥发性液体的相对分子质量。牢记：p 用 Pa，V 用 m³，T 用 K，R = 8.31 J mol⁻¹ K⁻¹。若压力为 kPa，R 可视为 8.31 × 10³，但更稳妥的方法是统一将 p 转为 Pa。此外，常温常压（RTP，20°C，101 kPa）下摩尔体积为 24.0 dm³ mol⁻¹；标准状况（STP，0°C，101 kPa）下为 22.4 dm³ mol⁻¹。务必辨别题目采用的条件。

Typical calculation: 0.200 mol of CO₂ is collected at 30°C and 100 kPa. Find volume. T = 303 K, p = 100 000 Pa. V = nRT/p = (0.200 × 8.31 × 303) / 100 000 = 0.00504 m³ = 5.04 dm³.

典型计算：0.200 mol CO₂ 在30°C、100 kPa下收集，求体积。T = 303 K, p = 100 000 Pa。V = nRT/p = (0.200 × 8.31 × 303) / 100 000 = 0.00504 m³ = 5.04 dm³。

5. Enthalpy Changes and Calorimetry | 焓变与量热计算

Energetics calculations in Paper 2 often involve q = mcΔT and then ΔH = -q/n. The exam may provide a temperature rise when a solid dissolves or a reaction occurs in a calorimeter. Students must correctly identify the mass used in m (usually the mass of water or solution; assume density 1 g cm⁻³ for dilute solutions). The specific heat capacity c is 4.18 J g⁻¹ K⁻¹ for water. Convert ΔT from °C to K as necessary (the numerical value is the same). Remember that ΔH has a negative sign if the temperature increases (exothermic) and positive if temperature decreases (endothermic).

试卷2的能量计算常涉及 q = mcΔT，而后 ΔH = -q/n。题目可能给出温度变化——固体溶解或反应在量热计中进行时的温升。学生须正确确定式中 m 所指的质量（通常是水或溶液的质量；稀溶液密度可视为 1 g cm⁻³）。水的比热容 c 为 4.18 J g⁻¹ K⁻¹。ΔT 数值在摄氏度和开尔文下相同。切记温度升高时 ΔH 为负（放热），降低时为正（吸热）。

Example: 2.00 g of NaOH is dissolved in 100 g of water; temperature rises from 22.0°C to 28.5°C. Mass = 100 g (ignore solid’s mass for simplicity). q = 100 × 4.18 × 6.5 = 2717 J. Moles NaOH = 2.00/40.0 = 0.0500 mol. ΔH = -2.717 kJ / 0.0500 mol = -54.3 kJ mol⁻¹.

示例：2.00 g NaOH 溶于 100 g 水；温度从22.0°C升至28.5°C。质量 m 取 100 g（忽略固体质量以简化）。q = 100 × 4.18 × 6.5 = 2717 J。NaOH 物质的量 = 2.00/40.0 = 0.0500 mol。ΔH = -2.717 kJ / 0.0500 mol = -54.3 kJ mol⁻¹。

6. Hess’s Law and Enthalpy Cycles | 盖斯定律与焓循环

Hess’s Law states that the total enthalpy change of a reaction is independent of the route. Paper 2 may ask you to construct an enthalpy cycle or use algebraic manipulation of given ΔH values. Typical routes involve formation or combustion data. For example, ΔfH° values can be used to find reaction enthalpy: ΔrH° = Σ ΔfH°(products) – Σ ΔfH°(reactants). Alternatively, using bond enthalpies: ΔrH = Σ bonds broken – Σ bonds formed. Remember, bond enthalpy data are averages and apply to gaseous species only.

盖斯定律指出，反应总的焓变与途径无关。试卷2可能要求你构建焓循环，或通过给出的 ΔH 数据进行代数运算。常用途径包括生成焓或燃烧焓数据。例如，利用标准生成焓 ΔfH° 计算反应焓：ΔrH° = Σ ΔfH°(产物) – Σ ΔfH°(反应物)。也可用键焓：ΔrH = Σ 断裂键焓 – Σ 形成键焓。记住，键焓数据为平均值，且仅适用于气态物种。

Practice: Find ΔH for 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l). Given: ΔcH of C(s) = -394, H₂(g) = -286, C₂H₅OH(l) = -1367 kJ mol⁻¹. Construct a cycle via combustion products (CO₂ and H₂O). ΔH = [2(-394) + 3(-286)] – (-1367) = (-788 – 858) + 1367 = -279 kJ mol⁻¹.

练习：求 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) 的 ΔH。已知：C(s) 的燃烧焓 = -394, H₂(g) = -286, C₂H₅OH(l) = -1367 kJ mol⁻¹。构建以燃烧产物（CO₂ 和 H₂O）为终点的循环。ΔH = [2(-394) + 3(-286)] – (-1367) = (-788 – 858) + 1367 = -279 kJ mol⁻¹。

7. Equilibrium Constant Kc | 平衡常数 Kc

Kc calculations are a major feature in Paper 2. You may be given initial and equilibrium concentrations (or moles) and need to deduce the value of Kc. The expression must be written first, e.g., for aA + bB ⇌ cC + dD, Kc = ([C]ᶜ[D]ᵃ) / ([A]ᵇ[B]ᵈ). Concentrations are in mol dm⁻³. Pay attention to heterogeneous equilibria – solids are omitted from the expression. The temperature must be stated because Kc is temperature dependent. A typical question gives the equilibrium amount of one substance in a vessel of known volume; you must work out the equilibrium moles of all species using the balanced equation and the initial amounts.

Kc 计算是试卷2的重要考点。可能给出初始浓度和平衡浓度（或物质的量），要求推导 Kc 值。首先写出表达式，如反应 aA + bB ⇌ cC + dD，Kc = ([C]ᶜ[D]ᵃ) / ([A]ᵇ[B]ᵈ)。浓度单位均为 mol dm⁻³。注意多相平衡——固体不在表达式中出现。必须注明温度，因 Kc 随温度变化。典型题目给出某物质在已知体积容器中的平衡物质的量；你需利用配平方程式及初始量求出各物种的平衡物质的量。

Example: 2.0 mol A and 1.0 mol B are mixed in a 2.0 dm³ vessel; at equilibrium, 0.4 mol of C is formed. Reaction: A + 2B ⇌ 3C. Calculate Kc. Initial: A=2.0, B=1.0, C=0. Change: A -x, B -2x, C +3x. Given 3x = 0.4, so x=0.1333. Eqm mol: A=1.867, B=0.733, C=0.4. Conc: [A]=0.934, [B]=0.367, [C]=0.2. Kc = (0.2³) / (0.934 × 0.367²) = 0.008 / (0.934 × 0.135) = 0.0634 mol dm⁻³. (Units depend on the equation.)

示例：2.0 mol A 与 1.0 mol B 在 2.0 dm³ 容器中混合；平衡时生成了 0.4 mol C。反应：A + 2B ⇌ 3C。计算 Kc。初始：A=2.0, B=1.0, C=0。变化：A -x, B -2x, C +3x。由 3x = 0.4，得 x = 0.1333。平衡物质的量：A = 1.867, B = 0.733, C = 0.4。浓度：[A]=0.934, [B]=0.367, [C]=0.2。Kc = (0.2³) / (0.934 × 0.367²) = 0.008 / (0.934 × 0.135) = 0.0634（单位视方程式而定）。

8. Rate Equations and Kinetics | 速率方程与动力学

Rate calculations involve using initial rates data to determine orders and the rate constant k. The common format in Paper 2: a table of initial concentrations and initial rates. By comparing experiments where only one concentration changes, you can deduce the order with respect to that reactant. Then write the rate equation: rate = k[A]ᵐ[B]ⁿ. Solve for k with units: mol dm⁻³ s⁻¹ / (mol dm⁻³)ˢ, where s is the overall order. The Arrhenius equation may also appear, linking k to temperature and activation energy: ln k = ln A – Ea/(RT).

速率计算涉及利用初始速率数据确定反应级数和速率常数 k。试卷2常见形式：给出初始浓度与初始速率的表格。比较仅一个浓度变化的实验，可推知该反应物的分级数。然后写出速率方程：r = k[A]ᵐ[B]ⁿ。求解 k 时注意单位，例如总级数为 s，则 k 的单位为 mol dm⁻³ s⁻¹ / (mol dm⁻³)ˢ。阿伦尼乌斯方程也可能出现，将 k 与温度、活化能关联：ln k = ln A – Ea/(RT)。

Worked example: Experiment [A] [B] Initial rate (mol dm⁻³ s⁻¹) 1: 0.1, 0.1, 2.0×10⁻⁴. 2: 0.2, 0.1, 8.0×10⁻⁴. 3: 0.1, 0.2, 2.0×10⁻⁴. Comparing 1 and 2: [A] doubles, rate ×4 → order w.r.t A = 2. Comparing 1 and 3: [B] doubles, rate unchanged → order w.r.t B = 0. Rate equation: rate = k[A]². k = rate/[A]² = 2.0×10⁻⁴ / (0.1)² = 0.020 dm³ mol⁻¹ s⁻¹.

例题：实验 [A] [B] 初始速率(mol dm⁻³ s⁻¹) 1: 0.1, 0.1, 2.0×10⁻⁴。2: 0.2, 0.1, 8.0×10⁻⁴。3: 0.1, 0.2, 2.0×10⁻⁴。比较1和2：[A]加倍，速率×4 → 对A为2级。比较1和3：[B]加倍，速率不变 → 对B为0级。速率方程：rate = k[A]²。k = rate/[A]² = 2.0×10⁻⁴ / (0.1)² = 0.020 dm³ mol⁻¹ s⁻¹。

9. Electrode Potentials and Cell EMF | 电极电势与电池电动势

Calculations in electrochemistry typically focus on E°cell = E°(right-hand electrode) – E°(left-hand electrode) under standard conditions. Paper 2 may also test the Nernst equation for non-standard conditions: E = E° – (RT/nF) ln Q. At 298 K, this simplifies to E = E° – (0.0592/n) log Q (using log₁₀). You must be able to calculate the cell potential when concentrations are not 1 mol dm⁻³. A common task is to predict whether a reaction is feasible under specified concentrations by calculating Ecell; if Ecell > 0, the reaction is thermodynamically feasible.

电化学计算通常围绕 E°cell = E°(右侧电极) – E°(左侧电极)（标准条件）。试卷2也可能考查非标准条件下的能斯特方程：E = E° – (RT/nF) ln Q。在298 K时简化为 E = E° – (0.0592/n) log Q（以10为底）。你必须能够计算浓度不为 1 mol dm⁻³ 时的电池电势。常见任务是计算指定浓度下的电池电动势，判断反应是否可行；若 Ecell > 0，则反应热力学可行。

Example: Zn(s) | Zn²⁺(0.010 mol dm⁻³) || Cu²⁺(0.10 mol dm⁻³) | Cu(s). E°(Zn²⁺/Zn) = -0.76 V, E°(Cu²⁺/Cu) = +0.34 V. E°cell = 0.34 – (-0.76) = 1.10 V. Reaction: Zn + Cu²⁺ ⇌ Zn²⁺ + Cu. n=2. Q = [Zn²⁺]/[Cu²⁺] = 0.010/0.10 = 0.10. Ecell = 1.10 – (0.0592/2) log(0.10) = 1.10 – 0.0296 × (-1) = 1.13 V. Feasible as Ecell > 0.

示例：Zn(s) | Zn²⁺(0.010 mol dm⁻³) || Cu²⁺(0.10 mol dm⁻³) | Cu(s)。E°(Zn²⁺/Zn) = -0.76 V, E°(Cu²⁺/Cu) = +0.34 V。标准电池电势 E°cell = 1.10 V。反应：Zn + Cu²⁺ ⇌ Zn²⁺ + Cu，n=2。Q = [Zn²⁺]/[Cu²⁺] = 0.010/0.10 = 0.10。Ecell = 1.10 – (0.0592/2) log(0.10) = 1.10 – 0.0296 × (-1) = 1.13 V。Ecell > 0，反应可行。

10. Redox Titration and Iodine-Thiosulfate | 氧化还原滴定与碘-硫代硫酸盐反应

Redox titrations are a classic quantitative technique. The iodine-thiosulfate titration (iodometry) is especially common: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. A known concentration of sodium thiosulfate is used to titrate iodine solution produced from an oxidising agent. The stoichiometry must be deduced from half-equations. Other oxidising agents like KMnO₄ (acidified) or K₂Cr₂O₇ often appear. Key steps: determine moles of titrant, use the balanced redox equation to find moles of analyte, then convert to mass, purity, or concentration.

氧化还原滴定是经典定量技术。碘-硫代硫酸盐滴定（碘量法）尤为常见：I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻。用已知浓度的硫代硫酸钠滴定由氧化剂产生的碘溶液。化学计量比需从半反应推导。其他如酸性 KMnO₄ 或 K₂Cr₂O₇ 也常见。关键步骤：计算滴定剂的物质的量，用配平的氧化还原方程式求分析物的物质的量，再换算为质量、纯度或浓度。

For instance, in a copper(II) determination: 2Cu²⁺ + 4I⁻ → 2CuI + I₂. The liberated I₂ is titrated with 0.100 mol dm⁻³ S₂O₃²⁻, and 24.50 cm³ is required. n(S₂O₃²⁻) = 0.100 × 0.0245 = 0.00245 mol. From the stoichiometry: 2Cu²⁺ ≡ I₂ ≡ 2S₂O₃²⁻, so n(Cu²⁺) = n(S₂O₃²⁻) = 0.00245 mol. Mass Cu = 0.00245 × 63.5 = 0.156 g.

例如，铜(II)离子测定：2Cu²⁺ + 4I⁻ → 2CuI + I₂。生成的 I₂ 用 0.100 mol dm⁻³ S₂O₃²⁻ 滴定，用去 24.50 cm³。n(S₂O₃²⁻) = 0.100 × 0.0245 = 0.00245 mol。由化学计量关系：2Cu²⁺ ≡ I₂ ≡ 2S₂O₃²⁻，故 n(Cu²⁺) = 0.00245 mol。铜的质量 = 0.00245 × 63.5 = 0.156 g。

11. Percentage Purity and Water of Crystallisation | 纯度与水合结晶计算

Questions on percentage purity often follow titration or mass loss experiments. The principle: % purity = (mass of pure substance / mass of impure sample) × 100. Similarly, the determination of water of crystallisation involves finding x in a hydrated salt formula such as MgSO₄·xH₂O. Heating a known mass to constant mass drives off the water; the mass loss corresponds to the water. Using moles, the value of x is calculated.

纯度计算常紧随滴定或失重实验。原理：纯度(%) = (纯净物质量 / 不纯样品质量) × 100。同理，结晶水测定涉及求水合盐如 MgSO₄·xH₂O 中的 x。将已知质量的样品加热至恒重，驱除结晶水；质量损失即为水的质量。通过物质的量可求出 x。

Example: 3.00 g of hydrated sodium carbonate, Na₂CO₃·xH₂O, is heated, leaving 1.27 g of anhydrous Na₂CO₃. Mass of water = 3.00 – 1.27 = 1.73 g. M(Na₂CO₃) = 106.0, n(Na₂CO₃) = 1.27/106.0 = 0.0120 mol. M(H₂O) = 18.0, n(H₂O) = 1.73/18.0 = 0.0961 mol. Ratio H₂O : Na₂CO₃ = 0.0961/0.0120 = 8.0, so x = 8.

例题：3.00 g 水合碳酸钠 Na₂CO₃·xH₂O 加热后残留 1.27 g 无水 Na₂CO₃。水的质量 = 3.00 – 1.27 = 1.73 g。M(Na₂CO₃) = 106.0，n(Na₂CO₃) = 1.27/106.0 = 0.0120 mol。M(H₂O) = 18.0，n(H₂O) = 1.73/18.0 = 0.0961 mol。比值 = 0.0961/0.0120 ≈ 8.0，故 x = 8。

12. Bringing It All Together – Exam Strategy | 综合运用——应试策略

Success in Paper 2 calculation questions requires a systematic approach: read the question carefully, identify what is asked, write down the relevant formulas and balanced equations, show all working steps, and include units. Check your answer for reasonableness (e.g., % yield cannot exceed 100% under normal circumstances). Manage your time – some calculations are multi-step and may carry 3-5 marks, so allocate sufficient time. Practising past papers, like the June 2018 Paper 2, under timed conditions is the best preparation.

攻克试卷2计算题需要系统的方法：仔细读题，明确所求，写下相关公式和配平方程式，展示所有计算步骤，并附带单位。检查答案的合理性（例如正常情况下产率不超过100%）。合理分配时间——部分计算题涉及多步、权重3-5分，务必保证足够时间。在计时条件下反复练习如2018年6月试卷2等真题，是最佳备考方式。

Remember, these calculations are not isolated – they often integrate concepts. A question might link a titration to a Kc determination or an enthalpy change. Building fluency in each type will equip you to handle any combination with confidence.

请记住，这些计算并非孤立——它们常交叉概念。一道题可能将滴定与 Kc 测定或焓变关联。对每一类型建立熟练度，将使你能自信地应对任何组合。

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A-Level Chemistry Calculation Questions in June 2018 Paper 2 | A-Level 化学：2018年6月试卷2计算题型精讲