📚 Answers to Test Yourself Questions: IB Chemistry Core Concepts | IB化学核心概念自测题答案详解
This article provides detailed worked answers to typical 'test yourself' questions that appear in IB Diploma Chemistry (Cambridge) textbooks. The questions span the core topics of Stoichiometric Relationships, Atomic Structure, Energetics, and Kinetics. Each answer is followed by a step-by-step explanation in both English and Chinese to reinforce key concepts and problem-solving techniques.
本文为IB文凭化学（剑桥版）教材中典型的“自我测试”题提供详细的解题答案。题目涵盖化学计量关系、原子结构、能量学和动力学核心主题。每个答案均附有中英双语的分步解析，以强化关键概念和解题技巧。
1. Mole Concept and Avogadro's Number | 摩尔概念与阿伏伽德罗常数
Test Yourself Question: Calculate the number of atoms present in 0.5 mol of helium gas. (He is monatomic.)
自测题：计算0.5 mol氦气中所含的原子数。（氦是单原子分子。）
Answer: Number of atoms = 0.5 mol × 6.02 × 10²³ atoms mol⁻¹ = 3.01 × 10²³ atoms.
答案：原子数 = 0.5 mol × 6.02 × 10²³ 个原子 mol⁻¹ = 3.01 × 10²³ 个原子。
Explanation: One mole of any substance contains Avogadro's number (6.02 × 10²³) of specified particles. For a monatomic gas like helium, 'particles' means individual atoms. Multiply the amount in moles by Avogadro's constant to obtain the number of atoms. Always write the unit 'mol⁻¹' to indicate 'per mole'.
解析：任何物质的一摩尔都含有阿伏伽德罗常数（6.02 × 10²³）个指定的粒子。对于氦这样的单原子气体，“粒子”就是指单个原子。将摩尔数乘以阿伏伽德罗常数即可得到原子数。务必写出单位“mol⁻¹”以表示“每摩尔”。
2. Empirical and Molecular Formulae | 经验式与分子式
Test Yourself Question: A compound is found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.
自测题：某化合物经测定含有40.0%碳、6.7%氢和53.3%氧（均为质量分数）。确定其实验式。
Answer: Empirical formula = CH₂O.
答案：实验式 = CH₂O。
Explanation: C: 3.33/3.33 = 1; H: 6.63/3.33 ≈ 2; O: 3.33/3.33 = 1. Hence the empirical formula is CH₂O.
解析：假设样品为100 g，则质量分别为40.0 g C、6.7 g H、53.3 g O。用各质量除以相对原子质量（Aᵣ）得到摩尔数：C: 40.0 / 12.01 = 3.33 mol；H: 6.7 / 1.01 = 6.63 mol；O: 53.3 / 16.00 = 3.33 mol。将所有摩尔值除以最小值（3.33），得到最简整数比：C: 3.33/3.33 = 1；H: 6.63/3.33 ≈ 2；O: 3.33/3.33 = 1。因此实验式为CH₂O。
3. Reacting Masses and Limiting Reactants | 反应质量与限量反应物
Test Yourself Question: g of hydrogen gas reacts with 32.0 g of oxygen gas, determine which reactant is limiting.
自测题：2H₂(g) + O₂(g) → 2H₂O(l)。如果4.0 g氢气与32.0 g氧气反应，确定哪种反应物是限量反应物。
Answer: Hydrogen (H₂) is the limiting reactant.
答案：氢气（H₂）是限量反应物。
Explanation: alanced equation shows a 2:1 mole ratio of H₂ to O₂. The available H₂ (1.98 mol) is slightly less than the 2.00 mol required to react completely with 1.00 mol O₂, so H₂ runs out first and is the limiting reactant. Consequently, O₂ is in excess.
解析：首先将质量换算为摩尔数：n(H₂) = 4.0 g / 2.02 g mol⁻¹ ≈ 1.98 mol；n(O₂) = 32.0 g / 32.00 g mol⁻¹ = 1.00 mol。配平方程式显示H₂与O₂的摩尔比为2:1。可用的H₂（1.98 mol）略少于与1.00 mol O₂完全反应所需的2.00 mol，因此H₂先耗尽，是限量反应物。O₂则过量。
4. Molar Volume of a Gas at STP | 气体在标准状况下的摩尔体积
Test Yourself Question: What volume (in dm³) does 0.250 mol of nitrogen gas (N₂) occupy at STP (273 K, 100 kPa)? Use the molar volume 22.7 dm³ mol⁻¹.
自测题：0.250 mol氮气（N₂）在标准状况（273 K, 100 kPa）下占有多少体积（dm³）？使用摩尔体积22.7 dm³ mol⁻¹。
Answer: Volume = 0.250 mol × 22.7 dm³ mol⁻¹ = 5.68 dm³.
答案：体积 = 0.250 mol × 22.7 dm³ mol⁻¹ = 5.68 dm³。
Explanation: At standard temperature and pressure (STP: 0 °C, 100 kPa), one mole of an ideal gas occupies 22.7 dm³. The volume of a gas is directly proportional to its amount in moles when temperature and pressure are constant (V ∝ n). Simply multiply the number of moles by the molar volume to find the volume.
解析：在标准温度和压力（STP: 0 °C, 100 kPa）下，一摩尔理想气体占据22.7 dm³。在温度和压力恒定时，气体的体积与摩尔数成正比（V ∝ n）。只需将摩尔数乘以摩尔体积即可得到体积。
5. Concentration of Solutions | 溶液浓度
Test Yourself Question: You dissolve 5.85 g of sodium chloride (NaCl) in water to make 250.0 cm³ of solution. Calculate its concentration in mol dm⁻³.
自测题：你将5.85 g氯化钠（NaCl）溶于水，配制成250.0 cm³溶液。计算其浓度（单位为mol dm⁻³）。
Answer: Concentration ≈ 0.400 mol dm⁻³.
答案：浓度 ≈ 0.400 mol dm⁻³。
Explanation: entration c = n / V = 0.1001 mol / 0.2500 dm³ = 0.400 mol dm⁻³ (to 3 significant figures).
解析：NaCl的摩尔质量 = 22.99 + 35.45 = 58.44 g mol⁻¹。NaCl的摩尔数 = 质量 / M = 5.85 g / 58.44 g mol⁻¹ = 0.1001 mol。将溶液体积换算为dm³：250.0 cm³ = 0.2500 dm³。然后，浓度c = n / V = 0.1001 mol / 0.2500 dm³ = 0.400 mol dm⁻³（保留三位有效数字）。
6. Acid-Base Titration Calculations | 酸碱滴定计算
Test Yourself Question: In a titration, 25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid (HCl) is neutralised by 20.0 cm³ of sodium hydroxide (NaOH) solution. Find the concentration of the NaOH solution.
自测题：在一次滴定中，25.0 cm³ 0.100 mol dm⁻³的盐酸（HCl）被20.0 cm³氢氧化钠（NaOH）溶液中和。求NaOH溶液的浓度。
Answer: c(NaOH) = 0.125 mol dm⁻³.
答案：c(NaOH) = 0.125 mol dm⁻³。
Explanation: 50 mol as well. Concentration of NaOH = n / V = 0.00250 mol / 0.0200 dm³ = 0.125 mol dm⁻³.
解析：写出配平方程式：HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)。反应物以1:1的摩尔比反应。所用HCl的摩尔数 = c × V = 0.100 mol dm⁻³ × 0.0250 dm³ = 0.00250 mol。因此，NaOH的摩尔数也为0.00250 mol。NaOH的浓度 = n / V = 0.00250 mol / 0.0200 dm³ = 0.125 mol dm⁻³。
7. Redox Titration Calculations | 氧化还原滴定计算
Test Yourself Question: centration of Fe²⁺ ions. The ionic equation is: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
自测题：25.0 cm³硫酸亚铁（Fe²⁺）溶液在酸性介质中完全氧化需要18.0 cm³ 0.0200 mol dm⁻³高锰酸钾（KMnO₄）。计算Fe²⁺离子的浓度。离子方程式为：MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O。
Answer: c(Fe²⁺) = 0.0720 mol dm⁻³.
答案：c(Fe²⁺) = 0.0720 mol dm⁻³。
Explanation: solution is 0.0250 dm³, so concentration = 1.80 × 10⁻³ mol / 0.0250 dm³ = 0.0720 mol dm⁻³.
解析：首先求出加入的MnO₄⁻摩尔数：n = 0.0200 mol dm⁻³ × 0.0180 dm³ = 3.60 × 10⁻⁴ mol。方程式显示1 mol MnO₄⁻与5 mol Fe²⁺反应。因此，Fe²⁺的摩尔数 = 5 × (3.60 × 10⁻⁴) = 1.80 × 10⁻³ mol。Fe²⁺溶液的体积为0.0250 dm³，因此浓度 = 1.80 × 10⁻³ mol / 0.0250 dm³ = 0.0720 mol dm⁻³。
8. Electron Configuration and Anomalies | 电子构型与反常情况
Test Yourself Question: r atom (Cu, Z = 29) and explain why it is considered an exception to the Aufbau principle.
自测题：写出铜原子（Cu，原子序数29）的完整电子构型，并解释为什么它被认为是构造原理的一个例外。
Answer: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ (or [Ar] 3d¹⁰ 4s¹). It is an exception because a fully filled 3d subshell (3d¹⁰) and a half-filled 4s subshell (4s¹) provide extra stability compared to the expected [Ar] 3d⁹ 4s².
答案：1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹（或[Ar] 3d¹⁰ 4s¹）。它之所以是一个例外，是因为全满的3d亚层（3d¹⁰）和半满的4s亚层（4s¹）相比预期的[Ar] 3d⁹ 4s²具有额外的稳定性。
Explanation: According to the Aufbau principle, electrons fill orbitals in order of increasing energy, predicting 3d⁹ 4s² for copper. However, the actual configuration is 3d¹⁰ 4s¹. This occurs because the 3d subshell is now completely filled, creating a symmetrical, lower-energy arrangement. The energy gained from a full d¹⁰ configuration outweighs the slight promotion energy required to move one 4s electron. Similar anomalies occur in chromium (Cr, [Ar] 3d⁵ 4s¹) where a half-filled d-subshell is favoured.
解析：根据构造原理，电子按照能量递增的顺序填充轨道，因此预期铜的构型为3d⁹ 4s²。但实际构型为3d¹⁰ 4s¹。这是因为3d亚层现在完全填满，形成了对称的、能量更低的排布。由全满d¹⁰构型获得的能量超过了将一个4s电子移至3d所需的微小激发能。类似的反常也出现在铬（Cr, [Ar] 3d⁵ 4s¹）中，此时半满的d亚层更为有利。
9. Enthalpy Changes and Bond Enthalpies | 焓变与键焓
Test Yourself Question: stimate the standard enthalpy change (ΔH°) for the reaction: H₂(g) + Cl₂(g) → 2HCl(g). E(H–H) = 436 kJ mol⁻¹, E(Cl–Cl) = 242 kJ mol⁻¹, E(H–Cl) = 431 kJ mol⁻¹.
自测题：使用下列平均键焓来估算反应H₂(g) + Cl₂(g) → 2HCl(g)的标准焓变（ΔH°）。E(H–H) = 436 kJ mol⁻¹，E(Cl–Cl) = 242 kJ mol⁻¹，E(H–Cl) = 431 kJ mol⁻¹。
Answer: ΔH° ≈ –184 kJ mol⁻¹.
答案：ΔH° ≈ –184 kJ mol⁻¹。
Explanation: alpy change for a reaction can be estimated using: ΔH = Σ(bond enthalpies of bonds broken) – Σ(bond enthalpies of bonds formed). Bonds broken: one H–H (436) and one Cl–Cl (242), total = +678 kJ (energy absorbed). Bonds formed: two H–Cl bonds (2 × 431) = –862 kJ (energy released). ΔH = 678 – 862 = –184 kJ mol⁻¹. The negative sign indicates the reaction is exothermic.
解析：反应的焓变可以使用下式估算：ΔH = Σ（断裂键的键焓） – Σ（形成键的键焓）。断裂的键：一个H–H（436）和一个Cl–Cl（242），总和 = +678 kJ（吸热）。形成的键：两个H–Cl键（2 × 431）= –862 kJ（放热）。ΔH = 678 – 862 = –184 kJ mol⁻¹。负号表示该反应为放热反应。
10. Reaction Rate and the Arrhenius Equation | 反应速率与阿伦尼乌斯方程
Test Yourself Question: nergy (Eₐ) of this reaction. Use the two-point Arrhenius equation: ln(k₂/k₁) = (Eₐ/R)(1/T₁ – 1/T₂), where R = 8.31 J K⁻¹ mol⁻¹.
自测题：某化学反应的速率常数在温度从298 K升高到308 K时增大了一倍。计算该反应的活化能（Eₐ）。使用两点阿伦尼乌斯方程：ln(k₂/k₁) = (Eₐ/R)(1/T₁ – 1/T₂)，其中R = 8.31 J K⁻¹ mol⁻¹。
Answer: Eₐ ≈ 52.9 kJ mol⁻¹ (or 5.29 × 10⁴ J mol⁻¹).
答案：Eₐ ≈ 52.9 kJ mol⁻¹（或5.29 × 10⁴ J mol
Published by TutorHao | IB Chemistry Revision Series | aleveler.com
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