📚 Year 10 Eduqas Chemistry: Unit Test Mock Paper Analysis | Year 10 Eduqas 化学：单元测试模拟卷解析
Mock paper analysis is one of the most effective ways to prepare for your Eduqas Chemistry unit test. By breaking down typical questions, we can pinpoint the key concepts, data interpretation skills, and common mistakes that examiners look for. This walkthrough mirrors the style of a Year 10 unit assessment, covering purity, atomic structure, bonding, quantitative chemistry and more. Use it to test your understanding and refine your exam technique.
模拟卷解析是备考 Eduqas 化学单元测试最高效的方法之一。通过拆解典型题目，我们能够精确锁定考官关注的核心概念、数据判读技能以及常见失分点。这篇解析模拟了一份 Year 10 单元评估的风格，涵盖物质纯度、原子结构、化学键、定量化学等重要模块。请用来自测理解、打磨应试技巧。
1. Identifying Pure Substances and Mixtures | 鉴别纯物质与混合物
A common question provides melting and boiling point data for two samples. Sample A melts sharply at 0 °C and boils at 100 °C, while Sample B melts over a range 95–103 °C. The sharp values indicate a pure substance, in this case pure water. A mixture, such as salt solution, melts and boils over a temperature range – the boiling point rises and the melting point is lowered due to the dissolved impurity.
一个常见题型会给出两个样品的熔点和沸点数据。样品 A 在 0 °C 尖锐熔化、在 100 °C 沸腾；样品 B 则在 95–103 °C 区间内熔化。锋利的数值表明是纯物质，这里是纯水。混合物（如盐水）会在一个温度区间内熔化与沸腾——由于溶解了杂质，沸点升高而熔点降低。
You may also be asked to interpret heating or cooling curves. A pure substance shows a horizontal plateau at its melting or boiling point because all the energy goes into changing state. A mixture lacks a perfectly flat plateau; instead the curve slopes gently. Remember: impurities broaden phase-change ranges.
你还可能被要求判读加热或冷却曲线。纯物质在熔点或沸点处会出现水平平台，因为所有能量都用于物态变化。混合物没有完全平坦的平台，曲线会轻微倾斜。牢记：杂质会让相变区间变宽。
2. Separation Techniques: Distillation and Chromatography | 分离技术：蒸馏与色谱法
In the mock, you might need to design a separation sequence for a mixture of sand, salt and water. First, filtration removes the insoluble sand. The filtrate is then heated by simple distillation: water evaporates, condenses and is collected as pure distillate, leaving salt crystals behind. For a mixture of coloured inks, paper chromatography is the method: spots rise at different rates depending on their solubility in the solvent and their attraction to the paper.
在模拟卷中，你可能需要为一个沙、盐和水的混合物设计分离流程。首先，过滤除去不溶的沙子。然后对滤液进行简单蒸馏：水蒸发、冷凝并被收集为纯蒸馏液，盐晶体则留在蒸馏瓶内。对于混合色素墨水，纸色谱是首选方法：不同色素因其在溶剂中的溶解度和对纸的吸附力差异而上升速率不同。
When interpreting a chromatogram, you may calculate the Rf value. The formula is given in the data sheet but must be applied:
当判读色谱图时，你可能需要计算 Rf 值。数据表中会提供公式，但必须会使用：
Rf = dspot / dsolvent
Measure the distance from the baseline to the centre of the spot and to the solvent front. A pure substance gives one spot; a mixture gives multiple. Rf values are dimensionless and always <1. This technique regularly appears in Year 10 unit tests.
测量从基线到斑点中心以及到溶剂前沿的距离。纯物质只呈现一个斑点；混合物则出现多个斑点。Rf 值无量纲且始终小于 1。这项技术经常出现在 Year 10 单元测试中。
3. Atomic Models through History | 历史上的原子模型
The mock typically traces how the atomic model evolved after new experimental evidence. Dalton imagined atoms as tiny, indivisible spheres. Thomson's discovery of the electron led to the 'plum pudding' model – a positive mass with embedded negative electrons. Rutherford's gold foil experiment showed that most alpha particles passed straight through, but a few were deflected, revealing a dense, positively charged nucleus. Bohr then refined the model by placing electrons in fixed energy levels or shells.
模拟卷通常追溯原子模型在实验证据推动下的演变。道尔顿设想原子是微小的、不可分割的球体。汤姆逊发现电子后提出了“葡萄干布丁”模型——正电荷团块中嵌着负电子。卢瑟福的金箔实验显示绝大多数 α 粒子径直穿过，但极少数被反弹，揭示了致密带正电的原子核。随后玻尔完善了模型，将电子安置在固定的能级或壳层上。
Chadwick's later discovery of the neutron explained why the mass number was greater than the proton number without altering the charge. Exam questions often link a diagram or a description to the correct scientist and ask you to explain why the model was changed. For instance: 'Beta particles were observed to be bent by an electric field towards the positive plate, showing they are negative.' This evidence led to Thomson’s plum pudding.
后来查德威克发现中子，解释了为什么质量数大于质子数而又不改变电荷。考题经常将示意图或描述与对应科学家关联，并要求解释模型为何被修正。例如：“观察到 β 粒子在电场中偏向正极板，表明它们带负电。”这一证据促成了汤姆逊的葡萄干布丁模型。
4. Subatomic Particles and Electronic Configuration | 亚原子粒子与电子排布
On the mock paper, you will be given a nuclide such as ²³Na (sodium) with atomic number 11. You must state that it contains 11 protons, 11 electrons and 12 neutrons (mass number 23 − 11). The relative charges are: proton +1, neutron 0, electron −1. All atoms are neutral because the numbers of protons and electrons are equal.
在模拟卷上，你会面对一个核素如 ²³Na（钠），原子序数为 11。你必须答出它含有 11 个质子、11 个电子和 12 个中子（质量数 23 − 11）。相对电荷分别为：质子 +1、中子 0、电子 −1。由于质子与电子数相等，所有原子都呈电中性。
The electronic configuration is written as 2,8,1 for sodium. You may be asked to draw a shell diagram or to match configurations to a Group. For example, an atom with configuration 2,8,7 belongs to Group 7 (the halogens). The number of electrons in the outermost shell determines the chemical properties and the group number (for Groups 1–2 and 13–18).
钠的电子排布写作 2,8,1。你可能被要求画出电子层示意图或将排布与族匹配。例如，排布为 2,8,7 的原子属于第 7 族（卤素）。最外层电子数决定了化学性质及族序数（对第 1-2 族和第 13-18 族而言）。
5. Group 1 and Group 7 Trends | 第 1 族和第 7 族元素趋势
ivity increases because the outer electron is further from the nucleus, with more shielding, so it is lost more easily. Rubidium would react violently, producing a hydroxide and hydrogen gas: 2Rb + 2H₂O → 2RbOH + H₂.
典型问题要求你根据锂与水平缓地嘶嘶冒气、钾会燃烧的事实，预测铷与水的反应。沿第 1 族向下，反应性增强，因为外层电子离核更远、屏蔽更多，更易失去。铷会剧烈反应，生成氢氧化物和氢气：2Rb + 2H₂O → 2RbOH + H₂。
use the atom gets larger and the outer shell attraction for an extra electron weakens. A displacement reaction, e.g. Cl₂ + 2KBr → 2KCl + Br₂, shows chlorine is more reactive than bromine. You must explain the colour change and write a balanced ionic equation.
对于第 7 族，卤素在室温下的状态从气体（F₂, Cl₂）到液体（Br₂）再到固体（I₂）变化。反应性沿族向下减弱，因为原子变大，对外来电子的吸引力减弱。置换反应如 Cl₂ + 2KBr → 2KCl + Br₂ 表明氯比溴更活泼。你必须解释颜色变化并写出配平的离子方程式。
6. Formation of Ionic Compounds | 离子化合物的形成
A dot-and-cross diagram question may ask you to show the bonding in sodium chloride. Sodium (2,8,1) loses its outer electron to become Na⁺ with a 2,8 stable configuration; chlorine (2,8,7) gains that electron to become Cl⁻ with a 2,8,8 stable octet. The oppositely charged ions attract strongly via electrostatic forces, forming a giant ionic lattice. You must only show the outer shell electrons, using crosses and dots for different atoms.
点叉图题可能要求你展示氯化钠中的成键。钠 (2,8,1) 失去外层电子变成 Na⁺，达到 2,8 稳定结构；氯 (2,8,7) 获得该电子变成 Cl⁻，达到 2,8,8 稳定八隅体。带相反电荷的离子通过静电引力强烈吸引，形成巨型离子晶格。你只需要画出最外层电子，用叉和点区分不同原子。
Ionic compounds have high melting and boiling points because a lot of energy is needed to break the strong electrostatic bonds throughout the lattice. They conduct electricity when molten or dissolved in water, as the ions become free to move, but not when solid. A typical 3-mark question links properties to structure.
离子化合物具有高熔点和沸点，因为需要大量能量来打破遍布整个晶格的强静电键。它们在熔融或溶于水时能导电，因为离子可以自由移动，但固态时不导电。典型的 3 分题就是把性质与结构联系到一起。
7. Covalent Bonding and Simple Molecules | 共价键与简单分子
Covalent bonding occurs between non-metal atoms, including hydrogen. In water, H₂O, each hydrogen shares one electron with oxygen to achieve a full outer shell. The molecule has a bent shape due to two lone pairs on oxygen. Dot-and-cross diagrams must show the shared pairs clearly, ideally with circles overlapping to indicate the bond.
共价键发生在非金属原子（包括氢）之间。在水分子 H₂O 中，每个氢原子与氧原子共用一对电子以达到满壳层。由于氧上有两对孤对电子，分子呈角形。点叉图必须清晰显示共用电子对，最好用重叠圆圈表示共价键。
Simple molecular substances like H₂O, CO₂, and CH₄ have low melting and boiling points because only weak intermolecular forces need to be overcome. They do not conduct electricity – there are no free ions or delocalised electrons. A common question asks you to compare the boiling points of ionic and simple molecular substances and account for the difference in terms of particles and forces.
像 H₂O、CO₂ 和 CH₄ 这样的简单分子物质具有较低的熔点和沸点，因为只需克服微弱的分子间力。它们不能导电——没有自由离子或离域电子。常见考题要求你比较离子化合物与简单分子物质的沸点，并从粒子和作用力的角度说明差异。
8. Balancing Chemical Equations | 配平化学方程式
a symbol equation that is either incorrectly balanced or completely missing. For example, '_Fe₂O₃ + _CO → _Fe + _CO₂' s, then non-metals, and oxygen last. Never change the subscript numbers inside the formula – only add big coefficients in front.
模拟卷常包含一个配平不当或完全缺失的符号方程式。例如“_Fe₂O₃ + _CO → _Fe + _CO₂”必须配平为 Fe₂O₃ + 3CO → 2Fe + 3CO₂。先配平金属原子，然后配非金属，最后配氧。切勿改动化学式内部的下标数字——只能在前面加上大的化学计量数。
+ CO₂(g), missing state symbols lose marks. Make sure you remember that aqueous solutions are written as (aq), and liquid water as (l). Balancing and state symbols together can be worth up to 4 marks.
状态符号也常被要求写出：(s)、(l)、(g)、(aq)。在反应 2HCl(aq) + MgCO₃(s) → MgCl₂(aq) + H₂O(l) + CO₂(g) 中，遗漏状态符号就会失分。务必记住，水溶液写 (aq)，液态水写 (l)。配平与状态符号合在一起最高可达 4 分。
9. Calculating Relative Formula Mass (Mr) | 计算相对分子质量 (Mr)
A straightforward calculation uses the relative atomic masses from the Periodic Table. For calcium carbonate, CaCO₃, Ar: Ca = 40, C = 12, O = 16. So Mr = 40 + 12 + (3 × 16) = 100. The unit is either omitted or given as 'no unit' because it is relative. For hydrated crystals such as CuSO₄·5H₂O, calculate Mr = 63.5 + 32 + (4 × 16) + 5 × (2 + 16) = 249.5.
一个直接的计算是利用周期表中的相对原子质量。对于碳酸钙 CaCO₃，Ar：Ca = 40, C = 12, O = 16。因此 Mr = 40 + 12 + (3 × 16) = 100。单位一般不写或注明“无单位”，因为它是相对值。对于水合晶体如 CuSO₄·5H₂O，计算 Mr = 63.5 + 32 + (4 × 16) + 5 × (2 + 16) = 249.5。
Mr (CaCO₃) = 40 + 12 + (3 × 16) = 100
Be careful with brackets in the formula: Mg(OH)₂ means one magnesium, two oxygens and two hydrogens. Mr Always show your working so that you can earn method marks even if the final answer slips.
注意化学式中的括号：Mg(OH)₂ 表示一个镁、两个氧和两个氢。Mr = 24.3 + 2 × (16 + 1) = 58.3。始终展示计算过程，这样即使最终答案出错，仍能拿到步骤分。
10. Mass Calculations in Reactions | 反应中的质量计算
Stoichiometry questions combine Mr with the mole concept. A typical problem: 'hat mass of magnesium oxide is produced when 4.8 g of magnesium burns completely in oxygen?' Equation: 2Mg + O₂ → 2MgO. Ar of Mg = 24, so Mr l. Mole ratio Mg : MgO = 2 : 2, i.e. 1 : 1, so 0.20 mol MgO forms. Mass of MgO = moles × Mr = 0.20 × 40 = 8.0 g.
化学计量计算题将 Mr 与摩尔概念结合。典型问题：“4.8 g 镁完全在氧气中燃烧能生成多少质量的氧化镁？”方程式：2Mg + O₂ → 2MgO。Mg 的 Ar = 24，因此 Mr(MgO) = 24 + 16 = 40。Mg 的物质的量 = 质量 / Ar = 4.8 / 24 = 0.20 mol。摩尔比 Mg : MgO = 2 : 2，即 1 : 1，所以生成 0.20 mol MgO。MgO 的质量 = 物质的量 × Mr = 0.20 × 40 = 8.0 g。
occupy 0.20 × 24 = 4.8 dm³. Always convert masses to moles first, use the balanced equation to find the mole ratio, then convert back to mass or volume. These calculation chains can appear as structured 6-mark questions.
若题目要求计算室温常压下的气体体积，记住 1 mol 任何气体占据 24 dm³。因此 0.20 mol CO₂ 会占据 0.20 × 24 = 4.8 dm³。始终先将质量转化为物质的量，利用配平方程式求摩尔比，再转换回质量或体积。这种计算链可能以结构化 6 分题形式出现。
One more high-value task is determining the limiting reactant. Calculate the moles of each reactant, compare the required ratio, and identify the one that runs out first – the amount of product is based on the limiting reactant. Show your comparison in a table for clarity.
另一个高分任务是判断限量反应物。计算各反应物的物质的量，比较所需配比，找出先耗尽的那一个——产物的量由限量反应物决定。为了清晰，可用表格展示比较过程。
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