📚 Year 10 Eduqas Chemistry: Unit Test Mock Paper Analysis | 10年级Eduqas化学：单元测试模拟卷解析
Are you preparing for the Year 10 Eduqas Chemistry end-of-unit test? This walkthrough covers a full mock paper, giving you detailed answers and key explanations for every question. Whether you are revising atomic structure, bonding, acids or electrolysis, this article will help you sharpen your exam technique and avoid common mistakes.
正在备考10年级Eduqas化学单元测试吗？这份模拟卷解析涵盖了一套完整的样题，为你提供每道题的参考答案和关键解释。无论你是在复习原子结构、化学键、酸还是电解，这篇文章都会帮你提升应试技巧，避开常见的丢分点。
1. Atomic Structure and Isotopes | 原子结构与同位素
Question: An atom contains 12 protons and 14 neutrons. (a) Calculate the mass number. (b) Identify the element and write its full atomic symbol, including the mass number and atomic number.
Answer and explanation: atomic number is 12, so the element is magnesium (Mg). Its symbol is written with the mass number as a superscript and the atomic number as a subscript on the left: ²⁶₁₂Mg. Always remember that the number of protons defines which element you have — 12 protons means magnesium, regardless of the neutron count.
题目：某原子含有12个质子和14个中子。(a) 计算它的质量数。(b) 指出该元素并写出它的完整原子符号，包括质量数和原子序数。
答案与解析：(a) 质量数 = 质子数 + 中子数 = 12 + 14 = 26。(b) 原子序数是12，因此该元素是镁（Mg）。其原子符号写作质量数在左上标、原子序数在左下标的形式：²⁶₁₂Mg。请务必记住，质子数决定了元素的种类——12个质子就是镁，与中子数无关。
2. Ionic Bonding in Sodium Chloride | 氯化钠中的离子键
Question: Use 'dot and cross' diagrams to show how the ions are formed when sodium (electronic configuration 2,8,1) reacts with chlorine (2,8,7). Clearly label the charges on the ions.
Answer and explanation: lorine has seven outer electrons; it gains one electron to complete its outer shell, forming a chloride ion with a 1− charge: Cl + e⁻ → Cl⁻. In a dot and cross diagram, the single electron transferred from sodium is drawn with crosses (or dots) and moves to the chlorine atom, which originally had seven electrons shown with dots (or crosses). The ions are then shown in square brackets with the charge written outside: [Na]⁺ and [Cl]⁻. The electrostatic attraction between these oppositely charged ions is the ionic bond.
题目：用“点叉图”表示钠（电子排布 2,8,1）与氯（2,8,7）反应时离子是如何形成的。请清晰地标出离子的电荷。
答案与解析：钠的最外层有1个电子。它失去这个电子以达到全满的最外层，变成带1+电荷的钠离子：Na → Na⁺ + e⁻。氯的最外层有7个电子，它得到1个电子以填满最外层，形成带1−电荷的氯离子：Cl + e⁻ → Cl⁻。在点叉图中，钠转移的那个电子用叉（或点）表示，并移动到氯原子上，而氯原本的7个电子用点（或叉）表示。离子写在方括号内，并把电荷标在括号外：[Na]⁺ 和 [Cl]⁻。这些带相反电荷的离子之间的静电吸引力就是离子键。
3. Covalent Bonding in Water | 水分子中的共价键
Question: Draw a dot and cross diagram for a molecule of water, H₂O. Use different symbols for the electrons from oxygen and from hydrogen, and explain how the covalent bonds are formed.
Answer and explanation: Oxygen has 6 valence electrons (electronic configuration 2,6), so it needs 2 more electrons to achieve a stable octet. Each hydrogen atom has 1 valence electron and needs 1 more to fill its shell (duplet). In the water molecule, oxygen shares one pair of electrons with each hydrogen atom. This gives two O–H single covalent bonds. The shared pairs are drawn as overlapping circles or simply as lines between the atoms. After bonding, oxygen has two bonding pairs and two lone pairs of electrons.
题目：画出水分子 H₂O 的点叉图。用不同的符号表示来自氧和氢的电子，并解释共价键是如何形成的。
答案与解析：氧有6个价电子（电子排布 2,6），因此它需要2个电子才能达到稳定的八隅体。每个氢原子有1个价电子，需要1个电子来填满它的壳层（双电子层）。在水分子中，氧与每个氢原子共用一对电子，从而形成两个 O–H 单共价键。共享电子对可以用重叠的圆圈或用原子之间的横线表示。成键后，氧有两对成键电子和两对孤对电子。
4. Balancing Chemical Equations | 配平化学方程式
Question: Write the word equation and the balanced symbol equation for the complete combustion of methane. Include state symbols where appropriate.
Answer and explanation: e 1 carbon, 4 hydrogen and 4 oxygen atoms; on the right we have 1 carbon in CO₂, 4 hydrogen in 2H₂O and 4 oxygen (2 from CO₂ and 2 from 2H₂O). The equation is balanced. Remember to use whole numbers as coefficients — never change the chemical formulas themselves.
题目：写出甲烷完全燃烧的文字方程式和配平的符号方程式。适当标注状态符号。
答案与解析：文字方程式为：甲烷 + 氧气 → 二氧化碳 + 水。配平的符号方程式：CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)。左边有1个碳、4个氢和4个氧原子；右边CO₂中有1个碳，2H₂O中有4个氢，氧原子共有4个（CO₂中的2个和2H₂O中的2个）。方程式已配平。务必使用整数作为系数，绝不能改变化学式本身。
5. Acids, Bases and pH | 酸、碱与pH
Question: Hydrochloric acid (HCl) is added to sodium hydroxide solution (NaOH) until the mixture is neutral. (a) Write the ionic equation for the neutralisation reaction. (b) Describe how the pH of the solution changes during the addition.
Answer and explanation: (aq) → H₂O(l). (b) Initially the sodium hydroxide solution is alkaline with a pH around 13– OH⁻) the pH becomes 7. If excess acid is added, the pH will fall below 7, becoming acidic.
题目：将盐酸（HCl）加入到氢氧化钠溶液（NaOH）中，直至混合物呈中性。(a) 写出该中和反应的离子方程式。(b) 描述在加入过程中溶液的pH如何变化。
答案与解析：(a) 旁观离子是 Na⁺ 和 Cl⁻，因此中和反应的离子方程式为：H⁺(aq) + OH⁻(aq) → H₂O(l)。(b) 最初氢氧化钠溶液呈碱性，pH 约在13–14之间。随着盐酸的加入，pH逐渐下降。在恰好完全中和时（H⁺ 与 OH⁻ 的物质的量相等），pH变为7。如果加入过量的酸，pH会降到7以下，溶液呈酸性。
6. Electrolysis of Molten Sodium Chloride | 电解熔融氯化钠
Question: Explain what happens during the electrolysis of molten sodium chloride. Include half‑equations for the reactions at each electrode and name the products formed.
Answer and explanation: um metal is produced at the cathode. At the positively charged anode, chloride ions lose electrons (oxidation): 2Cl⁻ → Cl₂(g) + 2e⁻. Chlorine gas forms at the anode. No water is present because the electrolyte is molten, so the products are simply sodium metal and chlorine gas.
题目：解释在电解熔融氯化钠的过程中会发生什么。写出每个电极反应的半方程式，并说出生成的产物。
答案与解析：熔融氯化钠中含有 Na⁺ 和 Cl⁻ 离子。在带负电荷的阴极，钠离子得到电子（还原反应）：Na⁺ + e⁻ → Na(l)。阴极生成金属钠。在带正电荷的阳极，氯离子失去电子（氧化反应）：2Cl⁻ → Cl₂(g) + 2e⁻。阳极生成氯气。由于电解质是熔融态，没有水存在，因此产物就是金属钠和氯气。
7. Diamond and Graphite: Giant Covalent Structures | 金刚石与石墨：巨型共价结构
Question: Both diamond and graphite are carbon allotropes. Compare their structures and explain why graphite can conduct electricity but diamond cannot.
Answer and explanation: In diamond, each carbon atom forms four strong covalent bonds in a tetrahedral arrangement, creating a rigid 3D giant covalent lattice. All valence electrons are localised in bonds, so there are no free charge carriers — diamond cannot conduct electricity. In graphite, each carbon atom bonds to three others in flat hexagonal layers. The fourth valence electron is delocalised between the layers. These delocalised electrons can move freely, allowing graphite to conduct electricity along the layers. Additionally, the weak forces between layers make graphite soft and slippery, while diamond is extremely hard.
题目：金刚石和石墨都是碳的同素异形体。比较它们的结构，并解释为什么石墨能导电而金刚石不能。
答案与解析：在金刚石中，每个碳原子以四面体方式形成四个牢固的共价键，构建出刚性的三维巨型共价网络。所有价电子都定域在键中，因此没有自由电荷载体——金刚石不能导电。在石墨中，每个碳原子与另外三个碳原子键合，形成扁平的六边形层状结构。第四个价电子在层间离域。这些离域电子可以自由移动，使得石墨能够沿层间导电。此外，层间的弱作用力使石墨柔软滑腻，而金刚石却极其坚硬。
8. Metallic Bonding and Conductivity | 金属键与导电性
Question: Explain, in terms of bonding, why metals are good conductors of electricity and why they are malleable.
Answer and explanation: Metals consist of a lattice of positive metal ions surrounded by a 'sea' of delocalised electrons. When a voltage is applied, these delocalised electrons can move through the structure, carrying charge — hence metals conduct electricity. The layers of positive ions can slide over each other while the delocalised electrons continue to hold them together. This allows metals to be bent or hammered into shape without breaking, which is why they are malleable. The metallic bond is not broken during this deformation because the electrons are free to move with the ions.
题目：从化学键的角度解释，为什么金属是电的良导体，以及它们为什么具有延展性。
答案与解析：金属由正金属离子晶格镶嵌在离域电子的“海洋”中构成。当施加电压时，这些离域电子可以在结构中移动，携带电荷——因此金属能导电。正离子层可以在彼此间滑动，而离域电子仍然将它们结合在一起。这使得金属能够被弯曲或锤打成各种形状而不断裂，因此金属具有延展性。在变形过程中金属键没有被破坏，因为电子能随离子一起移动。
9. Relative Atomic Mass Calculation | 相对原子质量计算
Question: Chlorine exists as two stable isotopes: chlorine‑35 (abundance 75%) and chlorine‑37 (abundance 25%). Calculate the relative atomic mass (Aᵣ) of chlorine, showing your working.
Answer and explanation: : Relative atomic mass is the weighted average mass of all isotopes. Aᵣ = (mass number of isotope 1 × % abundance + mass number of isotope 2 × % 100 = 35.5. This explains why the periodic table value for chlorine is not a whole number.
题目：氯有两种稳定同位素：氯‑35（丰度75%）和氯‑37（丰度25%）。计算氯的相对原子质量（Aᵣ），并展示计算过程。
答案与解析：相对原子质量是所有同位素的加权平均质量。Aᵣ = （同位素1的质量数 × 丰度% + 同位素2的质量数 × 丰度%）/ 100。因此 Aᵣ(Cl) = (35 × 75 + 37 × 25) / 100 = (2625 + 925) / 100 = 3550 / 100 = 35.5。这解释了为什么元素周期表中氯的数值不是整数。
10. Preparing a Pure Dry Salt: Copper Sulfate Crystals | 制备纯净干燥的盐：硫酸铜晶体
Question: Describe a method to prepare pure, dry crystals of copper(II) sulfate from insoluble copper(II) oxide and dilute sulfuric acid. Include the word equation and explain why filtration and crystallisation are necessary.
Answer and explanation: te sulfuric acid and stir until no more reacts. The excess solid is removed by filtration, leaving a blue copper(II) sulfate solution. The filtrate is then heated gently to evaporate some water, and then left to cool so that blue hydrated crystals form. Filtration separates the unreacted base, while crystallisation yields pure, dry crystals. The chemical equation is: CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l).
题目：描述一种从不溶于水的氧化铜和稀硫酸制取纯净干燥的硫酸铜晶体的方法。写出文字方程式，并解释为什么需要过滤和结晶。
答案与解析：反应方程式：氧化铜 + 硫酸 → 硫酸铜 + 水。在实验中，将过量的黑色氧化铜粉末加入到温热的稀硫酸中，搅拌至不再反应。多余固体通过过滤除去，得到蓝色的硫酸铜溶液。滤液再缓缓加热蒸发掉一部分水，然后冷却，蓝色水合晶体便会析出。过滤是为了分离未反应完的碱，而结晶则是为了获得纯净干燥的晶体。化学方程式为：CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)。
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