Secondary 4 Combined Science Chemistry Practice Paper 2

TuitionGoWhere Practice Paper - Combined Science Chemistry Secondary 4

Answer Key and Marking Scheme

Version: 2 Total Marks: 65

Section A: Structured Questions (20 marks)

1.

(a) pH range: 11–14 / strongly alkaline [1]

(b) Cation: Zn²⁺ / zinc ion [1]. Explanation: White precipitate forms with NaOH that dissolves in excess NaOH; white precipitate with ammonia that is insoluble in excess. These are characteristic reactions of zinc ions. [1]

(c) Formula: Zn(OH)₂ [1]

(d) Sodium salt: zinc sulfate / ZnSO₄ (accept any soluble zinc salt with a sodium counter-ion that would give a neutral or alkaline solution, e.g., zinc chloride) [1]

2.

(a) ZnO(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂O(l) [1 for correct formulae, 1 for correct state symbols; accept multiples]

(b) To ensure all the sulfuric acid reacts completely / to ensure the acid is fully neutralised. [1]

(c)(i) Unreacted / excess zinc oxide is removed. [1]

(c)(ii) To obtain a saturated solution so that crystals form on cooling / to concentrate the solution to the point of crystallisation. [1]

(c)(iii) Press between sheets of filter paper / leave in a warm place to dry / place in a desiccator. [1]

3.

(a) Iron / finely divided iron. [1]

(b) A lower temperature would give a higher yield of ammonia (equilibrium shifts right) [1], but the rate of reaction would be too slow. 450 °C is a compromise temperature that gives a reasonable rate while still producing an acceptable yield. [1]

(c) 2NH₃(g) + H₂SO₄(aq) → (NH₄)₂SO₄(aq) [1; accept state symbols]

(d) Molar mass of NH₄NO₃ = 14 + (4 × 1) + 14 + (3 × 16) = 80 g/mol [1]. Mass of nitrogen = 2 × 14 = 28 g. Percentage nitrogen = (28/80) × 100 = 35.0% [1].

4.

(a) Calcium chloride. [1]

(b) Curve should start at origin, rise more steeply than the original curve, and level off at the same final volume of CO₂. [1 for steeper gradient, 1 for same final volume; curve must be clearly labelled]

(c) Powdered marble chips have a larger surface area than marble chips of the same mass [1]. This means more particles are exposed, so the frequency of effective collisions between reactant particles increases, leading to a faster rate of reaction [1].

Section B: Data-Based and Application Questions (30 marks)

5.

(a) Oxide W has ionic bonding with a giant ionic lattice structure [1]. Evidence: High melting point (strong electrostatic forces between ions require large energy to overcome) [1]. Conducts electricity when molten (ions are free to move and carry charge). Reacts with water to form an alkaline solution (characteristic of soluble metal oxides / basic oxides) [1].

(b) Oxide X is likely carbon dioxide (CO₂) or sulfur dioxide (SO₂) [1]. Evidence: Low melting point/sublimes (simple molecular structure with weak intermolecular forces). Reacts with water to form an acidic solution (characteristic of non-metal oxides). Does not conduct electricity (no free ions or electrons) [1].

(c) Oxide Z is sodium chloride (NaCl) [1]. Its aqueous solution is neutral because it is a salt formed from a strong acid (HCl) and a strong base (NaOH). Neither ion hydrolyses significantly in water [1].

(d) In a giant covalent structure, all electrons are held in strong covalent bonds / there are no free ions or delocalised electrons to carry charge [1].

6.

(a) Titrations 2 and 4 should be used [1]. They are concordant (within 0.1 cm³ of each other). Titration 1 is a rough titration, and Titration 3 is anomalous (not concordant) [1].

(b) Average volume = (23.60 + 23.60) / 2 = 23.60 cm³ [1]

(c) Moles of H₂SO₄ = 0.100 × (23.60/1000) = 0.00236 mol [1]. From equation, 2 mol NaOH react with 1 mol H₂SO₄, so moles NaOH = 2 × 0.00236 = 0.00472 mol [1]. Concentration NaOH = 0.00472 / (25.0/1000) = 0.189 mol/dm³ (accept 0.1888 mol/dm³) [1].

(d) Molar mass NaOH = 23 + 16 + 1 = 40 g/mol. Concentration in g/dm³ = 0.189 × 40 = 7.56 g/dm³ (accept 7.55 g/dm³) [1].

(e) Yellow to orange / yellow to pink / yellow to red. [1]

7.

(a) Acidic. [1]

(b) Hydrochloric acid is a strong acid; it dissociates completely in water to produce a high concentration of H⁺ ions [1]. Ethanoic acid is a weak acid; it dissociates partially in water, producing a lower concentration of H⁺ ions at the same acid concentration. Lower [H⁺] means higher pH [1].

(c)(i) Calcium oxide (quicklime) / calcium hydroxide (slaked lime) / calcium carbonate (limestone). [1; accept any suitable base]

(c)(ii) The substance neutralises the excess acid in the soil [1]. The base/oxide/carbonate reacts with H⁺ ions in the soil, removing them and raising the pH [1]. Ionic equation (example using calcium carbonate): CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g) [1; accept other valid ionic equations depending on substance named in (c)(i)].

8.

(a) Substance A has ionic bonding with a giant ionic lattice structure [1]. It does not conduct when solid (ions fixed in lattice) but conducts when molten and in aqueous solution (ions are free to move) [1].

(b) Substance B has metallic bonding with a giant metallic lattice structure [1]. It conducts in both solid and molten states because of the presence of delocalised electrons.

(c) Substance C has covalent bonding with a simple molecular structure [1]. It does not conduct in any state (no ions or free electrons) and dissolves without forming ions [1].

(d) In graphite, each carbon atom is covalently bonded to three other carbon atoms, forming layers of hexagonal rings [1]. Each carbon atom has one delocalised electron that is free to move between the layers, carrying electrical charge [1].

9.

(a) CuO(s) + 2HNO₃(aq) → Cu(NO₃)₂(aq) + H₂O(l) [1 for correct formulae, 1 for correct balancing; state symbols not required but accept if correct]

(b) The residue is copper metal [1]. Copper metal does not react with dilute nitric acid under these conditions (or reacts very slowly), while copper(II) oxide reacts to form soluble copper(II) nitrate. Copper metal remains as an insoluble solid [1].

(c) Copper(II) nitrate. [1]

(d) Blue. [1]

10.

(a) Sulfuric acid is diprotic (produces 2 H⁺ ions per molecule), while hydrochloric acid is monoprotic (produces 1 H⁺ ion per molecule) [1]. At the same concentration, sulfuric acid has twice the concentration of H⁺ ions, so the frequency of effective collisions is higher, and the reaction is faster [1].

(b) Ethanoic acid is a weak acid; it dissociates only partially in water, producing a much lower concentration of H⁺ ions than hydrochloric acid at the same acid concentration [1]. The lower [H⁺] means a lower frequency of effective collisions, so the reaction is slower [1].

(c) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) [1]

(d) Moles of Mg = 0.50 / 24 = 0.0208 mol [1]. From equation, 1 mol Mg produces 1 mol H₂, so moles H₂ = 0.0208 mol [1]. Volume of H₂ = 0.0208 × 24 = 0.499 dm³ ≈ 0.50 dm³ (or 500 cm³) [1].

Section C: Free-Response Questions (15 marks)

11.

(a) Precipitation / double decomposition. [1]

(b) Any soluble lead(II) salt (e.g., lead(II) nitrate, Pb(NO₃)₂) [1] and any soluble sulfate (e.g., sodium sulfate, Na₂SO₄, or dilute sulfuric acid) [1].

(c) Procedure (5 marks):

• Mix aqueous solutions of the two starting materials in a beaker [1].
• A white precipitate of lead(II) sulfate forms immediately [1].
• Filter the mixture to separate the precipitate [1].
• Wash the residue with distilled water to remove any soluble impurities [1].
• Dry the residue between sheets of filter paper or in a warm oven [1].

(d) Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s) [1 for correct ionic equation, 1 for correct state symbols]

(e) Percentage yield = (actual yield / theoretical yield) × 100 = (2.85 / 3.20) × 100 = 89.1% (accept 89.0% or 89%) [2; 1 for formula, 1 for correct answer]

(f) Some precipitate was lost during filtration / some precipitate remained in the beaker / some precipitate passed through the filter paper / transfer losses [1; accept any reasonable answer].

(g) Wear safety goggles / gloves (lead compounds are toxic) / wash hands after experiment [1; accept any reasonable safety precaution].

(h) Add dilute hydrochloric acid followed by barium chloride solution. A white precipitate of barium sulfate confirms the presence of sulfate ions [1].

12.

(a) Table (4 marks):

(b) Sodium chloride has a giant ionic lattice structure with strong electrostatic forces of attraction between oppositely charged Na⁺ and Cl⁻ ions [1]. These forces extend throughout the lattice in all directions [1]. A large amount of energy is required to overcome these strong forces, resulting in a high melting point [1].

(c) Diamond has a giant covalent structure in which each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement [1]. These strong covalent bonds extend throughout the entire structure in three dimensions [1]. A very large amount of energy is required to break these bonds, giving diamond a very high melting point and extreme hardness [1].

(d) Copper has a giant metallic lattice structure with a sea of delocalised electrons surrounding positive metal ions [1]. The delocalised electrons are free to move throughout the structure, allowing copper to conduct electricity [1]. When a force is applied, the layers of metal ions can slide over each other without breaking the metallic bonds (the delocalised electrons continue to hold the ions together), making copper malleable [1].

(e) Carbon dioxide has a simple molecular structure consisting of discrete CO₂ molecules [1]. The intermolecular forces between CO₂ molecules are weak van der Waals' forces. Very little energy is required to overcome these weak forces, so carbon dioxide is a gas at room temperature [1].

END OF ANSWER KEY