Compassvale Sec 4Exp Prelim Chem P2 2024 MS

COMPASSVALE SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY FOUR EXPRESS CHEMISTRY 6092 Page 1 of 8 Mark Scheme Paper 2 Theory Section A 1 (a) (i) CH4 / CO2 [1] (ii) CO2 [1] (iii) Ar [1] (iv) Cl2 [1] (v) Al3+ [1] (vi) O2‒ [1] (b) acidic oxide: any non-metal oxides (except H2O, NO, CO) It reacts with alkalis to form salt and water. / It dissolves in water to form acids. neutral oxide: H2O, NO, CO It does not react with acids or alkalis. [1] [1] 2 (a) NH3 + HCl ⇌ NH4Cl [1] (b) Universal Indicator paper colour pH 1 blue / indigo 11 / 12 2 red 1 / 2 / 3 [1] [1] (c) • Ammonia (Mr = 17) has a lower relative molecular mass than hydrogen chloride (Mr = 36.5) • hence it would diffuse faster • and the white powder forms closer to the cotton wool soaked in concentrated hydrochloric acid. [1] [1] [1] 3 (a) • similarity: good conductors of electricity / good conductors of heat / malleable • difference: sodium has a low melting point while niobium has a high melting point / sodium has a low density while niobium has a high density / sodium is soft and can be easily cut with a knife while niobium is hard / sodium forms compounds that are white in colour while niobium forms coloured compounds / sodium has a fixed oxidation state while niobium has a variable oxidation state [1] [1] (b) (i) Niobium is a transition metal hence it should react with chlorine to form an ionic compound with a giant ionic lattice structure however, it formed a covalent molecule with a simple molecular structure instead. [1] (ii) NbCl5 [1]

COMPASSVALE SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY FOUR EXPRESS CHEMISTRY 6092 Page 2 of 8 (c) Sodium chloride: • Sodium chloride is an ionic compound with a giant ionic lattice structure. • A lot of energy is required to overcome strong electrostatic forces of attraction between oppositely charged sodium ions and chloride ions. • Hence, sodium chloride has a high boiling point. Niobium chloride: • Niobium(V) chloride is a covalent compound with a simple molecular structure. • Little energy is required to overcome weak intermolecular forces of attraction between niobium(V) chloride molecules. • Hence, niobium(V) chloride has a low boiling point. [1] - correct structure and bonding [1] – correct amount of energy and description of bonding [1] – correct comparison of boiling point [3] 4 (a) (i) At ‒10°C, chlorine exists as a gas. • Chlorine molecules are very far apart in a random arrangement • and they move freely / in all directions at high speeds. [1] [1] (ii) • The relative molecular mass of chlorine is 71 whereas the relative molecular mass of bromine is 160 . The relative molecular mass of bromine is nearly double the relative molecular mass of chlorine. However, the density of bromine is nearly a hundred times more than the density of chlorine. • The volume of the two substances needs to also be considered to account for the differences in densities. [1] [1] (b) (i) • Fluorine / chlorine, • being more reactive than bromine, can displace bromine from potassium bromide to form potassium fluoride / potassium chloride and bromine. • Solution turns from colourless to red-brown. • F2 + 2Br‒ → 2F‒ + Br2 / Cl2 + 2Br‒ → 2Cl‒ + Br2 [1] [1] [1] [1] (ii) • test: Add dilute nitric acid and aqueous silver nitrate into seawater. • observation: A yellow precipitate is observed if iodide ions are present. [1] [1] 5 (a) All the dyes must have different solubilities in the same solvent / a mixture of water and ethanol. [1]

COMPASSVALE SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY FOUR EXPRESS CHEMISTRY 6092 Page 3 of 8 (b) • The Rf value for the yellow dye in pure water / 100% water / 0% by volume of ethanol is zero, which implies that the yellow dye is insoluble in pure water. • The Rf value for the red dye in pure ethanol / 100% ethanol / 0% by volume of water is zero, which implies that the red dye is insoluble in pure ethanol. • The black dye would not be completely separated into its components if either pure water or pure ethanol is used / the results of the separation would be affected. [1] [1] [1] (c) Percentage of ethanol = 32 200 ×100% = 16 % Hence, Rf value of blue dye is 0.72. [1] [1] (d) (i) No, he cannot conclude that the black dye is a pure substance. • The Rf value of the blue, red and yellow dye is the same at 0.4 • when there is 80% ethanol as solvent, hence the student only observed one spot when the black dye is actually a mixture of blue, red and yellow dyes. [1] [1] (ii) • The start line must be drawn using a pencil. • The level of solvent must be below the start line. [1] [1] 6 (a) • Comparing experiments 1 and 3, when the concentration of Br‒ in experiment 3 at 2.00 mol/dm 3 is twice that of experiment 1 at 1.00 mol/dm 3, the initial rate of reaction in experiment 3 at 0.032 moldm‒3/s was four times the initial rate of reaction in experiment 1 at 0.008 moldm‒3/s. • Comparing experiments 2 and 4, when the concentration of H+ in experiment 4 at 2.00 mol/dm 3 is twice that of experiment 2 at 1.00 mol/dm 3, the initial rate of reaction in experiment 4 at 0.032 moldm ‒3/s was twice the initial rate of reaction in experiment 2 at 0.016 moldm‒3/s. [1] [1] (b) • As the concentration increases, there are more particles per unit volume. • This increases the frequency of effective collisions between reacting particles, resulting in an increase in the initial rate of reaction. [1] [1] (c) (i) species BrO3‒ Br‒ Br2 oxidation state +5 ‒1 0 [1] – 2 correct [2] – all correct [2]

COMPASSVALE SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY FOUR EXPRESS CHEMISTRY 6092 Page 4 of 8 (ii) • The oxidation state of Br decreases from +5 in BrO3‒ to 0 in Br 2, hence reduction occurred. • The oxidation state of Br increases from ‒1 in Br ‒ to 0 in Br 2, hence oxidation occurred. This reaction is a disproportionation reaction as Br is simultaneously reduced and oxidised in a single reaction. [1] [1] 7 (a) (i) gain in mass at cathode = loss in mass at anode = 1.44 – 1.20 = 0.24 g mass of anode after electrolysis = 1.45 – 0.24 = 1.21 g [1] [1] (ii) gain in mass at cathode after electrolysis at 8.0A for 90 s = gain in mass at cathode after electrolysis at 4.0A for 180 s = 0.24 g mass at cathode = 1.51 + 0.24 = 1.75 g [1] (iii) electrode anode cathode electrolyte carbon observation: effervescence of a colourless (odourless) gas / size of anode remains the same explanation: hydroxide ions are selectively discharged at the anode and are oxidised to form oxygen gas observation: pink- brown solid forms at the cathode / cathode increases in size explanation: copper is less reactive than hydrogen hence copper(II) ions are selectively discharged at the cathode and are reduced to form copper metal observation: blue solution fades / turns light blue / becomes colourless OR becomes acidic explanation: there is a net loss of copper(II) ions, concentration of copper(II) ions decreases OR hydroxide ions are selectively discharged to form oxygen gas; less hydroxides ions [1] – correct observation and explanation for anode [1] – correct observation and explanation for cathode [1] – correct observation and explanation for electrolyte [3]

COMPASSVALE SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY FOUR EXPRESS CHEMISTRY 6092 Page 5 of 8 (b) (i) most reactive C B A