Zhonghua Prelim P2 Ans

2024 ZHSS Pure Chem (6092) Preliminary Examination Section A (70 m) 1 (a) Put a tick (√) if a redox reaction has occurred. (i) a copper strip added to zinc nitrate solution (ii) aqueous chlorine added to potassium iodide solution √ (iii) hydrogen gas passed over heated copper(II) oxide √ (iv) Dilute sulfuric acid reacts with lithium hydroxide ( 1m each) 2 (b) Difference: Reaction rate is slower when used with aqueous bromine. Explain: Chlorine is more reactive than bromine/ gains electrons more readily (and displaces iodine from iodide solution faster); 1 1 (c) CuO + H2  Cu + H2O 1 [5] 2 (a)(i)  Silicate has an orderly arrangement, while soda-lime glass does not have an orderly arrangement.  Each oxygen atom is covalently bonded to 2 Si atoms in silicate, but not all oxygen atoms are covalently bonded to 2 Si atoms in soda-lime glass;  Silicate contains only covalent bonds, but soda-lime glass contains covalent and ionic bonds; Any two points 2 (ii) Conducts electricity only in molten state. There are more free moving/mobile ions that act as charge carriers 1 1 (b)(i) Potassium hydroxide has a giant lattice structure; A large amount of energy is needed to overcomes the strong forces of attraction between the oppositely charged ions; Chlorine has a simple molecular/ covalent structure where a small amount of energy is needed to overcome weak intermolecular forces; structure 1m, bonding & energy 2 m 1 1 1 (ii) hydroxide ion, OH- O 2.6, H 1 2 9m

3 (a) The greater the solubility of a solute in the mobile solvent, the faster the solute moves up the paper, the higher the spot is on the paper. 1 (b)(i) 2 spots (no mark awarded) Pigments X and Y have the same Rf value of 0.6 and thus travelled the distance up the chromatogram appearing as one spot. Pigment Z has smaller Rf value of 0.1 and thus travelled a shorter distance up the chromatogram, appearing as another different spot. 1 1 (ii) Pigment Z and Y would not have been separated from the mixture as they are insoluble in pure water and pure ethanol respectively. 1 1 5m 4 (a) Concentration of CuSO4 decreases from 1.00 mol/dm3 until there is no more Cu2+ ions in solution as Cu2+ ions is reduced to form Cu at the cathode. Cu2+(aq) +2e  Cu(s) 1 1 (b) Use aqueous sodium hydroxide, blue ppt insoluble in excess if there is copper(II) ions present OR Use aqueous ammonia solution, blue ppt soluble in excess to form a dark blue solution. 1 (c) Amount of Cu deposited = 200/1000 X 1.00 = 0.200 mol (1) Mass of Cu = 0.200 X 64 =12.8 g [1] Increase in mass of the iron object = 12.8 g 2 (d)(i) 1

(ii) solution name of products of electrolysis ionic equation for reaction at each electrode silver nitrate solution at anode silver ions Ag(s)  Ag+(aq) + e at cathode Silver coated on iron object Ag+ (aq)+ e  Ag(s) 1 1 (iii) Ag+ + e  Ag The same quantity of electricity passes through the circuit as the previous set-up and since Ag+ requires half the no off mole of electrons to be reduced as compared to Cu2+. No of mole of Ag+ =0.200 X 2 = 0.400 mol. Increase in mass of in cell B = 0.400 X108 =43.2 g 1 1 [10] 5 (a) Percentage yield decreases with increase in temperature (1) reaction is exothermic / increasing temperature favours reaction which absorbs heat (1) 2 (b) Any mistakes -1m / wrong curve- no mark 3 (c) Comparing the equation in steps 1 and 2 and cancelling the same components on the left and right of the equations. NH3 + NaOCl  NH2Cl + NaOH NH3 + NH2Cl + NaOH  N2H4 + NaCl + H2O (+ 2NH3 + NaOCl  N2H4 + NaCl + H2O 1 1

(d) 2NH3 + NaOCl  N2H4 + NaCl (allow ecf if the equation is wrong) No of moles of N2H4 = 75/ Mr N2H4 = 75/2(14) + 4(1) =2.34 mol Mole ratio: N2H4 : NH3 1 : 2 2.34 : 2.34 x 2 = 4.68 mol Mass of ammonia = 4.68 X Mr NH3 = 4.68 x 17 = 79.56 g % purity of ammonia = 79.56/ 100g x 100% =79.6% (3sf) 1 1 (e)(i) NH3 + 3HOCl  NCl3 + 3H2O 1 (ii) N is oxidized as the oxidation state of N increases from-3 in NH3 to +3 in NCl3. Cl is reduced as the oxidation state of Cl decreases from +1 in Cl to -1 in NCl3 Since oxidation and reduction occur, it is a redox reaction. 1 [11] 6 (a)(i) [1] [1] 2 (ii) Catalyst/enzyme 1 (b)(i) name - butanal formula - C3H7CHO 1 1 (ii) CnH2n+1CHO / Cn-1H2n-1CHO 1 (iii) Length of carbon chain/number of carbon atoms, increase in the intermolecular forces of attraction between the molecules ACCEPT percentage by mass of carbon 1 1 (d)(i) 2 repeat units of poly(propene) 1

(ii) Waxworms can remove non-biodegradable plastic from the environment, reducing pollution. 1 10 7 (a)(i) 1 (ii) Chlorine gas is soluble in water so the mass loss is less than expected 1 (iii) Change: use a gas syringe to measure the volume of chlorine gas at regular intervals. Explanation: Chlorine gas is collected once is produced. Less gas will be soluble in the water. 1 1 (b) Conclusion 1: increase in the surface area of contact between particles [1] Conclusion 2: MnO2 is a catalyst + no change in the amount/ mole of limiting reactant / no change in yield. [1] 2 (c) Catalyst provides an alternative pathway with a lower activation energy. Particles possesses energy greater or equal to than the activation energy thus increases the frequency of effective collision. 1 1 8m Data-based question 8 a(i) NOx, rated C2 1 (ii) With the worst performing pollutant at C2, there is a surcharge of $25000. 1 (b) Carbon dioxide is a greenhouse gas results in global warming which causes an increase in the average temperature of the Earth's surface Any valid description of effect of increased global temperature. i.e heat waves/droughts, melting of polar caps etc 1 1 c(i) Nitrogen oxides which has an increase increased by 2.5 g. (carbon monoxide decreases by 1g; hydrocarbon remains the same) For the speed of the car to increases, more fuel are combusted and thus the engine gets hotter / increase in temperature/ high temperature in car engine causes more nitrogen and oxygen in the air to react to form nitrogen oxides 1 1 (ii) As the speed increase from 50 km/h to 80km/h, the mass of carbon monoxide decreases as the volume of air entering the car engine

increases with speed/ increase in air/ fuel ratio, more oxygen is drawn into the car, allowing a greater percentage of fuel to undergo complete combustion. However, as the speed increase from 80 km/h to 120km/h, a larger mass of fuel is required for combustion as indicated in the graph for hydrocarbon This results in incomplete combustion due to limited amount of oxygen being drawn in which results in an increase in carbon monoxide produced. 1 1 1 d 2CO + 2NO  2CO2 + N2 2NO2  N2 + 2O2 At the end of the reactions, NO and NO2 are converted by the catalytic converter into harmless gaseous products as shown in the redox equations above. 1 1 1 12

Section B 9 (a) MgCO3 + H2SO4  MgSO4 + H2O + CO2 Mr MgCO3= 24 +12 +16(3) =84 No of mole of MgCO3 = 4.33/ Mr MgCO3 = 4.33