2024 Zhenghua Prelim MS

ZHENGHUA SECONDARY SCHOOL PRELIMINARY EXAMINATION 2024 SECONDARY FOUR EXPRESS CHEMISTRY 6092/02 MARKS SCHEME 1 a combustion b precipitation c electrolysis d rusting 1 1 1 1 2ai in correct order: methane nitrogen oxygen argon 1 2aii gases with lowest atomic/molecular mass diffuse fastest. (OR gases with highest atomic/molecular mass diffuse slowest) argon will diffuse the slowest since its Ar is the lowest at 40, followed by oxygen (Mr = 32) and nitrogen (Mr = 28) and methane will diffuse the fastest since it has the lowest M r of 16 (A: shows trend without stating Ar or Mr) 1 1 2b when temperature is constant, particles of all gases have the same average kinetic energy (OR at different temperatures, particles will have different average kinetic energy) constant temperature ensures results of the experiment is dependent solely on the molecular mass of the gases. 1 1 2c helium / hydrogen 1 2d ammmonia is very soluble in water, hence would dissolve instead of diffuse towards the diffusion plug. 1 3a carbon can reduce silicon oxide to silicon, hence, silicon is less reactive than carbon. aluminium oxide cannot be reduced to aluminium by carbon, hence aluminium is more reactive than carbon hence aluminium is more reactive than silicon. 1 1 3b make references about the (correct) type of bonding & structure present in each molecule link to amt of energy needed and hence, the melting point make references about presence of (correct) mobile charged particles link to ability to conduct electricity 1 1 1 1

CO is a covalent compound with simple molecular structure. smaller amount of energy is needed to overcome the weak intermolecular forces of attraction between the molecules, this its low melting point. SiO2 is a covalent compound with giant molecular / covalent structure. large amount of energy is needed to overcome the strong covalent bonds between silicon atoms and oxygen atoms thus, its very high melting point. Al2O3 is an ionic compound with giant ionic / crystal lattice structure. Large amount of energy is needed to overcome strong electrostatic forces of attraction between aluminium cations and oxide anions. Thus, its high melting point. Both CO and SiO2 cannot conduct electricity as both have no mobile charged particles as all the valence electrons are used up for bonding. In molten state, the giant ionic / crystal lattice structure of Al 2O3 breaks down, hence there are free moving aluminium cations and oxide anions to conduct electricity. 4a calculate no. of moles of K, Cu and Cl no. of mole K = 31.5 ÷ 39 no. of mole Cu = 25.6 ÷ 64 no. of mole Cl = 42.9 ÷ 35.5 calculate simplest ratio by dividing with smallest no. of moles (Cu) empirical formula: K2CuCl3 1 1 1 4b 27; 29; 37 1 4c F2 + CuCl2 → CuF2 + Cl2 fluorine is more reactive than chlorine; displaces chlorine from copper(II) chloride 1 1 5ai No. of moles of zinc = 1.65 ÷ 65 mole ratio= 1:1 vol. of H2 = (1.65 ÷ 65) x 24 = 0.609 dm3 1 1 5aii different atomic mass of Zn and Fe; different number of moles for the same mass 1 5b add a few drops of aq. sodium hydroxide / aq. ammonia until in excess. white precipitate of zinc(II) hydroxide is observed. 1 1

6ai energy absorbed in bond breaking of O2 is less than the energy released in the bond forming of O3. 1 6aii no of moles of oxygen = 48 x 16 × 2 = 1.5 mol energy released = 1.5 x 392 = 588 kJ 1 1 6aiii 7a 7b universal indicator turns blue as sodium oxide is an alkali / basic oxide 1 1 7c 4Na + O2 → 2Na2O 1 7di no. of moles of Na = 0.8 ÷ 23 no. of moles of Na2O = = (0.8 ÷ 23) ÷ 2 Mass of Na2O = 0.4 g 1 1 7dii actual mass of Na2O = 10.55 – 10.2 = 0.35g percentage yield = (0.35 – 0.4) x 100 % = 87.5% 1 1 8a partial dissociation of weak acid, ethanoic as compared to full dissociation of strong acid, hydrochloric higher concentration of H+ ions per unit volume in hydrochloric than in ethanoic acid leading to increased frequency of effective collisions, higher speed of reaction in experiment 1 than in experiment 3 1 1 1 progress of reaction energy O2 + O draw 2 ions with correct charges 1 correct number of electrons 1 correct shape of graph 1 products shown 1 Ea and ∆H labelled 1

8b experiment 1, time taken for experiment 4 will be less than 3 minutes. at higher temperature, reacting particles in experiment 4 has higher kinetic energy and move at higher speeds. leading to increased frequency of effective collisions, higher speed in reaction experiment 4 than experiment 1. 1 1 1 9a A reaction in which monomers react with each other and in the process, release a small molecule like water or alcohol 1 9b 1 9c saturated as it does not have any carbon – carbon double bond. add a few drops of aq. bromine; bromine remains reddish-brown in a sample of siloxanes 1 1 10a One molecule contains three ester linkages. 1 10bi 1 10bii carbon dioxide released by combustion of biodiesel is offset by the carbon dioxide absorbed by the plants in which the cooking oil is made from. 1

10c average vol. of KOH used = 21.55 cm3 Mass of KOH used for 10.0 g of oil = 1 x (21.55 / 1000) = 0.02155 g Mass of KOH needed for 1 kg of oil = 0.02155 x 100 = 2.155 g 1 1 1 11a Helium is a noble gas with a stable electronic configuration of 2. OR Helium has a fully filled valence electron shell. Hence, helium does not need to gain, lose or share electrons and will not react with other substances. (r: answers w/o explaining electrons) 1 1 11b Any two of the following: • Heating is involved in both separation methods. • Both separation methods separate miscible liquids. • The substances with the lower boiling point will be separated out / collected first for both methods. OR Both separation methods are based on difference in boiling points of the substances. 1 1 11ci Signal A – toluene Signal B – mesitylene 1 11cii amount of toluene : amount of mesitylene = Signal A height : Signal B height = 3 : 5 Percentage amount of toluene = 3 8⁄ x 100% = 37.5 % Percentage amount of mesitylene = 5 8⁄ x 100% = 62.5 % 1 1 11di A higher flow rate will result in shorter retention time. For example, with hexane when the flow rate increases from 20 cm/min to 35 cm/min, the retention time decreases from 5.54 min to 3.99 min. a: other relevant data from question 1 1 11dii The temperature is above the boiling points of both substances, hence both substances evaporated and are carried by the carrier gas together. 1

Section B 12a • describe reactions at both cathode and anode, includes correct half- equations and products at cathode, hydrogen formed. 2H+ (aq) + 2e- → H2 (g) at anode, chlorine formed. 2Cl- (aq) → Cl2 (g) +2e- • explained changes to pH of the solution during electrolysis pH of the solution increases. sodium ions and hydroxide ions remain in the solution resulting in an alkaline solution. • explained change in the difference in vol. of gases initially, less chlorine than hydrogen as chlorine is more soluble in water as compared to hydrogen, hence chlorine gas dissolved over time, chlorine stops dissolving as the solution is already saturated with it. 1 1 1 1 1 1 12bi The voltmeter reading will decrease as the difference in the reactivity between iron and copper is lower than the difference in reactivity between zinc and copper. 1 1 12bii The direction of the flow of electron is reversed as the electrons flowed f