BBSS Prelim P2 Ans

1 Bukit Batok Secondary School Sec 4 Express Chemistry (6092) Prelim Exam 2024 Answers 1a The molecules lose energy to the surroundings and slow down in their movement hence coming closer together making the air more dense / denser. (More collisions between air molecules on the bottom of aeroplane caused more lift) 1m 1m 1bi Flooding of low-lying areas leading to loss of crops/destruction of animal habitats. 1m 1bii Increase in usage of carbon containing fuels Higher usage or presence of cars Deforestation either 1m 1biii CS is expensive as it takes a lot of equipment to capture, purify, liquefy, transport and bury CO2. Safety concerns because it is uncertain what the consequences will be if there is leakage of large amounts of CO2. One might imagine that large amounts of CO2 could escape and incapacitating humans and animals. CO2 is a gas that occurs in nature, but can be dangerous in large, concentrated quantities. accept answers relevant to context of question either 1m 2a Titanium, carbon, magnesium 1m 2b The magnesium chloride produced in reactor 2 is collected and electrolysed to produce magnesium metal which is then used to react with titanium chloride again. Chlorine produced by electrolysis is used to react with rutile to produce titanium chloride in reactor 1. either 1m 2c Argon is used to provide an unreactive/ inert atmosphere to prevent the oxidation of / reaction of oxygen with magnesium / titanium / titanium chloride. 1m 2d Wash the mixture with water to dissolve the soluble magnesium chloride. 1m Filter the mixture to obtain titanium as the residue. 1m

2 3a true false An iron catalyst is used. √ The reaction is reversible and so the conditions must be chosen carefully to give the best yield. √ A good yield of ammonia is obtainable even without the use of high pressure. √ Nitrogen from the air is converted into a substance from which nitrogenous fertilisers can be easily produced. √ 4√ - 2m 2-3√ - 1m 1√ - 0m 3b (i) sodium hydroxide is added to A and warm 1m (ii) (dilute) nitric acid and barium nitrate is added to A 1m (iii) blue precipitate formed is insoluble in excess sodium hydroxide 1m (iv) sodium hydroxide and aluminium foil are added to A and warm 1m (v) ammonium iodide 1m 4a R f value of yellow dye is 0 in pure water. Thus yellow dye does not dissolve in pure water and be separated. 1m R f value of red dye is 0 in pure ethanol. Thus red dye is insoluble in pure ethanol and unable to be separated. 1m 4b All 3 dyes have same Rf value of 0.4 when solvent has 80% ethanol. 1m 4c Total volume = 80 + 120 = 200 cm3 1m %ethanol = (80/200) x 100 = 40% R f value of blue dye 0.6 Rf value of red dye 0.6 Rf value of yellow dye 0.04 1m 4d No. When solvent has 40% ethanol, blue and red dyes have the same Rf value thus will not be separated. 1m 5ai CH 3COOH + NaOH  CH3COONa + H2O 1m 5aii No of moles of NaOH = 10/1000 x 0.1 = 0.00200 mol (3sf) 1mol of acid reacts with 1mol of NaOH 0.00200 mol acid reacts with 0.00200 mol NaOH Concentration of ethanoic acid = 0.002/ (10/1000) = 0.200 mol/dm3 1m 1m

3 5aiii pH > 2 Ethanoic acid is a weak acid that dissociates partially in water to produce low concentration of H+ ions in water. Thus, in 0.200 mol/dm3 acid, concentration of H ions is lesser than 0.200 mol/dm3. pH = – lg [<0.2] = >2 partial dissociate 1m conc 1m 5bi 1) Initial pH for (1) is lower than (2) 2) At the start of (1), pH stays constant until 20cm3,but in (2) pH increases gradually. 3) Extent of pH change at endpoint is greater for (1), from 2 to 12, but is lesser for (2), from 7 to 11. either 2 each 1m total 2m 5bii The salt formed from a strong acid-strong alkali reaction is neutral (pH7). The salt formed from a weak acid-strong alkali reaction is alkaline/basic (pH= 8.5). both 1m 6a A : ethanoic acid 7√ - 4m 5-6√ - 3m 3-4√ - 2m 1-2√ - 1m B : ethyl ethanoate C : ethene D : carbon dioxide E : dibromoethane F : ethane L : ethanol 6b C 2H5OH + CH3COOH  CH3COOC2H5 + H2O 1m 7a 2.8.18.8.2 1m 7b Student A. As atomic radius of Group 2 elements increases, the distance between negatively charge valence electrons and positively charge nucleus increases. 1m Electrons are more readily lost as they are less strongly attracted to nucleus. 1m 7ci Hg 2O and HgO both 1m 7cii basic 1m 7d shared e- 1m valence e- & charge 1m

4 8a Oxidising agent : S2O82– Reducing agent : Fe2+ both 1m 8b S2O82–(aq) + 2I–(aq)  2SO42–(aq) + I2(aq) 1m 8c A catalyst speeds up a reaction by providing an alternative pathway with lower activation energy. 1m More particles have required activation energy to react. Frequency of effective collisions increases and reaction is faster. 1m 8d Uncatalysed reaction is slow as it involves collisions of (three) negatively charged ions which will repel one another. 1m 8e Fe 3+ ion used up in Step 1 of is regenerated in Step 2. 1m 8f Iron forms compounds with variable oxidation states (+2 and +3) but magnesium only has one oxidation state (+2). 1m Reject : magnesium does not / does not have variable oxidation state 9ai When the distance between the two hydrogen atoms increases, the bond energy decreases to a minimum energy of – 432 kJ/mol at 74pm When the distance is greater than 74pm, the H-H bond energy increases to almost 0 kJ/mol. 1m 1m 1m Answers that only mention positive/negative energy will be awarded max. 2 marks 9aii At 74pm, there is the greatest attraction of the negatively charged electrons in one atom to the positively-charged protons in the other atom. 1m 1m If no mention of negative/positive charges of electrons/protons, or the “greatest” attraction, max. 1 mark 9b Do not agree. H-H is a single bond / 1 bond, but the bond length is the lowest at 74pm compared to N≡N, where there is a triple bond but the bond length is longer at 110pm. OR compare no of bond 1m bond length 1m Agree. O=O has a double bond / 2 bonds and a longer bond length of 121pm compared to N≡N at 110pm. Any combination is fine, best is to compare between same type of atoms

5 9c There is more energy taken in to break (N≡N and O=O) bonds than energy given out to form the (N-O/N=O) bonds. more 1m break/form 1m 9di Silicon tetrahydride has a simple molecular structure / is a simple molecule. 1m Weak intermolecular force between molecules requires little amount of energy to overcome. Thus boiling point is low. 1m 9dii Hydrogen bonds are stronger than weak intermolecular force between molecules. More energy required to break the bonds. 1m 10a A : hydrogen (gas) B : oxygen (gas) both 1m 10b Insert a glowing splint, it will relight. 1m 10c Cathode : 4H+ (aq) + 4e–  2H2 (g) Anode : 4OH– (aq)  2H2O (l) + O2 (g) + 4e– eqn 1m With same no. of moles of electrons flow in circuit, 2 moles of hydrogen is produced at cathode and 1 mole of oxygen produced at anode. explain 1m OR Overall equation : 2H2O (l)  2H2 (g) + O2 (g) Mole ratio of hydrogen to oxygen produced is 2:1. Volumes of gases produced is also in ratio of 2:1. 1m 10di Aluminium anode reacts with oxygen produced, forming a layer of aluminium oxide, causing mass to increase. 10dii Solid aluminium oxide does not conduct electricity. When the anode is fully coated, electrolysis stops as circuit is broken. 1m 10diii Mass increase is due to oxygen Mass of O 2 = 27.8 – 15.0 = 12.8 g No of moles of O 2= 12.8 ÷ 2(16) = 0.400 mol mass & mol 1m 1mol of O 2 is produced with 2mol of H2 0.400 mol of O 2 is produced with 0.800 mol of H2 Volume of H 2 = 0.800 x 2