SJI 2020 Year 3 EOY Chemistry 6092 ANSWERS

1 Y3 OP END OF YEAR EXAMINATION 2020 (ANSWERS) Paper 1 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 D C A B C B A C D D Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 C A B D D A A C B B Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 D B A A C C C B A C Paper 2 Section A A1(a)(i) Propanol is a flammable liquid. Or More even heating around the flask (ii) When evaporation occurs in figure 1.2, the vapours escape into the surroundings so less reactants are available for reaction. In figure 1.1, the liebig condenser condenses the vapours back into liquid state so more reactants are able to react in the flask to form more product. (b) (i) to react with the colourless / invisible (spots) to make them visible / coloured / seen (ii) The spot of sample 2 is more soluble as it has moved higher up the paper than sample 1. (iii) Rf = distance travelled by sample / distance travelled by solvent front (iv) sample 1 Rf = 0.20 to 0 .24 (show working) tartaric acid sample 2 Rf = 0.44 to 0.48 (show working) malic acid A2(a) The particles roll and slide over each other initially and vibrate at fixed positions thereafter The particles were closely packed and disorderly arranged initially and become closely packed, orderly arranged. (b) Temperature / ⁰C Time / s 80 69.3 30

2 (c) The liquid turns into a gas. Heat energy is absorbed and overcomes the intermolecular forces of attraction between the particles, causing the particles to move further apart from each other. A3 (a) 40 2,8,8 A3 (b) R has a stable electronic configuration, with eight electrons in its valence shell/has an octet electronic configuration, causing R to be inert. A3 (c) Atom Q loses two electrons to atom P, causing the electronic configuration of Q to change from 2,8,2 to 2,8. A3 (d) A3 (e)(i) Isotopes are atoms of t he same element with the same number of protons but different number of neutrons. Reject: same neutron number/nucleon number A3 (e)(ii) Isotopes of the same element possess the same number of valence electrons. A4 (a) CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) (b) Clear presentation/Calculating number of moles of both reactants Comparing molar volume or number of moles Correct identity of limiting reagent based on working/explanation Vol ratio of CH4 : O2 = 1:2, 10 cm3 of CH4 requires 20 cm3 of O2 for complete reaction. 20 cm3 of CH4 requires 40 cm3 of O2 which is less than the 60 cm3 of O2 available. O2 is in excess and CH4 is limiting reagent. XX P S XX XX OO O X O X X X S XX OO X X

3 (c) Vol ratio of CH4 : CO2 = 1:1 Volume of CO2 produced = 20.0 cm3 Volume of O2 remaining = 60 - 40 = 20.0 cm3 Total volume = 20 + 20 = 40.0 cm3 (d) Actual volume of CO2 produced = 0.7 * 20cm3 = 14.0 cm3 A5(a) Ba(NO3)2 + K2SO4 → BaSO4 + 2KNO3 Ba2+ + SO42- → BaSO4 A5(b) Sulfuric acid/ sodium sulfate solution (any soluble sulfate solution) A6 (a) P: Copper / Silver / Gold (Any unreactive metal) Q: Hydrogen R: Carbon dioxide S: Sodium chloride T: Iron(II) chloride (b) 1. Add excess iron(II) oxide to 25 cm3 of hydrochloric acid in a beaker. 2. Carry out filtration to remove the excess iron(II) oxide and collect the filtrate in an evaporating dish. 3. Heat the filtrate to obtain a saturated solution. Allow the saturated solution to cool and form crystals. 4. Carry out filtration to collect the crystals and wash them with a little cold distilled water. The crystals are then dried by pressing them between sheets of filter paper.

4 Section B B7(a)(i) Similarity: Each silicon atom in silicon carbide is bonded to four carbon atoms while each carbon atom in diamond is bonded to four carbon atoms. Accept: both are tetrahedral in structure. Difference: Silicon carbide consists of two elements – carbon and silicon while diamond consists of only carbon atoms. (a)(ii) Each silicon atom in silicon carbide is bonded to four carbon atoms by strong covalent bonds throughout the structure. A lot of energy is required to break these strong bonds / the structure is rigid, making silicon carbide hard. All four valence electrons of silicon & carbon atoms are used up in covalent bonding. The structure does not possess mobile valence electrons, and cannot conduct electricity. (a)(iii) Used in drills or cutting tools. (b)(i) Simple covalent structure / simple molecular structure (b)(ii) There are weak intermolecular forces of attraction between silicon tetrachloride molecules Little energy is required to overcome these weak forces. B8 (a) (i) A substance which dissociates/ionises partially in water to form a low concentration of hydrogen ions. (a)(ii) Reagent (reactive metals such as Mg/carbonate such as Na2CO3) [1m] Observation (More/Less vigorous bubbling / time taken for solid to disappear is shorter/longer) Conclusion (strong/weak acid is used => more/less vigorous bubbling) or (shorter/longer time for solid to disappear) [1m] (b)(i) 3NaOH (aq) + H3PO4 (aq) → Na3PO4(aq) + 3H2O (l) 1m for balanced equation 1m for state symbols (b)(ii) Sodium hydrogen phosphate

5 (b)(iii) Volume of NaOH/cm3 Colour of UI Species/particles present 0 red H+, H2PO4- , OH- 20 Orange/yellow H2PO4-, Na+ OH-/H+ 40 Violet/purple HPO42-, Na+ OH-/H+ B9 (a) element no. of electron shells in the atoms atomic radius / pm no. of electron shells when it forms ion ionic radius / pm lithium 2 152 1 68 sodium 3 185 2 98 potassium 4 227 3 133 element no. of electron shells in the atoms atomic radius / pm no. of electron shells when it forms ion ionic radius / pm fluorine 2 71 2 133 chlorine 3 99 3 181 bromine 4 115 4 196 (Note: 1 pm = 10-12 m) (b) (i) The radius of lithium ion is smaller than the radius of lithium atom. Reason: The number of electron shell for lithium ion is lesser than lithium atom. Hence the valence electrons are closer to the nucleus. (ii) The radius of Li is larger than the radius of F. Reason: Fluorine has a more protons in the nucleus and there is a stronger attractive forces of attraction between the valence electrons and the nucleus of Fluorine. Hence atomic radius is smaller. (c) (i) The reactivity decreases down Group VII elements.

6 (ii) positions appearance P Reddish brown vapour formed Q Dark purple vapour formed Accept: purple vapour R Black solid formed

7 Paper 3 Marking Scheme Question 1: (a) test no. test observations 1 (a) To a portion of P, add an equal volume of aqueous barium nitrate. White ppt formed 1 (b) Add dilute nitric acid to the mixture from 1(a). White ppt dissolves [to form colourless solution Effervescence observed which forms a white ppt in limewater Presence of CO2 gas 2(a) To a portion of P, add a few drops of aqueous silver nitrate. White ppt formed 2(b) Add dilute nitric acid to the mixture from 2(a). White ppt is insoluble in nitric acid Bonus: - Decreased in quantity OR - Effervescence observed which forms a white ppt in limewater OR - Presence of CO2 gas The formula of the impurity present in P is NaCl……………………. (i) Cl-(aq) + Ag+(aq) -> AgCl(s) (this answer should be based on the unknown anion and not the known) (b) Results:

8 Titration number Final burette reading / cm3 Initial burette reading / cm3 Volume of Q used / cm3 Best titration results ( ) Summary Tick () the best titration results. Using these results, the average volume of Q required was ………….. cm3. (2 dp) Volume of solution P used was …………… cm3. (1 dp) (c) No of moles of HCl = conc x dm3 use mole ratio to find No. moles of Na2CO3 = ½ x no of moles of HCl Concentration of Na2CO3 = concentration of sodium ca