2021 SASS 4E Chem MYE P2 Ans

1 CHEMISTRY 6092/2 Paper 2 80 Marks ______________________________________________________________________________________________________________________________________________________________________________________ MARKING SCHEME ______________________________________________________________________________________________________________________________________________________________________________________ SECTION A [50 Marks] Qn Answer Mark Remarks 1(a)(i) G 1 1(a)(ii) F 1 1(a)(iii) H 1 1(a)(iv) F 1 1(b) test tube 1 – D test tube 2 – G 1 1 2(a)(i) same element forming different structures / different forms of the same element 1 2(a)(ii) octasulfur larger relative molecular mass (Mr) for a larger intermolecular force of attraction, requiring more heat energy to overcome 1 1 Relation must be evident 2(b)(i) contains different number of neutrons 1 2(b)(ii) (0.9401 × 32.0) + (0.0076 × 33.0) + (0.0523 × 34.0) = 32.11 M1 A1 Rej. for wrong sig.fig. 2(b)(iii) difference in the abundance / %composition of isotopes 1 3(a)(i) B does not attain / have / acquire stable noble gas electronic configuration 1 Accept if mentioned of only 6 outershell e– instead of 8 for EC 3(a)(ii) gas simple covalent structure weak intermolecular forces of attraction present 1 1 1 Rej. explanations if incorrect physical state provided 3(b)(i) Mg atom does not share electrons with H Mg atom loses two valence electrons to form Mg2+ 1 1 Ignore any explanation wrt H 3(b)(ii) correct number of electrons for each ion correct charge annotated for each ion 1 1 3(b)(iii) free moving / mobile ions (Mg2+ & H–) as charge carriers that conduct electricity 1 Rej. electrons – ╳ ● H 4 © UCLES 2019 5070/12/M/J/19 6 A m i n e r a l d e p o s i t i s f o u n d t o c o n t a i n s m a l l grains made entirely of the element carbon. Which property will definitely be true of the grains of carbon? A T h e y w i l l b e m a d e o f a t o m s a r r a n g e d i n l a y e r s . B T h e y w i l l b e s o f t . C T h e y w i l l b u r n t o g i v e c a r b o n d i o x i d e . D T h e y w i l l c o n d u c t e l e c t r i c i t y . 7 W h i c h d i a g r a m s h o w s t h e o u t e r e l e c t r o n a r r a n g e m e n t i n c a l c i u m f l u o r i d e ? F FCa F– Ca2+ an electron from calcium an electron from fluorine key A B C D F– F– Ca2– F– F FCa 8 H o w m a n y s h a r e d p a i r s o f e l e c t r o n s a r e t h e r e i n o n e c a r b o n d i o x i d e m o l e c u l e ? A 2 B 4 C 8 D 1 2 2+ Mg 2 ST ANDREW’S SECONDARY SCHOOL MID-YEAR EXAMINATION 2021 SECONDARY 4 EXPRESS GD Updated: 07/05/2023

2 4(a) Cu2+ , Fe2+ , Fe3+ all give coloured solutions instead of colourless observed 1 4(b) Pb(NO3)2 either Pb2+ or Ag+ will result in ppt formed require half of the amount of T added for mole ratio of 2:1 1 1 1 Rej. Pb2+ for chemical formula Rej. explanations that do not relate to Pb2+ 4(c) Any 1 • use a burette instead of measuring cylinder to measure Q • stir the contents in each test tube using glass rod • measure mass of ppt. formed 1 4(d) white ppt. formed, instead of yellow ppt. equal volume of K2SO4 solution used, instead of using only half 1 1 5(a)(i) moles of NH4NO3 used = moles of N2O produced = 4.40 g ÷ 80 g mol–1 = 0.055 mol mass of N2O produced = 0.055 mol × 44 g mol–1 = 2.42 g M1 A1 5(a)(ii) %yield = 2.50 g ÷ 2.42 g × 100% = 103.3% A1 Ignore inaccurate sig.fig. 5(a)(iii) impurities in the product obtained 1 Rej. impurities in reactant 5(b) additional O2 utilised in combustion to provide more energy exothermic reaction loses energy to the engine / engine gained energy from exothermic reaction 1 1 6(a) absorbs excess UV rays from the Sun (entering earth) 1 6(b) UV breaks down CFCs resulting in Cl radicals / atoms (Cl•) Cl• reacts with O3 to liberate O2 1 1 6(c)(i) A 1 Accept CF3CCl3 6(c)(ii) no, as the use of this flammable compound will be dangerous 1 Both ans and explanation must be provided 7(a) hydrogen (gas) extinguishes a lighted splint with a ‘pop’ sound 1 1 Rej. if chemical formula is provided instead Test result must be provided 7(b) different atomic masses for the different metals reacted different number of reacting particles 1 1 Explanation wrt the no. of moles 7(c)(i) A – Fe B – Zn 1 Both must be correct 7(c)(ii) larger Ar results in a smaller number of moles reacted leading to a smaller molar volume 1 1 7(d) exothermic; enthalpy during bond breaking is smaller than that for bond forming ; 1 1 Vice versa 7(e) M + 2H+ ® M2+ + H2 (equimolar for the metal and hydrogen gas) volume of H2 by Fe = 64 cm3 moles of H2 = 2.667 × 10–3 mol moles of Fe = 0.15 g ÷ 56 g mol–1 = 2.667 × 10–3 mol since moles of Fe = moles of H2 produced, metals are the LR A1 1 Accept if HCl proved to be the ER

3 SECTION B [30 Marks] Qn Answer Marks Remarks 8(a)(i) 4OH– (aq) ® O2 (g) + 2H2O (l) + 4e– 1 Rej. if missing/ incorrect ss 8(a)(ii) Ag+ (aq) + e– ® Ag (s) / 4Ag+ (aq) + 4e– ® 4Ag (s) 1 8(b) all points plotted graph of 1.0A drawn graph of 1.2A drawn 1 1 1 8(c) graph of 1.2A has a higher rate / gradient same mass of Ag deposited / same increase in mass 1 1 Vice versa 8(d) moles of Ag+ deposited = 2.68 g ÷ 108 g mol–1 = 2.48 × 10–2 mol concentration of AgNO3 = 2.48 × 10–2 mol ÷ 2 dm3 = 1.24 × 10–2 mol/dm3 M1 A1 8(e) no change in mass of cathode as selective discharge of H+ (aq) instead 1 Explanation must be provided 8(f)(i) formation of O2 at graphite electrode will lead to its oxidation and hence, depletion 1 8(f)(ii) Any 1 • Platinum • Palladium • Titanium 1 Rej. if chemical symbol given 9(a) KBr Br in KBr increases in oxidation states from –1 to 0 1 1 9(b)(i) reddish-brown liquid formed in solution 1 9(b)(ii) Cl2 > Br2 Cl2 results in the oxidation of Br– 1 1 9(c) Any 1 • Cl2 readily available • No further separation or purification required • Less toxic (compared to method 2 with the production of SO2) 1 9(d) Br in Br2 increases its oxidation state from 0 to +1 in HBrO Br in Br2 also decreases its oxidation state from 0 to –1 in HBr 1 1 EITHER 10(a)(i) regular arrangement of the metal particles present easily slide over when force is applied 1 1 10(a)(ii) delocalised electrons result in the conduct of electricity 1 10(b) chromium forms an oxide layer that functionalised as a barrier for iron to be in contact with O2 and H2O 1 1 10(c)(i) increasing rate of reaction due to the increased concentration of acid increase in the number of successful collisions per unit time 1 1 10(c)(ii) constant mass of iron, particle size, temperature of the reaction measure the gas evolved at regular intervals using gas syringe / water displacement OR measure the loss in mass at regular intervals using electronic mass balance 1 1* 1* 1** 1** Either set of (*) OR (**) provided in answer.

4 OR 10(a)(i) high melting point / boiling point 1 10(a)(ii) variable oxidation states in the compounds formed / coloured compounds 1 10(b) diagram increase strength by preventing the iron atoms to slide over easily due to the regular arrangement 1 1 10(c)(i) haematite coke added into the blast furnace gets oxidised in hot air to form CO2 then CO reduces the haematite Fe2O3 to obtain Fe 1 1 1 10(c)(ii) more particles will possess energy that is greater than EA particles will collide more frequently due to increased kinetic energy 1 1 10(c)(iii) very high melting point 1 10 © UCLES 2018 5070/11/O/N/18 27 T h e d i a g r a m s h o w s t h e s t r u c t u r e o f a n a l l o y . Which statement about alloys is correct? A