NYJC 2020 Prelim 9729 P1 worked solutions

Prelim Exam 2020 H2 Chemistry Paper 1 Answers and CommentsPage 1 of 6 NYJC 2020 H2 Chemistry 9729 P1 worked solutions Nanyang JC J2 Preliminary Examination 2020 H2 Chemistry 9729/01 Paper 1 MCQ Answers and Comments Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans 1 B 6 C 11 C 16 D 21 C 26 B 2 C 7 D 12 A 17 D 22 A 27 A 3 D 8 C 13 A 18 D 23 B 28 B 4 B 9 D 14 D 19 D 24 A 29 D 5 B 10 B 15 A 20 B 25 B 30 D 1 B 3 15n(N) in 100g fertiliser 14.0 15n(N) in 14g fertiliser 100 14 0.150 mol14.0 0.150 mol is dissolve 15 d in 5 dm 100 . 0.150Hence conc of N atoms = 0.030 mol 5 .         There is g of N per g of fertiliser -3dm 2 C For protons, argch e mass = 1 angle of deflection = argch ek mass +15 = k(1) k = 15 For particle Y which is deflected by an angle of +5o, argch e mass = + 5 15 = + 1 3 p n e charge mass argch e mass A 1 2 1 0 3 no deflection B 3 3 5 2 6 1 3 C 4 5 1 +3 9 + 1 3 D 4 5 3 +1 9 + 1 9 Hence ans is C. 3 D BeCl2 and CO2 (2 bp and 0 lp  linear) BH4– and NH4+ (4 bp and 0 lp  tetrahedral) OF2 and SCl2 (2 bp and 2 lp  bent) PH3 (3 bp and 1 lp  trigonal pyramidal); SO3 (3 bp and 0 lp  trigonal planar) 4 B Energy released in forming the stronger intermolecular forces in liquid Z is greater than the energy absorbed in overcoming the intermolecular forces in liquid X and liquid Y. (option 1 is correct) Since the boiling point of Z is higher, the vapour pressure of liquid Z will be less than that of either liquid X or liquid Y. (option 2 is correct) More energy is required to overcome the stronger intermolecular forces in liquid Z. Hence, the boiling point of liquid Z is higher than that of either liquid X or liquid Y. (option 3 is incorrect) 5 B pV = nRT  mRTpV = M  mRp = ( )TVM For graph of p against T, gradient = mR VM Since P has a larger Mr than Q, the graph for P has a less steep gradient. T / oC p Q P –273

Prelim Exam 2020 H2 Chemistry Paper 1 Answers and CommentsPage 2 of 6 NYJC 2020 H2 Chemistry 9729 P1 worked solutions For a fixed mass of gas at constant T, mRTpV = M = constant Hence, the graph of pV against p is a horizontal straight line (option D is incorrect). Since P has a larger Mr than Q, the pV value for P is smaller than that for Q (option C is incorrect). 6 C Energy absorbed by water = 250 × 4.18 × (100 – 12) × 10–3 = 91.96 kJ Energy evolved by combustion of butane = 91.96 × 100 47 = 195.6 kJ 195.6 = 2877 × m(butane) 58 m(butane) = 3.944  3.94 g 7 D Hr 2SO2(g) + O2(g)  2SO3(g) 2S(s) + 3O2(g) 2(–296.5) + Hr = –791.4 Hr = –198.4 kJ mol–1 8 C When [1 -bromobutane] doubles 0.2( 2)0.1 , initial rate also doubles 5 5 3.0 10( = 2) 1.5 10     . Therefore, order of reaction with respect to 1-bromobutane is 1. When [HS –] increased by 1.5 times 0.3( 1.5)0.2  , initial rate also increased by 1.5 times 5 5 4.5 10( = 1.5) 3.0 10     . Order of reaction with respect to HS– is 1. Hence, rate = k[1 -bromobutane][HS–], where both 1 -bromobutane and HS – are involved in the rate determining step. Using Expt 1, 1.5 × 10–5 = k (0.1)(0.1) k = 1.5 × 10–3 mol–1 dm3 s–1 9 D When the temp erature increased, the Maxwell Boltzmann graph flattens out with the peak lower than T1. With the addition of catalyst, Ea(cat) is lower as more particles with energy  Ea. 10 B Pressure/ atm 2NO2 ⇌ 2NO + O2 Initial a 0 0 Change y +y +0.5y Eqm a  y y 0.5y p = a  y + y + 0.5y = 1.3a  y = 0.6a mole fraction of oxygen = 2O T p p = 0.5y p = 0.5(0.6 ) 1.3 a a = 0.23 11 C Option Change made 1 Increase in temperature  Increases both forward and backward rate  Favors end othermic reaction, so backward reaction is favoured, percentage yield decreases 2 Add a catalyst  Increases both forward and backward rate  No effect on POE, hence percentage yield remains unchanged 3 Increase in pressure  Increases both forward 2(–296.5) –791.4

Prelim Exam 2020 H2 Chemistry Paper 1 Answers and CommentsPage 3 of 6 NYJC 2020 H2 Chemistry 9729 P1 worked solutions and backward rate  Favors reaction that produces lesser moles of gas, hence forward reaction favoured, yield increases 12 A Brønsted-Lowry base is a proton acceptor A NH3 is a proton acceptor where it accepts H+ from water. B NH3 is a nucleophile where it attacks the electron deficient C attached to Br. C NH3 is a reducing agent . Oxidation no. of nitrogen decreases from +3 in HNO 2 to 0 in N2. D NH3 is proton donor as it donates H+ to H– which forms H2. 13 A [H+] = 10–pH = 10–6 = 1.0 × 10–6 mol dm–3 Since [H+] is not the same [ X] (= 2 mol dm – 3), X undergoes partial dissociation. It is a weak acid. Degree of dissociation = –61.0 10 2  = 5.0 × 10–7 [OH–] = 10–pOH = 10–(14–9) = 1.0 × 10–5 mol dm–3 Since [OH–] is not the same [Y] (= 2 mol dm–3), Y undergoes partial dissociation. It is a weak base. Degree of dissociation = –51.0 10 2  = 5.0 × 10–6 Extent of dissociation of X is less than Y. Hence, stu dent R and S gave the correct statements. 14 D Zn2+ + 2e Ý Zn Eʅ = –0.76 V VO2++ 2H+ + e Ý H2O + VO2+ Eʅ = +1.00 V yellow blue VO2++ 2H+ + e Ý H2O + V3+ Eʅ = +0.34 V blue green V3+ + e Ý V2+ Eʅ = –0.26 V green purple Eʅ cell = (+0.34) – (–0.76) = +1.10 V Eʅ cell = (–0.26) – (–0.76) = +0.50 V Colour change will be from blue to green to purple (final colour). 15 A 2HOCl + 2H+ + 2e Ý Cl2 + 2H2O +1.64 V I2 + 2e Ý 2I– +0.54 V 1 Eʅ cell = Eʅ (HOCl/Cl2) – Eʅ (I2/I–) = +1.64 – (+0.54) = +1.10 V 2 The size of the electrode has no impact on Eʅ cell 3 G = –nFEʅ cell = –2(96500)(1.10) = –212300 J mol–1 = –212.3 kJ mol–1 4 AgI is precipitated, lowering [I–]. POE for I2 + 2e Ý 2I– shifts right to increase [I–], Eʅ (I2/I–) becomes more positive. Since Eʅ cell = Eʅ (HOCl/Cl2) – Eʅ (I2/I–), Eʅ cell becomes less positive 16 D In general, across a period,  atomic size decreases due to increasing nuclear charge and constant shielding effect. (So A is WRONG)  Ionic radii decreases from Na+, Mg2+, Al3+, Si4+ and N3–, O2– , F– due to increasing nuclear charge and constant shielding effect. (So B & C are WRONG)  Anions are larger than cations because anions contain one more electron shell. F- should be larger than Ne or Na+. D is the answer. (Note this is an exact CIE qn and D is the best option) 17 D Mg(NO3)2 MgCO3 MgCl2 Mg(OH) 2 sodium carbonate solution excess HCl(aq) then boil excess NaOH(aq) Ba(NO3)2 no ppt BaSO 4 BaSO 4 remains excess ammonia excess H2SO4(aq) excess NaOH(aq) white ppt white ppt white pptcolourless soln white ppt Students should refer to the Data Booklet for reactions of Mg and Ba ions with alkali

Prelim Exam 2020 H2 Chemistry Paper 1 Answers and CommentsPage 4 of 6 NYJC 2020 H2 Chemistry 9729 P1 worked solutions 18 D Property Q of chlorine is larger than that of iodine for 1,2 & 4. 1 oxidising ability of the element (True, oxidising power of halogens decrease down the group) 2 solubility of the silver halide in NH 3(aq) (True, AgCl is soluble in NH3(aq) while AgI is only soluble in conc NH3) 3 strength of intermolecular forces between molecules of the element (False, iodine has a bigger e cloud to be polarized, hence id -id forces of attraction are stronger between iodine molecules) 4 thermal decomposition temp of the hydrogen halide (True, H -X bond increases in length down the group, hence HCl is more thermally stable , and thermal decomposition tem