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