RI 2021 Year 6 June TP Suggested Solutions

2021 H2 Chemistry Y6 Term 3 Common Test Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C B C D B A C D D A A A D C B Question 1 (C) nestradiol = 76/1000 (18×12.0+24×1.0+2×16.0) = 2.794 × 10−4 mol ROH + Na → RO–Na+ + ½H2 1 mol of alcohol reacts with Na to produce 0.5 mol of H2. Since there are 2 –OH groups in estradiol, 1 mol of estradiol produces 1 mol of H2. amt of H2 formed = 2.794 x 10–4 mol volume of H2 formed at s.t.p. = 2.794 x 10–4 x 22.7 = 0.00634 dm3+ = 6.34 cm3 Question 2 (B) no. of protons no. of electrons no. of neutrons 2 1D 1 1 1 11 5B 5 5 6 12 6C 6 6 6 14 7N 7 7 7 16 8O 8 8 8 species no. of electrons no. of neutrons A 11BD4– 10 10 B CD3– 10 9 C ND4+ 10 11 D D3O+ 10 11 Question 3 (C) butan-1-ol (C4H10O) 2-methylpropan-2-ol (C4H10O) Butan-1-ol is a straight-chain isomer of C4H10O and has a greater surface area for instantaneous dipole–induced dipole (id -id) interactions to occur compared to 2 -methylpropan-2-ol which is a branched-chain isomer of C4H10O. Thus, id -id interactions are stronger in butan-1-ol than in 2-methylpropan-2-ol. (Option 1 is correct) The strength of the hydrogen bonds in both molecules are the same as the hydrogen atom involved in hydrogen bonding is bonded to the same element, oxygen. The extent of hydrogen bonding is also the same as they have same number of lone pairs and number of H atoms attached to O. (Options 2 and 3 are incorrect) Question 4 (D) The enthalpy change of the neutralisation reaction between KOH and HNO3 is −57.3 kJ mol−1. Since KOH and HNO 3 have the same concentrations and KOH has a lower volume, KOH is the limiting reagent. Hence, 57300 2( )1000 114.6 J qqH yn qy  =− − =− = Also, q = mcT = (25.0 + y)(4.18)(10.0) Therefore, (25.0 + y)(4.18)(10.0) = 114.6y y = 14.35 cm3 Question 5 (B) By combining the 2 equations, you can obtain the overall equation for the decomposition of 1,2-dioxetanedione (C2O4). C2O4 → 2CO2 + light ∆H : (–). Due to formation of the more stable CO2, the reaction is exothermic. ∆S : (+). Due to the formation CO 2 gas from liquid 1,2-dixetanedione, there is an increase in entropy. ∆G: ( –). Since the reaction occurs when the chemicals are mixed, it is spontaneous. Alternatively, you can derive the sign of ∆G using ∆G = ∆H – T∆S and the signs for ∆H and ∆S deduced previously. This document is copyrighted, please do not reproduce it without permission

Question 6 (A) Since the decomposition of X follows first -order kinetics, t½ = ln 2 / k. t½ = ln 2 / k = ln 2 / (7  10–2) = 9.90 years Let n be the number of half-lives in 3 years. n  t½ = 3 years n = 3 / 9.90 = 0.303 [X] = (½)n [X]initial [X] / [X]initial = (½)0.303 = 0.811 Hence 81.1% of X remains after 3 years. Question 7 (C) A Incorrect. The chemical equation does not provide any information about the order of a reaction. No other information is provi