NJC Prelim P1 Detailed solutions

1 NJC Preliminary Examination 9647/01/15 [Turn Over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet Data Booklet 9647/01 Thu 17 Sep 2015 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Write your name, subject class and registration number on the Optical Answer Sheet (OAS) in the spaces provided unless this has been done for you. There are 40 questions in this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Optical Answer Sheet. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. This booklet consists of 19 printed pages and 1 blank page.

2 NJC Preliminary Examination 9647/01/15 [Turn Over Section A For each question there are four possible answers, A, B, C, and D. Choose the one you consider to be correct. 1 When scientists research on atomic nuclei, they often correlate the stability of isotopes with the proton to neutron ratio. Which isotope has the same proton to neutron ratio as 18 8 O? A 9 5 B B 26 12 Mg C 36 16 S D 52 24 Cr Given: n p ratio of 18 8 O = 8 - 18 8 = 5 4 n p ratio of 9 5 B = 4 5  5 4 ; n p ratio of 26 12 Mg = 14 12 = 7 6  5 4 ; n p ratio of 52 24 Cr = 28 24 = 7 6  5 4 n p ratio of 36 16 S = 20 16 = 5 4  ans is C 2 AxOy, a gaseous oxide of an element A, gives an acidic gas AO2 on combustion with excess oxygen. When 10 cm 3 of AxOy is burnt in 40 cm 3 oxygen (in excess), the total volume of gas after cooling to the original temperature is unchanged . On shaking with excess aqueous potassium hydroxide, the volume is reduced to 20 cm 3. All volumes are measured at the same temperature and pressure. What are the values of x and y? x y A 1 1 B 2 3 C 3 2 D 3 4 After burning, total volume = 50 cm3 = vol of unreacted O2 + vol of AO2 Since O2 does not react with KOH(aq), AO2 is the acidic gas that reacts with KOH(aq) Vol. of unreacted O2 = 20cm3 Vol. of AO2 = 30cm3 Eqn for reaction: AxOy + (x – ½y)O2  xAO2 Initial vol / cm3 10 40 0 Final Vol after burning / cm3 0 20 30 Vol reacted/produced/ cm3 10 20 30 Reacting mol ratio 1 2 3 x=3 x½ y = 2, ½ y = 1, y = 2

3 NJC Preliminary Examination 9647/01/15 [Turn Over 3 The diagram below shows the first ionisation energy of some element s with atomic numbers less than 20. Which is the most likely formula of the chloride of L and M? A LCl B LCl5 C MCl2 D MCl5 Based on the trend of the 1st IE of elements given; L must be N (in Group V, Period 2) and M must be P (in Group V, Period 3). Both N and P form trichlorides. P can form pentachloride but not N as the latter cannot expand its octet configuration when forming compounds. Thus, answer is D. 4 The figure below shows the stable oxidation numbers of five consecutive elements, P to T, plotted against their atomic numbers. What are the likely atomic numbers of P to T? A 1 to 5 B 3 to 7 C 11 to 15 D 19 to 23 first ionisation energy L M atomic number

4 NJC Preliminary Examination 9647/01/15 [Turn Over Consider atomic no given in option A: Atomic no 1 2 3 4 5 element H He Li Be B Known stable O.N. of atom in compounds +1 or 1 does not form compound +1 +2 +3 Thus option A is not the answer. Consider atomic no given in option B: Atomic no 3 4 5 6 7 element Li Be B C N Known stable O.N. of atom in compounds +1 +2 +3 from 4 to +4 from 3 to +5 Thus option B is not the answer. Consider atomic no given in option C: Atomic no 11 12 13 14 15 element Na Mg Al Si P Known stable O.N. of atom in compounds +1 +2 +3 +4 +3, +5 Thus option C is not the answer. Consider atomic no given in option D: Atomic no 19 20 21 22 23 element K Ca Sc Ti V Known stable O.N. of atom in compounds +1 +2 +3 +2, +3, +4 +2, +3, +4, +5 Thus option D is the answer. 5 When a sample of gas is compressed at constant temperature from 20 atm to 80 atm, its volume changes from 67.0 cm3 to 15.5 cm3. Which is a possible explanation of this behaviour? A The gas particles are adsorbed onto the vessel walls. B The gas particles dimerised. C The gas particles dissociated. D The gas is liquefied at 80 atm. For the same amount of an ideal gas at constant T, P2V2 = P1V1 = nRT From given info, P2V2 = 80 ×15.5 = 1240; P1V1 = 20 × 67.0 = 1340 P2V2 < P1V1  gas is non-ideal For A: If the gas is adsorbed onto the vessel walls, 2 < 1 as the amount of free moving gas particles decreases  P2V2 < P1V1; thus option A is the ans For B: If the gas dimerised completely, 2 = ½1  P2V2 = ½P1V1; thus option B is not the ans For C: If the gas dissociates, 2 > 1  P2V2 > P1V1; thus option C is the not ans For D: If the gas is liquefied, P2 will be very low  P2V2 << P1V1; thus option D is not the ans

5 NJC Preliminary Examination 9647/01/15 [Turn Over 6 The shapes of two species, D and E, are linear and T-shaped respectively. What could D and E be? D E A SO2 BH3 B I3  BrCl3 C CS2 AsCl3 D H2S ICl3 Based on VSEPR theory, SO2 is bent in shape (2 bp’s + 1 lp around S) BH3 is trigonal planar (only 3 bp’s around B) I3  is linear (2 lp’s and 3 bp’s around I) (lp e’s occupy equatorial positions so as to minimise their mutual repulsion due to lp-lp > lp – bp > bp – bp); BrCl3 is T-shaped (3 bp’s + 2 lp’s around Br) (lp e’s occupy equatorial positions so as to minimise their mutual repulsion due to lp- lp > lp – bp > bp – bp) CS2 is linear (only 2 bp’s around C) AsCl3 is trigonal pyramidal (3 bp’s + 1 lp around As) H2S is bent (2 bp’s + 2 lp’s around S) ICl3 is T-shaped (3 bp’s + 2 lp’s around I) (lp e’s occupy equatorial positions so as to minimise their mutual repulsion due to lp- lp > lp – bp > bp – bp)

6 NJC Preliminary Examination 9647/01/15 [Turn Over 7 Some enthalpy changes of combustion are given below. Hc / kJ mol1 CO(g) + 2 1 O2(g)  CO2(g) 283 H2(g) + 2 1 O2(g)  H2O(l) 286 CH3OH(l) + 2 3 O2(g)  CO2(g) + 2H2O(l) 715 What is the enthalpy change of the following reaction? CO(g) + 2H2(g)  CH3OH(l) A 146 B 140 C +140 D +146 2 3 O2(g) + CO(g) + 2H2(g)   rΔΗ CH3OH(l) + 2 3 O2(g) 283 2(286) 715 CO2(g) + 2H2O(l) By Hess’ law: Hr = 283 + 2(286)  (715) = 140 kJ mol1 (B) 8 When a large current was passed through acidified aqueous nickel(II) sulfate, there was a simultaneous liberation, at the cathode, of x mol of a grey solid and y dm3 of a diatomic gas (measured at s.t.p.). How many moles of electrons passed through the cell? A 𝑥 + 𝑦 22.4 B 𝑥 + 𝑦 11.2 C 2𝑥 + 𝑦 22.4 D 2𝑥 + 𝑦 11.2 Eqn for reactions at cathode: [R] Ni2+ + 2e  Ni and [R] 2H+ + 2e  H2 Total no of mol of e passed through solution = 2(no of mol of Ni2+ + no of mol of H2) = 2x + 2( 22.4 y ) = 2x + 11.2 y (molar gas vol at stp = 22.4 dm3) Thus ans is D.

7 NJC Preliminary Examination 9647/