AJC_H2_CHEM_P1_detailed_soln

H2 Chemistry 9647 2014 JC2 Prelim P1 Solutions 1 As dilute H 2SO4 is added to aqueous Ba(OH) 2, all Ba 2+(aq) ions present in solution will precipitate out as BaSO 4(s), a white precipitate, while the OH – is neutralised by H + to form H 2O. Hence, the number of ions present in solution decreases. H 2SO4(aq) + Ba(OH)2(aq)  BaSO4(s) + 2H2O(l) When dilute H 2SO4 is added in excess, the number of ions present in solution increases as H2SO4(aq) ionises completely in solution. H2SO4(aq)   2H+(aq) + SO4 2–(aq) A 2 Oxidation: Fe Fe2+ + 2e– Reduction: Fe3+ + e– Fe2+   Overall: Fe + 2Fe3+  3Fe2+ Fe + 2Fe 3+ 3Fe2+ initial (mol) 1 5 0 change (mol) –1 –2 +3 remaining (mol) 0 3 3 Given that reaction goes to completion, you may complete the above table using each of the options to come up with the remaining values below. n(Fe) n(Fe 3+) n(Fe 2+) A 0 0 3 B 0 1 3 C 0 3 3 D 0.5 0 4.5 Alternatively, for n(Fe2+) formed = 3 (start with the smallest whole no.) = n(Fe3+) left initial n(Fe) : n(Fe 3+) 1 (reacted) 2 (reacted) + 3 (left) 1 5 (option C) C 3 mass charge deflection of angle α angle of deflection of 241Am+ = +2 ( 241 1 ) angle of deflection of 32S– = 2)( 32 241  = –15.1 B 4 Each carbon atom in a molecule of ethene is sp2 hybridised.                                                                                                                                                                                      D 2s 2p 2s 2p excitation hybridisation sp2 2p

5 A At constant T, pV = nRT  )V 1a( p where a = nRT = constant A graph of p against V 1 is a straight line that passes through the origin. B pV = nRT pV = aT where a = nR = constant A graph of pV against T is a straight line that passes through the origin. C At constant T, pV = nRT  pV = a where a = nRT = constant A graph of pV against V is a horizontal line with y–intercept = a D At constant p, pV = nRT  V = aT where a = p nR = constant A graph of V against T is a straight line that passes through the origin. D 6 A   COO     C O OO   180  120  B B H HH B H HH H 120 109.5  C N HH N HH H 105 107  D Xe FF Xe FF FF 180 90  C 7 C CH3 CH2Cl H Cl and C CH3 ClCH2 H Cl     are a pair of optical isomers with identical phys ical properties (except their interactions with plane–polarised light) A

8 Na+(g) + H-(g) NaH(s)L.E Na(g) + H(g) 1st I.E 1st E.A Na(s) + 1/2 H2(g) A 9 [Cu(H2O)6]2+ + EDTA4– [Cu(EDTA)]2– + 6H2O A There is an increase in the number of particles (from 2 to 7) hence S for the forward reaction is positive. B The oxidation state of Cu remains unchanged at +2. C Since the coordination number of Cu in [Cu(H 2O)6]2+ and that in [Cu(EDTA)] 2– is 6, 6 coordinate (dativ