AJC H2 Chemistry 9647 N2012 P1 Suggested Solutions

©2013AndersonJC/CHEM 1 H2 Chemistry 9647 N2012 P1 Suggested Solutions 1 Molecular formula of glucose = C6H12O6 (from the diagram given, you should be able to observe that there are 4 hydroxyl groups, which will be replaced by chlorine atoms) C6H8O2(OH)4 C6H8O2Cl4 (hence, empirical formula = C3H4OCl2) C 2 Fe + 2Fe 3+ 3Fe2+ initial moles 1 2 0 change in moles –1 –2 +3 moles left 0 0 3 Given that reacti on goes to completion, you may comp lete the above table using each of the options to come up with the 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 2 + 3 1 5 (option C) C 3 Cr [Ar] 3d 5 4s1 Ge [Ar] 3d 10 4s2 4p2 S [Ne] 3s 2 3p4 S– [Ne] 3s 2 3p5 Sc [Ar] 3d 1 4s2 D 4 From the large increase in IE when the 3rd electron is removed, it can be deduced that there are only 2 val ence electrons present in X. Hence it is in Group II of the Periodic Table. B 5 Factors affecting strength of hydrogen bonds  extensiveness of HB  magnitude of polarity (i.e. di fference in electronegativ ity between F/O/N–H bond) Each water molecule can form, on average, 2 hydrogen bonds as compared with 1 hydrogen bond per HF molecule. C ANDERSON JC

©2013AndersonJC/CHEM 2 6 Due to the formation of hydrogen bonds between CO 2 and H 2O molecules, heat is evolved and H is < 0. As the forward reaction results in a decrease in no. of gaseous molecules, there will be less ways of arranging the particles and S is < 0. A 7 Electrolysis of molten CuCl2 Anode: 2Cl– Cl2 + 2e Electrolysis of aq. H2SO4 Anode: 2H2O O2 + 4H+ + 4e (preferentially discharged over SO42–) Since 2 1 V V 2 2 O Cl  for the same duration, 8 2 n n 2 2 O for passed e Cl for passed e  (4H2O 2O2 + 8H+ + 8e) and hence the current used in electrolysis 2 will be 4I (n  I as nF = It). D 8 2H 2(g) + CO(g) CH3OH(g) initial moles 2.0 1.0 0 change in moles –x – ଵ ଶ x + ଵ ଶ x moles left 2.0 – x 1.0 – ૚ ૛ x ૚ ૛ x C 9 Pyruvic acid is a weak acid wh ile NaOH is a strong alkali. The equivalence point of the titration will be above pH 7 because the salt formed CH3COCO2– hydrolyses in water to give a basic solution. CH3COCO2– + H2O CH3COCO2H + OH– A 10 When aspirin (HA) enter s the stomach where [H +] is very high (10–1 mol dm –3), it will remain effectively as HA due to the suppression of its ionisation in the presence of high [H +]. Hence, [HA] > [A–]. HA H+ + A– D 11 no. of moles of excess HCl = 0.0040 – 0.0025 = 0.0015 mol [H+] = [HCl] = 0.0015 1 = 0.00150 mol dm–3 pH = 2.82 D 12 When molar proportion of uranium to lead is 1 : 3, it can be deduced that the [uranium] has decreased to ¼ of its initia l concentration. Hence 2 half–lives have occurred. C ANDERSON JC

©2013AndersonJC/CHEM 3 13 rate = k [H2O2] (1st order reaction) If [H2O2] is doubled,