2019 Paper 4 Ans

2019 A level H2 Chem Paper 4 Answer 2019 A level H2 Chem Paper 4 Answer

Brown ppt formed in colourless solution. Brown ppt dissolved in NH3(aq) to give a colourless solution. Brown ppt formed in colourless solution. Silver mirror is formed in the test tube

Oxidation. COHH C CH2OH C H OH C HHO COOH OHH Note : Oxidation of aldehyde to carboxylic acid. Typically, a carboxylate ion RCOO− is formed. However the molecular formula given suggest that a carboxylic acid is produced. The reagent undergoes reduction. The oxidation state of Ag decreases from +1 (in [Ag(NH3)2]+) to 0 (in Ag(s))

Carbonyl (aldehyde and ketone) As stated in page 2, less than 1 % of D−glucose exist in the linear form with the aldehyde functional group. Hence the concentration of aldehyde could be too low to have a positive reaction with 2,4−DNPH. However, propanone fully exist s in the ketone form and can react readily with 2,4−DNPH.

CO32−, SO42−, Cl−, Br−, I− To test for CO32−, add HCl(aq) to FA2 solution. CO32− will give effervescence of CO2 that gives a white ppt with limewater. To test for SO42−, add Ba(NO3)2(aq) to FA2 solution, followed by excess HNO3(aq). SO42− will give white ppt that is insoluble in excess HNO3(aq). To test for Cl−, Br−, I−, add AgNO3(aq) to FA2 solution, followed by excess NH3(aq). Cl− will give white ppt that is soluble in excess NH3(aq). Br− will give pale cream ppt that is partially soluble in excess NH3(aq). I− will give yellow ppt that is insoluble in excess NH3(aq).

No effervescence observed. Solution remains colourless No ppt formed. Solution remains colourless. White ppt formed in colourless solution. Upon adding excess NH3(aq), white ppt dissolves to give a colourless solution. Cl−

Mass of FA 2 + weighing bottle / g 13.846 Mass of residual FA2 + weighing bottle / g 5.888 Mass of FA 2 used / g 7.958 Time/ min T / C 0.0 29.8 1.0 29.4 2.0 29.2 3.5 23.4 4.0 22.0 5.0 21.6 6.0 21.6 7.0 21.9 8.0 22.0 9.0 22.4

20.8 C 29.0 C −8.2 C q = maqc∆T = 50 × 4.18 × (−8.2) = −1713.8 = −1710 J (3s.f.) −1710 J Amount of KCl dissolved = 7.958  (39.1+35.5) = 0.1067 mol ∆Hsol = −(−1713.8)  0.1067 = +16062 Jmol−1 = +16.1 kJmol−1 (T is -ve, endothermic reaction, H is +ve) +16.1 kJmol−1 The value of ∆T will be halved as the same amount of heat change is distributed over twice the volume of solution. ∆Hsol calculated would be the same, as the heat change, q for the reaction would be the same and mol of solid dissolved remains the same.

∆Hsol = −L.E + ∆Hhyd (cations + anions) L.E. = (−322) + (−364) – 16.1 = −702 kJ mol−1 −702 kJ mol−1