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White precipitate dissolves √½
                                                                      -1
                        on boiling and re-appears √½               Cl  √1
                        on cooling

                 Q.3    a)     Observations                               Inferences

                        Burns with yellow √1 smoky/                C = C  or -C≡C-, Long chain
                        sooty  flame                                              hydrocarbon, unsaturated
                                                                   organic compound, hydrocarbon with high
                                                                   C : H ratio.   Any 1 = 1 mk

                 b)     Observations                               Inferences

                        Dissolves√1 to form a               Polar organic √1 compound / polar substance
                        colourless  solution.

                 i)     Observations                               Inferences
                        KMnO 4√1 decolorized / changes             C = C             -C ≡C-
                        from  purple to colourless.
                                                                   2 = 1 mk         1 = ½ mk


                 ii)    Observations                               Inferences
                                                                                     O

                                                                         +
                                                                                +
                        Methyl Orange turns √1                     √1 H  / H 3O  / - C
                        pink  / red.                                                    O - H


             Question 1.
                 Table 1
                 Distributed as follows:
                 (i) Complete table
                        Values must be ±0.2 of each other
                 (ii) Decimal place
                      Values should be n 1d.p or 2d.p consistently used.
                 (iii) Accuracy
                        Compare the school value to any of the readings and award as follows:
                     If ±0.1 award 1mk
                 ± 0.2 award ½mk
                 Outside 0.2 award 0mk
             (iv) Principle of averaging
                 -  Award 1mk for consistent value only.
                 -  Penalize ½mk for rounding of the answer to 1d.p unless it divides exactly.
                 -  In consistent values averaged award 0mk
                 (v) Final answer value to the school to compare the average value to the school value:-
                       If ±0.1   award 1mk
                     If ±0.2 award ½mk
                   If outside award 0mk

                 Calculations
                 (a) Titre 1 + Titre II + Titre III = Answer
                                   3

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