W07 Q 9, 10
W06 Q 4
W04 Q 9, 10, 23
Any help would be humbly appreciated Thank you very much
Nov 07 p19. From the information above we can note that ratio of metal to SO
32- is 2 : 1.
So now we calculate the oxidation number of S in SO
32- and SO
42-. You'll see that it is +4 and +6 respectively.
So we can observe that there is a change of +2 in oxidation number. Or you could have observed that by the 2 electrons emitted. So the metal will on the contrary decrease in oxidation number by 2. But since the ratio of metal was twice compared to the sulfite, change in oxidation number for each metal cation will be -2/2 = -1.
SO it will decrease from +3 to +2.
(+3 - 1)Answer is
B10.
Initial number of moles of NO
2 : 4 while initial number of the products are nil.
At equilibrium number of moles of oxygen is given to be 0.8. So that of the other product NO will be 1.6
Now we can determine the number of moles of NO2 at equilibrium : (4 - 2(0.
) = 2.4
Now K
c will be 1.6
2 x 0.8 / 2.4
2So answer is
DNov 06 No 410CH
4 + 20O
2 -----> 10CO
2 + 20H
2O
So residual gas at r.t.p will exclude volume of H
2O but include the 50cm3 excess of oxygen.
Hence for CH
4, total volume of residual gas will be 60cm
3. This eliminates A, B and C.
Answer is
DNov 04 p19. The alkane will release most energy since the others already carry an oxygen atom, meaning that they are already a bit oxidised, while the alkane needs to be completely burned.
So answer is
B23. A mixture of bromine is allowed to react with ethane. ---> Free radical substitution reaction.
I believe the bromine will be in excess, so it can be bromoethane, dibromoethane(2 types), tribromoethane(2 types), tetrabromoethane(2 types), pentabromoethane and heptabromoethane.
So answer is
D.
NOTE : AM sorry but i've not had time to check marking schemes, so forgive me if some of the answers are wrong. Do tell me and i'll try my best to correct my mistakes.