paper 1 Chemistry 2006

[1992]

I have few doubts from M/J 2006: Q 10 and 16
also O/N 2006: Q 1, 4 , 9,11 and 35
thanks in advance :d

 

[musa]

M/J 06
Q10 simple suppose there are 2 moles initially of N2O4 when 50% disassociate the moles of NO2 received will be 2 as 1 mole of N2O4 is diassociated
now at equilibrium you have
N2O4=1 mole
NO2=2 moles
total no of moles= 3
so partial pressures are N2O4=1/3
NO2=2/3
so apply Kp =partial pressure of products/partial pressure of reactants
answer is C
M/J 06 Q16
2Ca(NO3)2----2CaO + 4NO2 + O2
so solid residue is only CaO so 4.1/164=0.025
0.025*56=1.4g

 

[Amanina]

ON 06 Q1

CaSO4 + 2KOH ---> K2SO4 + CaO + H2O

no of mole of KOH = (25/1000) X 1.0 X 10^-2 = 2.5 X 10^-4
no of mole of CaSO4 is half KOH = 1.25 X 10^-4 mol

concentration of CaSO4 = 1.25 X 10^-4 / (50/1000) = 2.5 X 10^-3

 

[1992]

M/J 06
Q10 simple suppose there are 2 moles initially of N2O4 when 50% disassociate the moles of NO2 received will be 2 as 1 mole of N2O4 is diassociated
now at equilibrium you have
N2O4=1 mole
NO2=2 moles
total no of moles= 3
so partial pressures are N2O4=1/3
NO2=2/3
so apply Kp =partial pressure of products/partial pressure of reactants
answer is C
M/J 06 Q16
2Ca(NO3)2----2CaO + 4NO2 + O2
so solid residue is only CaO so 4.1/164=0.025
0.025*56=1.4g

i still dont understand the one of presure :S can u explain again or another form?

 

[1992]

can anyone explain the remain of O/N 2006