Couldn't do Q11.
Q14:
2 X(NO3)2 → 2 XO + 4 NO2 + O2
The gases formed are NO2 and O2. 3.29g are formed. So moles = 3.29 (2x (14+16+16) + 16) = 0.03046 Moles [2 Moles of nitrate gives 4 moles of NO2 and 1 mole of O2. 1 Mole of Nitrate would give 2 mole of No2 and half mole of O2.]
We know that n=m/Mr
0.03046 = 5.00/Mr
Mr = 164.13
Now we open the periodic table and find the Mr of the given nitrates. To make this easier, we can subtract the Mr of (NO3)2 from 164.13. 164.13-124 = 40.1
Calcium is 40.1 so answer is (B).
W07-
Q8:
Can't do this.