Ans.1) - Can't think of a better way, myself. Sorry.
Ans.2) you get the answer by simply subtracting 10 from 60. The 10 cm3 of methanethiol is used up/exploded with 60cm3, so the final volume of the resultant mixture is 50.
Ans 3) - Are you sure the answer is A?, I'm getting it as C.
Each calcium ion in the sample was exchanged for two hydrogen ions.
Therefore, half the concentration of Potassium Hydroxide was required to neutralize the reaction.
(1.0x10-2)X2 = 2.0x10-3
It's alright, I've contacted my teacher at school and she's clarified my doubts then!
For 1)
We know that 5g of the Metal nitrate gives 3.29 grams of the mass lost [i.e. nitrogen dioxide and oxygen], so we find the molar mass of the 2 gases which is 216g and then using ratio and proportion we find the molar mass of the metal nitrate which is 328g.
This molar mass is for 2 moles [if you write the equation: 2M(NO
3)
2 ----> 2MO + 4NO
2 + O
2 ], so we'll find the molar mass for one mole by dividing 328/2 and we get 164.
So now we know that the Mr of the metal + Mr of the nitrate => 164. We'll get the Mr of the nitrate which is 124, we subtract that from 164, we'll get Mr of the metal which is 40. Calcium which is Answer B has an Mr of 40!
For 2)
CH
3SH + 3O
2 ----> CO
2 + SO
2 + 2H
2O
1 mole of CH
3SH require 3 moles of 3O
2 to produce 1 mole of CO
2 + 1 mole of SO
2 which can be grouped into 2 moles of gases.
They're using 10cm
3 of CH
3SH so by ration, only 30cm
3 of O
2 will be required out of the given 60cm
3, so the remaining O
2 would be 30cm
3, in addition to that, since 2 moles of gases will be produced, and they are in the ratio 1:2, so 20cm
3 of the gases will be produced, and hence 20+30=50cm
3For 3) It's pretty long, I'll write it down here later!