2003 paper. Help!! (1 Viewer)

[mattstonestreet]

I am having trouble with Q6 in the 2003 paper. I understand the answer is propanol but I am having difficulty understanding the process in getting the answer! Was wondering if someone could walk me through the process??????

 

[pLuvia]

From the heat of combustion in kJ g^-1, to get the heat of combustion in kJ mol^-1. You mulitply the HOC in kJ g^-1 by the molar mass of the fuel, hence propanol is the closest to the answer.

I.e.

33.6 kJ g^-1 x 60.11 g mol^-1
=2019.6..

 

[mattstonestreet]

AHHHH yes I was using the molar mass of propane not propanol. Thanks for setting me straight..

 

[mattstonestreet]

Sorry to be a dumb ass but I would also like an explanation for question 12 on the 2003 paper if any1 could. thanx

 

[tunawalker]

The molecular formula is the number of atoms of each element.

n(C)=24/12=2
n(F)= 76/19=4
n(Cl)=71/35.45=2

It can be read off the graph.

Therefore C2F4Cl2

Geez I hope that was the right question. lol.