Answer:
The heat of vaporisation of methanol is "3.48 KJ/Mol"
Explanation:
The amount of heat energy required to convert or transform 1 gram of liquid to vapour is called heat of vaporisation
When 8.7 KJ of heat energy is required to vaporize 2.5 mol of liquid methanol.
Hence, for 1 mol of liquid methanol, amount of heat energy required to evaporate the methanol is = 
= 3.48 KJ
So, the heat of vaporization 
Therefore, the heat of vaporization of methanol is 3.48KJ/Mol
Answer:
It sounds fine, but it may be a bit too long. It's difficult to shorten things like this, but getting more straight to the point would give it that "catchy" feel.
Explanation:
Answer: False
Explanation:
Since the given equation is not balanced properly.
Since oxygen and hydrogen atoms are not balanced.
There should be 6 H2O (g) molecules and 14 mol H2 (g)