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
<span>In physics, the law of conservation of energy states that the total energy of an isolated system remains constant—it is said to be conserved over time. Energy can neither be created nor destroyed; rather, it transforms from one form to another.</span>
Moles of phosphoric acid would be needed : 0.833
<h3>Further explanation</h3>
Given
15 grams of water
Required
moles of phosphoric acid
Solution
Reaction(decomposition) :
H3PO4 -> H2O + HPO3
mol water (H2O :
= mass : MW
= 15 g : 18 g/mol
= 0.833
From the equation, mol ratio H3PO4 = mol H2O = 1 : 1, so mol H3PO4 = 0.833
According to the valence shell electron pair repulsion (VSPER) theory, an ammonia molecule <span> has a </span>trigonal pyramidal<span> shape with an experimental bond angle measure of 106.7 degrees. This is why it is difficult to accurately represent ammonia two-dimensionally because the molecular structure entails a 3-D projection with angles in it unlike the linear structure.</span>
Lead fluoride hope this helps
