I need help understanding this problem
1 answer:
You might be interested in
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:
Look at the picture.
Explanation:
(2S,3S)-2-Bromo-3-phenylbutane will undergo E2 reaction and form trans product of elimination due to its thermodynamic stability.
