Answer:
8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.
Explanation:
Moles of ethanol of ethanol = 0.200 mol
Heat required to heat 0.200 moles of ethanol = Q = 1.05 kJ
Enthalpy of vaporization of ethanol = 
Heat required to vaporize 0.200 moles of ethanol = Q'

Total heat required to fore heating and the vaporizing :
= Q + Q' = 1.05 kJ + 7.72 kJ = 8.77 kJ
8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.
Answer: The volume for 0.850 mol of
from a
solution is 1700 mL.
The volume of 30.0 g of LiOH from a 2.70 M LiOH solution is 464 mL.
Explanation:
Molarity is the number of moles of solute present in a liter of solution.
- As given moles of
are 0.850 mol and molarity of
solution is 0.5 M. Hence, its volume is calculated as follows.

Therefore, the volume for 0.850 mol of
from a
solution is 1700 mL.
- As given mass of LiOH are 30.0 g from a 2.70 M LiOH (molar mass = 23.95 g/mol) solution. Hence, its number of moles are calculated as follows.

So, volume for LiOH solution is calculated as follows.

Therefore, volume of 30.0 g of LiOH from a 2.70 M LiOH solution is 464 mL.
573 degrees F
hope this helped
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