Respuesta :
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 = [tex]\Delta H_{vap}=38.6 kJ/mol[/tex]
Heat required to vaporize 0.200 moles of ethanol = Q'
[tex]Q'=\Delta H_{vap}\times 0.200 mol=38.6 kJ/mol\times 0.200 mol=7.72 kJ[/tex]
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.
