The heat of vaporization of benzene c6h6 , is 30.8 kJ/mol at its boiling point of 80.1 degrees Celsius. How much energy in the form of heat is required to vaporize 128 g benzene at its boiling point?
STEP 1 - convert the mass of benezene in moles of benzene.
Moles of benzene = mass of benzene / molar mass of benezene
= 128 g/ 78 gmol^-1
= 1.64 mol
STEP 2 multiply this mole number to heat of vaporization
Heat required to vaporize 128g benzene
= (1.64 mol)× ( 30.8 kj/mol)
= 50.5 kj (Answer)
The heat of vaporization of benzene c6h6 , is 30.8 kJ/mol at its boiling point of...
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calculate the heat required to melt 9.73 g of benzene at it's normal melting point nd Phases (References] a. Calculate the heat required to melt 9.73 g of benzene at its normal melting point. Heat of fusion (benzene) 9.92 kJ/mol Heat = kJ b. Calculate the heat required to vaporize 9.73 g of benzene at its normal boiling point. Heat of vaporization (benzene) = 30.7 kJ/mol Heat = kJ Submit Answer Try Another Version 2 item attempts remaining arch о...