How much energy is required to vaporize 185 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol. express your answer to three significant figures and include the appropriate units.

73.5 kJ

Step-by-step explanation:

Mass of butane = 185 g

Heat of vaporization for butane = 23.1 kJ/mol

Molar mass of butane = 58.12 g/mol

Number of moles of butane =

Energy required for burning 185 g of butane = 3.18×23.1 = 73.5 kJ

∴ Energy is required to vaporize 185 g of butane at its boiling point is 73.5 kJ

all but the second one

step-by-step explanation:

cuz that one goes over 180

mark brainliest