Answer:
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Answer:
9.21 kJ is the enrgy required to vaporize 11.0 grams of ethanol.
Explanation:
Energy of vaporization can be stated as 38.6 kj/mole for ethanol. [<u>"It takes 38. 6 kj of energy to vaporize 1. 00 mol of ethanol"</u>]
11.0 grams of ethanol is not 1 mole. Calculate the moles of ethanol by dividing t=it's mass by its molar mass:
(11.0 grams)/(46.07 g/mole) = 0.2388 moles ethanol
(0.2388 moles ethanol)*(38.6 kj/mole) = 9.21 kJ is the enrgy required to vaporize 11.0 grams of ethanol.