Answer :
This is a chemistry application question that can be solved by understanding the molar heat of vaporization. 995.3 grams of ethanol can be boiled with 835.3 kJ of heat energy.
The subject of this question falls under Chemistry. To solve it, we need to know the molar heat of vaporization of ethanol which is given as 38.6 kJ/mol. From this, we can calculate the number of moles of ethanol that can be vaporized or boiled with the provided energy (835.3 kJ).
Firstly, we calculate the number of moles. The number of moles is obtained by dividing the heat energy provided by the molar heat of vaporization. This gives
835.3 kJ / 38.6 kJ/mol = 21.63 moles.
Secondly, knowing that the molecular weight of ethanol is approximately 46.07 g/mol, we can convert these moles into grams. This is achieved through simple multiplication:
21.63 moles * 46.07 g/mol = 996.5 g.
996.5 grams of ethanol can be boiled using 835.3 kJ of heat energy. This is a direct application of thermodynamics in Chemistry.
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