The warmth of vaporization the water in ~ 373 K is 40.7 kJ/mol. Findq, w, E, and also H because that the evaporation of 479g of water in ~ thistemperature. I already found q, it"s(479g/18g/mol)*40.7kJ/mol=1080kJ how do I find w, deltaE, anddeltaH?
The warm of vaporization of water at 373


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Answer: (479g/(18.004g/mol))*(40.7kJ/mol) = 1080 kJ
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Transcribed picture text: The heat of vaporization that water at 373 K is 40.7 kJ/mol. Find q, w, E, and also H for the evaporation that 479g of water at this temperature. I already found q, it's (479g/18g/mol)*40.7kJ/mol=1080kJ exactly how do I discover w, deltaE, and deltaH? The warmth of vaporization the water at 373Delta H = ?g ofwater in ~ this temperature.Answer: H for the evaporation that 479Delta E = ?Find g of water at this temperature.Answer: E because that the evaporation that 479w = ?Find g of water at this temperature.Answer: w because that the evaporation the 479g that water at this temperature.Answer: (479g/(18.004g/mol))*(40.7kJ/mol) = 1080 kJFind q for the evaporation of 479kJ/mol.Find K is 40.7