How many kJ of heat are needed to completely vaporize 43.2 g of H2O? The heat of

Question

How many kJ of heat are needed to completely vaporize 43.2 g of H2O? The heat of vaporization for water at the boiling point is 40.6 kJ/mole.

How many kJ of heat are needed to completely melt 3.30 moles of H2O, given that the water is at its melting point? The heat of fusion for water is 6.02 kJ/mole.

How many kJ of heat are needed to completely vaporize 1.30 moles of H2O? The heat of vaporization for water at the boiling point is 40.6 kJ/mole.

How many joules of heat are needed to completely vaporize 43.2 grams of water at its boiling point?

Given ?Hvap = 40.6 kJ/mol

Explanation / Answer

a.Mass of water=43.2g

Moles of water=43.2/18=2.4mol

Heat required=2.4*40.6=97.44 kJ

b.Heat req=moles *heat of fusion=3.30*6.02=19.866 kJ

c.Heat req=1.30*40.6=52.78 kJ

d.Mass of water=43.2g

Moles of water=43.2/18=2.4mol

Heat required=2.4*40.6=97.44 kJ

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