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The heat vaporization of water is 40.66 kj/mol . .? The heat vaporization of water is 40.66 kj/mol. How much heat is absorbed when 3.45g of water boils at atmospheric pressure?

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To answer the question above, first convert the given amount of water from grams to mol by dividing it with its molar mass which is 18 g/mol. 
                           amount in mol = 3.45 g x (mol/ 18 g) = 23/120 mol
Then, multiply this amount in mol with the heat of vaporization given.
                        (23/120) mol x (40.66 kj/mol) = 7.793 kJ
Thus, the heat absorbed is approximately equal to 7.793 kJ. 

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