How much heat, in joules and in calories, must be removed from 3.42 mol of water to lower its temperature from 25.2 to 10.9°C?
回答 (1)
2016-02-08 2:47 pm
(75.37 J/(mol·°C)) x (3.42 mol) x (25.2 - 10.9)°C = 3686 J
(3686 J) / (4.184 J/cal) = 881 cal
(3686 J) / (4.184 J/cal) = 881 cal
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