Boiling point: 100.0°C Heat of vaporization: 540 cal/g Melting point = 0.0°C Heat of fusion = 79.7 cal/g How many calories of energy are needed to vaporize a 25.3 g sample of liquid water at its norm2022-05-24 04:42:53GraduateWriterhelp.com Answers
Boiling point: 100.0°CHeat of vaporization: 540 cal/g Melting point = 0.0°CHeat of fusion = 79.7 cal/gHow many calories of energy are needed to vaporize a 25.3 g sample of liquid water at its normal boiling point of 100.0°C?
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