Chemistry, asked by adhyapratiksha, 1 year ago

The enthalpy of vaporization of liquid water using the date
H2(g)+ 12O2(g) H2O(l), ∆H = −285.77 kJ/mol
H2(g)+ 12O2(g) H2O (g) ∆H = −241.84 kJ/mol

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Answered by sarthakkhurana8
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