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What is the difference between lower heating value and higher heating value if the product has 2 moles of H2O and Enthalpy of vaporization is 40 J/mole at standard conditions?(a) -40 J(b) 40 J(c) -80 J(d) 80 J |
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Answer» Right answer is (c) -80 J The explanation is: LHV – HHV = -2*40 = -80 J. |
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