Question

Q20. 18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol-1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalpy of vapourisation for water?​

Answer 1

Explanation:

Sol: Enthalpy of a reaction is the energy change per mole for the process.

18 g of H20 = 1 mole (∆Hvap = 40.79 kJ moE1)

Enthalpy change for vapourising 2 moles of H20 = 2 x 40.79 = 81.58 kJ ∆H°vap = 40.79 kJ mol -1

Answer 2

Answer:

carbocations in the order of the site CH3 CHỊCH-CCH, CH..OHCS i ham coniugate acid and ...

1 vote

Explanation:Sol: Enthalpy of a reaction is the energy change per mole for the process.18 g of H20 = 1 mole (∆Hvap = 40.79 kJ moE1)Enthalpy change