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The molar enthalpies of fusion and vaporization of water at 298 K298 \mathrm{~K} are 6.016.01 and 44.01 kJ mol144.01 \mathrm{~kJ} \mathrm{~mol}^{-1}, respectively. Use these values to estimate the molar enthalpy of sublimation of ice.

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Given:

The molar enthalpy of fusion of water=6.01kJmol\mathrm{The \ molar \ enthalpy \ of \ fusion \ of \ water= 6.01 \dfrac{kJ}{mol}} The molar enthalpy of vaporization of water=44.01kJmol\mathrm{The \ molar \ enthalpy \ of \ vaporization \ of \ water= 44.01\dfrac{kJ}{mol}}

The molar enthalpy of sublimation of water=?\mathrm{The \ molar \ enthalpy \ of \ sublimation \ of \ water=?}

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