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What is the standard enthalpy change (ΔH°) for the sublimation of iodine (I2) at 298 K, given that the heats of fusion and vaporization of iodine are 15.5 kJ/mol and 41.7 kJ/mol respectively?

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The standard enthalpy change  H  for the sublimation of iodine  I2  can be calculated by adding the heats of fusion and vaporization together. This is because sublimation is the process of a solid converting directly into a gas, which involves both fusion  solid to liquid  and vaporization  liquid to gas .H sublimation  = H fusion  + H vaporization Given the heats of fusion and vaporization of iodine are 15.5 kJ/mol and 41.7 kJ/mol respectively, we can calculate the standard enthalpy change for sublimation:H sublimation  = 15.5 kJ/mol + 41.7 kJ/mol = 57.2 kJ/molSo, the standard enthalpy change for the sublimation of iodine  I2  at 298 K is 57.2 kJ/mol.

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