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Calculate the standard enthalpy change for the vaporization of 25 grams of water at 100 degrees Celsius and 1 atmosphere of pressure, given that the molar enthalpy of vaporization of water is 40.7 kJ/mol.

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To calculate the standard enthalpy change for the vaporization of 25 grams of water, we can use the following formula:H = n * Hvapwhere H is the enthalpy change, n is the number of moles of water, and Hvap is the molar enthalpy of vaporization.First, we need to find the number of moles of water  n . The molar mass of water  H2O  is 18.015 g/mol.n = mass / molar massn = 25 g / 18.015 g/moln  1.387 molesNow, we can calculate the enthalpy change:H = n * HvapH = 1.387 moles * 40.7 kJ/molH  56.5 kJThe standard enthalpy change for the vaporization of 25 grams of water at 100 degrees Celsius and 1 atmosphere of pressure is approximately 56.5 kJ.

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