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What is the standard enthalpy change for the vaporization of 10 g of water at a temperature of 100°C and atmospheric pressure?

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To calculate the standard enthalpy change for the vaporization of 10 g of water at 100C and atmospheric pressure, we need to use the following formula:H = m  Hvapwhere H is the enthalpy change, m is the mass of water, and Hvap is the molar enthalpy of vaporization.The molar enthalpy of vaporization for water at 100C and atmospheric pressure is approximately 40.7 kJ/mol.First, we need to convert the mass of water  10 g  to moles. The molar mass of water is 18.015 g/mol.moles of water = mass / molar massmoles of water = 10 g / 18.015 g/molmoles of water  0.555 molNow, we can calculate the enthalpy change:H = m  HvapH = 0.555 mol  40.7 kJ/molH  22.6 kJThe standard enthalpy change for the vaporization of 10 g of water at 100C and atmospheric pressure is approximately 22.6 kJ.

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