Course Home P Acceptable units x + courseld=16709491&OpenVellumHMA C=f5c9929f4e4da0b5529475e262c91d79= 10001 1 Review art A alculate the heat change in calories for condensation of 11.0 g of steam at 100°C. xpress your answer as a positive value using three significant figures and inc 2 MIKIN M HA Value CS​

The heat change in calories for condensation of 11.0 g of steam at 100°C is 5930 calories

Explanation:

Latent heat of condensation is the heat released when one mole of steam or water vapor condenses to form liquid droplets. The heat of condensation of water at 100° C is about 2,260 kJ/kg, which is equal to 40.68 kJ/mol. Since condensation of steam and vaporization of water occur at the same temperature and require the same amount of energy to occur, the heat of condensation is exactly equal to the heat vaporization, but has the opposite sign. In the vaporization, heat energy is absorbed by the substance, whereas in condensation heat energy is released by the substance.

The specific latent heat of vaporization of steam at 100° C = 40.68 kJ/mol

Number of moles of moles of water in 11.0 g of steam = mass/ molar mass

Molar mass of water = 18.0 g/mol

Number of moles of steam = 11.0 g / 18.0 g/mol = 0.61 moles

Heat released = 40.68 K/mol × 0.61 moles = 24.815 kJ

Converting to kcal by dividing 24.815 kJ by 4.184 = 5.93 kcal or 5930 calories

Therefore, the heat change in calories for condensation of 11.0 g of steam at 100°C is 5930 calories

c. 75.7

i think that is the right