At 1 atm pressure,the heat of sublimation of gallium is 277 kJ/mol and the heat of vaporization is 271 kJ/mol.To the correct number of significant figures,how much heat is required to melt 4.50 mol of gallium at 1 atm pressure?

Question

Quizplus · Accepted Answer

The heat of fusion is the difference between the heat of sublimation and the heat of vaporization. Therefore, the heat required to melt 4.50 mol of gallium is:$$\Delta H_{fusion} = \Delta H_{sublimation} - \Delta H_{vaporization}$$$$\Delta H_{fusion} = 277 kJ/mol - 271 kJ/mol = 6 kJ/mol$$Therefore, the heat required to melt 4.50 mol of gallium is:$$\Delta H_{fusion} = 6 kJ/mol * 4.50 mol = 27 kJ$$

At 1 Atm Pressure,the Heat of Sublimation of Gallium Is