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------------------------------------------------ How much heat must be removed to freeze a tray of ice cubes if the water has a mass of 40.0 g? (The molar heat of fusion of water is 6.01 kJ/mol.)

A. 36.6 kJ
B. 13.4 kJ
C. 25.1 kJ
D. 15.9 kJ

To freeze a tray of ice cubes, we need to remove the heat from the water until it reaches the freezing point and then convert it into ice. the heat required to freeze a substance is given by the equation Q = m × ΔHf, where Q is the heat, m is the mass, and ΔHf is the molar heat of fusion.

Explanation:

To freeze a tray of ice cubes, we need to remove the heat from the water until it reaches the freezing point and then convert it into ice. the heat required to freeze a substance is given by the equation Q = m × ΔHf, where Q is the heat, m is the mass, and ΔHf is the molar heat of fusion.

From the given information, the molar heat of fusion of water is 6.01 kJ/mol. We need to convert the mass of water from grams to moles before calculating the heat. the molar mass of water is approximately 18.0 g/mol, so the number of moles of water is calculated as 40.0 g ÷ 18.0 g/mol = 2.22 mol.

Therefore, the heat required to freeze the water is Q = 2.22 mol × 6.01 kJ/mol = 13.39 kJ. So, the answer to the question is b. 13.4 kJ.

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