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Vlad1618 [11]
3 years ago
8

The molar heat of fusion for water is 6.01 kJ/mol. How much energy must be added to a 75.0-g block of ice at 0°C to change it to

75.0 g of liquid water at 0°C?
Chemistry
2 answers:
andreyandreev [35.5K]3 years ago
7 0
Answer is: 25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.
bearhunter [10]3 years ago
4 0

Answer:

25.0 kJ

Explanation:

This is correct on edge.

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