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2. In an experiment similar to the procedure described in Part B of the lab procedure, 2.500 g of MgO were combined with 125

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Answer #1

The absorbed heat is calculated:

q = m * cp * ΔT = 125 g * 4.18 J / g * ° C * 9.6 ° C = 5016 J

Enthalpy change is calculated:

ΔH = - q * MM / m = - 5.02 kJ * 40.3 g / mol / 2.5 g = - 80.92 kJ / mol

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