Verified answer

Use the table in Appendix 4 of your textbook to find the corresponding values of each compound.

Compound = standard molar enthalpy of formation kJ/mol

MgO(s) = -601.8 kJ/mol

H2O(l) = -285.8 kJ/mol

Mg(OH)2(s) = -924.7 kJ/mol

ΔHf° = ∑ΔHf°products - ∑ΔHf°reactants

ΔHf° = [ΔHf° Mg(OH)2] - [ΔHf° MgO + ΔHf° H2O]

ΔHf° = [(1 mol Mg(OH)2)(-924.7 kJ/mol Mg(OH)2] - [(1 mol MgO)(-601.8 kJ/mol) + (1mol H2O)(-285.8 kJ/mol)]

ΔHf° = -924.7 kJ - (-601.8 kJ + -285.8 kJ)

ΔHf° = -924.7 kJ - (-887.6 kJ)

ΔHf° = -37.1 kJ

good luck!~

hope i helped :)