Answers by AmieBoyland

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Calculate the standard enthalpy change for the following reaction at 298 K: 2NaHCO3(s) + MgCl2(aq) → MgCO3(s) + 2NaCl(aq) + H2O(l)Given the following standard enthalpies of formation: ΔHf°(NaHCO3) = -950.7 kJ/molΔHf°(MgCl2) = -641.8 kJ/molΔHf°(MgCO3) = -1128.2 kJ/molΔHf°(NaCl) = -411.2 kJ/molΔHf°(H2O) = -285.8 kJ/mol

answered 2 days ago in Chemical thermodynamics

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