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Calculate the standard enthalpy change for the neutralization of HCl with NaOH, given that the heat of formation of water is -285.8 kJ/mol, the heat of formation of NaCl is -411.1 kJ/mol, and the initial concentrations of HCl and NaOH are 1.0 M.

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To calculate the standard enthalpy change for the neutralization of HCl with NaOH, we can use the following balanced chemical equation:HCl aq  + NaOH aq   NaCl aq  + H2O l We can use Hess's Law to determine the enthalpy change for this reaction. Hess's Law states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps that make up the overall reaction.In this case, we can break the reaction down into two steps:1. Formation of water from H2 and O2:H2 g  + 1/2 O2 g   H2O l   Hf = -285.8 kJ/mol2. Formation of NaCl from Na and Cl2:Na s  + 1/2 Cl2 g   NaCl aq   Hf = -411.1 kJ/molNow, we can use the enthalpy changes for these two steps to determine the enthalpy change for the overall reaction. Since the reaction involves the formation of 1 mole of water and 1 mole of NaCl, we can simply add the enthalpy changes for the two steps:H = Hf H2O  + Hf NaCl H =  -285.8 kJ/mol  +  -411.1 kJ/mol H = -696.9 kJ/molTherefore, the standard enthalpy change for the neutralization of HCl with NaOH is -696.9 kJ/mol.

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