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What is the standard enthalpy change for the neutralization of 25.0 mL of 0.100 M hydrochloric acid with 25.0 mL of 0.100 M sodium hydroxide, given that the specific heat capacity of the solution is 4.18 J/g°C and the temperature change is 5.0°C?

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To calculate the standard enthalpy change for the neutralization reaction, we need to first determine the heat released during the reaction  q  and then divide by the moles of the limiting reactant.The balanced equation for the neutralization reaction is:HCl aq  + NaOH aq   NaCl aq  + H2O l Since both reactants have the same concentration and volume, they will react in a 1:1 ratio, and neither is in excess.First, let's find the heat released  q  using the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity, and T is the temperature change.The total volume of the solution is 25.0 mL + 25.0 mL = 50.0 mL. Assuming the density of the solution is approximately 1 g/mL  similar to water , the mass of the solution is 50.0 g.Now, we can calculate q:q =  50.0 g  4.18 J/gC  5.0C  = 1045 JNext, we need to determine the moles of the limiting reactant. We can use the volume and concentration of either HCl or NaOH since they react in a 1:1 ratio.moles =  volume  concentration  =  0.025 L  0.100 mol/L  = 0.00250 molFinally, we can calculate the standard enthalpy change  H  by dividing the heat released by the moles of the limiting reactant:H = q / moles = 1045 J / 0.00250 mol = 418000 J/mol = 418 kJ/molThe standard enthalpy change for the neutralization of 25.0 mL of 0.100 M hydrochloric acid with 25.0 mL of 0.100 M sodium hydroxide is 418 kJ/mol.

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