First, we need to calculate the heat q released during the reaction using the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity, and T is the temperature change.Since the density of the solutions is 1.00 g/mL, the mass of the 50.0 mL HCl solution and the 50.0 mL NaOH solution are both 50.0 g. When combined, the mass of the resulting solution is:m = 50.0 g HCl + 50.0 g NaOH = 100.0 gThe specific heat capacity c of the solution is given as 4.18 J/gC, and the temperature change T is -5.01C. Now we can calculate the heat q released:q = 100.0 g 4.18 J/gC -5.01C = -2095.18 JSince the reaction is exothermic, the heat released is negative.Next, we need to determine the moles of HCl and NaOH reacting. We can use the formula:moles = volume concentrationFor HCl:moles = 50.0 mL 0.100 mol/L = 5.00 mmolFor NaOH:moles = 50.0 mL 0.100 mol/L = 5.00 mmolSince the balanced chemical equation shows a 1:1 ratio between HCl and NaOH, both reactants will be completely consumed in the reaction, and 5.00 mmol of NaCl and H2O will be produced.Finally, we can calculate the standard enthalpy change H for the reaction by dividing the heat released q by the moles of HCl and NaOH reacted:H = q / moles = -2095.18 J / 5.00 mmol = -419.04 J/mmolThe standard enthalpy change for the reaction between 50.0 mL of 0.100 M HCl and 50.0 mL of 0.100 M NaOH is -419.04 J/mmol.