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What is the standard enthalpy change for the reaction between 50.0 mL of 0.100 M HCl(aq) and 50.0 mL of 0.100 M NaOH(aq) at 25 °C, given that the temperature change of the mixture was 10.0 °C and the specific heat capacity of the solution is 4.18 J/(g·K)?

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To calculate the standard enthalpy change for the reaction, we first need to determine the heat released or absorbed during the reaction  q . We can do this using the formula:q = mcTwhere m is the mass of the solution, c is the specific heat capacity, and T is the temperature change.First, let's find the mass of the solution. Since the density of water is approximately 1 g/mL, we can assume that the density of the solution is also approximately 1 g/mL. Therefore, the mass of the solution is:m =  50.0 mL HCl + 50.0 mL NaOH  * 1 g/mL = 100.0 gNow, we can calculate the heat released or absorbed during the reaction:q =  100.0 g  *  4.18 J/ gK   *  10.0 C  = 4180 JNext, we need to determine the moles of the limiting reactant. Since both HCl and NaOH have the same concentration and volume, they will react in a 1:1 ratio. Therefore, we can calculate the moles of either reactant:moles =  0.100 mol/L  *  0.050 L  = 0.005 molFinally, we can calculate the standard enthalpy change for the reaction  H  by dividing the heat released or absorbed by the moles of the limiting reactant:H = q / moles = 4180 J / 0.005 mol = 836000 J/molThe standard enthalpy change for the reaction between 50.0 mL of 0.100 M HCl aq  and 50.0 mL of 0.100 M NaOH aq  at 25 C is approximately 836 kJ/mol.

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