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What is the standard enthalpy change for the reaction between 50 mL of 0.1 M HCl(aq) and 50 mL of 0.1 M NaOH(aq) at 25°C, given that the resulting solution has a density of 1.05 g/mL and a heat capacity of 4.184 J/goC?

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To calculate the standard enthalpy change for the reaction between HCl and NaOH, we need to first determine the amount of heat released during the reaction and then divide it by the moles of the reaction.The reaction between HCl and NaOH is an acid-base neutralization reaction, which can be represented as:HCl aq  + NaOH aq   NaCl aq  + H2O l Since both solutions have the same concentration and volume, the reaction will go to completion, and all of the HCl and NaOH will react.First, let's find the moles of HCl and NaOH:moles = Molarity  Volumemoles of HCl = 0.1 M  0.05 L = 0.005 molesmoles of NaOH = 0.1 M  0.05 L = 0.005 molesNow, let's find the mass of the resulting solution:mass = density  volumemass = 1.05 g/mL  100 mL = 105 gNext, we need to determine the heat released during the reaction. We can use the heat capacity and the temperature change to find this. Since the reaction is exothermic, the temperature of the solution will increase. Let's assume the temperature change is T.q = mass  heat capacity  Tq = 105 g  4.184 J/goC  TNow, we need to find the standard enthalpy change  H  for the reaction. We can do this by dividing the heat released  q  by the moles of the reaction:H = q / molesSince we have one mole of HCl reacting with one mole of NaOH, the moles of the reaction are equal to the moles of HCl or NaOH:H =  105 g  4.184 J/goC  T  / 0.005 molesTo find the standard enthalpy change, we need to know the temperature change  T  during the reaction. This information is not provided in the problem, so we cannot calculate the exact value of H. However, the equation above can be used to find H if the temperature change is known.

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