Assume you mix (1.25x10^2) mL of 0.400 M HCl with the same volume of 0.400 M...

Assume you mix (1.25x10^2) mL of 0.400 M HCl with the same volume of 0.400 M NaOH in a coffee-cup calorimeter (as will be done in the lab). The temperature of the solution before the mixing was 25.10 deg. C; after mixing and allowing the reaction to occur, the temperature is 27.78 Deg. C. What is the enthalpy of neutralization of the acid. Assume the densities of all the solutions are 1.00 g/mL and their specific heat capacities are 4.18 J/(g*K)

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Expert Solution

no of moles of HCl = molarity * volume in L

= 0.4*0.125 = 0.05 moles

no of moles of NaOH = molarity * volume in L

= 0.4*0.125 = 0.05 moles

Total volume of solution = 125 + 125 = 250ml

mass of solution = volume of solution * density

= 250*1 = 250g

q = mcT

= 250*4.18*(27.78-25.1)

= 2800.6J

H = -2800.6J/0.05moles = -56012J/mole = -56.012KJ/mole >>>>>answer

queen_honey_blossom answered 1 year ago

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