How much heat is absorbed, in kJ, by 100.0 mL of solution of NaCl which undergoe

Question

How much heat is absorbed, in kJ, by 100.0 mL of solution of NaCl which undergoes a temperature change from 23.5 degree C to 28.7 degree C? The density of the solution is 1.02 g/mL, and its specific heat capacity is 4.03 J/(g middot degree C). If 50.0 mL of 1.0 M HCl are mixed with 50.0 mL of 1.0 M NaOH [as in section (a) of part B], how many moles of each reactant were present? How many moles of water will be formed by this reaction? If you have determined the amount of heat given off in this process, q, and wish to convert it to Delta H on a molar basis, what value of n should be used in the equation Delta H_m = q/n? Show specifically how reactions a and c in Part B of this experiment can be combined in such a way that they add up to reaction d.

Explanation / Answer

1)

volume = 100 mL

density = 1.02 g / mL

mass = density x volume

         = 1.02 x 100

        = 102 g

Cp = 4.03 J / oc g

dT = 28.7 - 23.5 = 5.2 oC

Q = m Cp dT

Q = 102 x 4.03 x 5.2

Q = 2138 J

Q = 2.14 kJ

heat absorbed = 2.14 kJ

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