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My question is for the part circled in pencil, how do we get -12,970.4 and then -25,940.8 j/mole. Are we normalizing twice? If so, why?

My question is for the part circled in pencil, how do we get -12,970.4 and then -25,940.8 j/mole. Are we normalizing twice? If so, why?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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My question is for the part circled in pencil, how do we get -12,970.4 and then -25,940.8 j/mole. Are we normalizing twice? If so, why? 

Chemistry I Science Supplemental Problems from Chapter 7 CCC - Dr. Feudale 4. When 500 mL of 1.00 M barium nitrate solution at 25°C is mixed with 500 mL of 1.00 M sodium sulfate solution at 25°C, a white solid forms and the temperature of the mixture increases to 28.1°C. Calculate AH for this precipitation reaction per mole of BaSO4. Net Ionic reaction: Ba2+(aq) + SO, (aq) → BaSO4(s) Assume no heat lost to the surroundings: Heat released by the reaction = Heat absorbed by solution Assume Csoln = Cwater = 4.184 J/g-°C and dsoln = dwater = 1.00 g/mL Because the temperature of the solution increased, the reaction must be exothermic -AHreaction = qsolution = msolnCsolnATsoln -AHreaction = -12,970.4 J Moles of Ba2+ = Moles of SO42- = Moles of BaSO4 = (0.500 L)(1.00 M) = 0.500 moles %3D %3D %3D -AHreaction = -12,970.4 J/0.500 moles =-25,940.8 J/mole = -25.9 kJ/mole CCC - Dr. Feudale Chemistry I Science Supplemental Problems from Chapter 7 4. When 500 mL of 1.00M barium nitrate solution at 25°C is mixed with 500 mL of 1.00 M sodium sulfate solution at 25°C, a white solid forms and the temperature of the mixture increases to 28.1°C. Calculate AH for this precipitation reaction per mole of BaSO4. 1.00g 4e184y 3.1°C I mi 19°C 6485.2 -12,970.43=-AH %3D 500ml %3D 0.5
Transcribed Image Text:Chemistry I Science Supplemental Problems from Chapter 7 CCC - Dr. Feudale 4. When 500 mL of 1.00 M barium nitrate solution at 25°C is mixed with 500 mL of 1.00 M sodium sulfate solution at 25°C, a white solid forms and the temperature of the mixture increases to 28.1°C. Calculate AH for this precipitation reaction per mole of BaSO4. Net Ionic reaction: Ba2+(aq) + SO, (aq) → BaSO4(s) Assume no heat lost to the surroundings: Heat released by the reaction = Heat absorbed by solution Assume Csoln = Cwater = 4.184 J/g-°C and dsoln = dwater = 1.00 g/mL Because the temperature of the solution increased, the reaction must be exothermic -AHreaction = qsolution = msolnCsolnATsoln -AHreaction = -12,970.4 J Moles of Ba2+ = Moles of SO42- = Moles of BaSO4 = (0.500 L)(1.00 M) = 0.500 moles %3D %3D %3D -AHreaction = -12,970.4 J/0.500 moles =-25,940.8 J/mole = -25.9 kJ/mole CCC - Dr. Feudale Chemistry I Science Supplemental Problems from Chapter 7 4. When 500 mL of 1.00M barium nitrate solution at 25°C is mixed with 500 mL of 1.00 M sodium sulfate solution at 25°C, a white solid forms and the temperature of the mixture increases to 28.1°C. Calculate AH for this precipitation reaction per mole of BaSO4. 1.00g 4e184y 3.1°C I mi 19°C 6485.2 -12,970.43=-AH %3D 500ml %3D 0.5
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