50.0 mL 2.00M HCI and 50.0 mL 2.00 M NaOH mixed: HCI(aq) + NaOH(aq) → H,0() + NaCI(aq) - Csoln = Cwater = 4.184 J/(g °C) dsoln = dwater = 1.00 g/mL - mgoln = (50 mL + 50 mL) = 100 mL = 100 g - ΔΤ, %3D %3D %3D %3D AT 13.8 °C (same mass water, twice the moles) soln Asoin = (100 g)(4.184 J/(g °C)(13.8 °C) = +5773.92 J - Moles HCI: (M*L) = (2.00 M)(0.050 L) = 0.100 moles - Moles NaOH = (2.00 M)(0.050 L) = 0.100 moles %3D %3D %3D %3D %3D 9soln = +5773.92 J/0.100 moles = +57,739.2 J J/mole AHn = -57.7 kJ/mole %3D rxn Pearson Education, Inc.
50.0 mL 2.00M HCI and 50.0 mL 2.00 M NaOH mixed: HCI(aq) + NaOH(aq) → H,0() + NaCI(aq) - Csoln = Cwater = 4.184 J/(g °C) dsoln = dwater = 1.00 g/mL - mgoln = (50 mL + 50 mL) = 100 mL = 100 g - ΔΤ, %3D %3D %3D %3D AT 13.8 °C (same mass water, twice the moles) soln Asoin = (100 g)(4.184 J/(g °C)(13.8 °C) = +5773.92 J - Moles HCI: (M*L) = (2.00 M)(0.050 L) = 0.100 moles - Moles NaOH = (2.00 M)(0.050 L) = 0.100 moles %3D %3D %3D %3D %3D 9soln = +5773.92 J/0.100 moles = +57,739.2 J J/mole AHn = -57.7 kJ/mole %3D rxn Pearson Education, Inc.
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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For the part circled, what if the molarity is different and one has a molarity of 3 and one has M=2. When we normalize, are we dividing by the limiting reactant?
Transcribed Image Text:Calorimetry Example: Understanding AHn
If 50.0 mL 2.00 M HCI and 50.0 mL 2.00 M NaOH mixed:
HCI(aq) + NaOH(aq) →
H20(1) + NaCl(aq)
- Csoln = Cwater = 4.184 J/(g °C)
– dooln = dwater = 1.00 g/mL
- msoln = (50 mL + 50 mL) = 100 mL = 100 g
%3D
%3D
- ΔΤ,
ATSoln = 13.8 °C (same mass water, twice the moles)
%3D
9soln = (100 g)(4.184 J/(g °C)(13.8 °C) = +5773.92 J
Moles HCI: (M*L) = (2.00 M)(0.050 L) = 0.100 moles
Moles NaOH = (2.00 M)(0.050 L) = 0.100 moles
geoln = +5773.92 J /0.100 moles = +57,739.2 J J/mole
%3D
%3D
%3D
%3D
%3D
%3D
AHn =-57.7 kJ/mole
rxr
© 2014 Pearson Education, Inc.
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