B. Heat of Reaction Trial I Trial 2 volume of HCl, mL volume of NaOH, mL volume of solution, mL mass of solution, g = V x d 51.0mL 49.8 mL 48.4mL 50.5 mL 99.4mL 100.3mL (devolution T = 1.00 g/mL) solution 99.4 mL 100.3mL = T i, solution ATal i, cal Tf, cal = Tf, solution AT 'solution' (from your plot), °C 9cal, J = 21.0 J/°C XAT cal (from your plot), °C 22.8 °C 23.2°6 29.6°C 29.9°C °C = T - -T f, solution i, solution 6.8° 6-7°C 142.8 J 140.7J 9solution' J = m solution X 4.184 J/g °C XAT solution greaction' J = − (9cal + a solution) greaction, kJ moles of HCl used, mol (V x M = moles) AHDD, kJ/mol rxn' greaction, kJ == moles of HCI Average AH kJ/mol rxn'

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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Please help with filling in these missing pieces from trial 1 and 2. Thank you! 

B. Heat of Reaction
Trial I
Trial 2
volume of HCl, mL
volume of NaOH, mL
volume of solution, mL
mass of solution, g = V x d
51.0mL
49.8 mL
48.4mL
50.5 mL
99.4mL
100.3mL
(devolution
T
=
1.00 g/mL)
solution
99.4 mL
100.3mL
=
T
i, solution
ATal
i, cal
Tf, cal = Tf, solution
AT
'solution'
(from your plot), °C
9cal, J = 21.0 J/°C XAT
cal
(from
your plot), °C
22.8 °C
23.2°6
29.6°C
29.9°C
°C = T
-
-T
f, solution
i, solution
6.8°
6-7°C
142.8 J
140.7J
9solution' J = m solution
X 4.184 J/g °C XAT
solution
greaction' J = − (9cal + a solution)
greaction, kJ
moles of HCl used, mol (V x M = moles)
AHDD, kJ/mol
rxn'
greaction, kJ
==
moles of HCI
Average AH
kJ/mol
rxn'
Transcribed Image Text:B. Heat of Reaction Trial I Trial 2 volume of HCl, mL volume of NaOH, mL volume of solution, mL mass of solution, g = V x d 51.0mL 49.8 mL 48.4mL 50.5 mL 99.4mL 100.3mL (devolution T = 1.00 g/mL) solution 99.4 mL 100.3mL = T i, solution ATal i, cal Tf, cal = Tf, solution AT 'solution' (from your plot), °C 9cal, J = 21.0 J/°C XAT cal (from your plot), °C 22.8 °C 23.2°6 29.6°C 29.9°C °C = T - -T f, solution i, solution 6.8° 6-7°C 142.8 J 140.7J 9solution' J = m solution X 4.184 J/g °C XAT solution greaction' J = − (9cal + a solution) greaction, kJ moles of HCl used, mol (V x M = moles) AHDD, kJ/mol rxn' greaction, kJ == moles of HCI Average AH kJ/mol rxn'
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