Hint: This is the diluted : heat capacity = 3.184 J / g °C and D = 1.00 g/ moleHeat of reaction = - (heat of solution) - heat of calorimetryc. Calculate the heat of reaction for the following chemical reaction:NaOH (s) + HC1(aq) --------→ NaC(aq) + H2O(1)Using the following reactions:NaOH(s) --→ NaOH(aq).AH = -102 kj / moleNaOH(aq) + HCl(aq) --calculations.-→ NaCl(aq) + H2O(1). use the heat of reaction from your- - Chem. 1A : Thermochemistry1. Determination of heat capacity of calorimeter:50.0 mL. H2O at the initial temperature of 23.5 °C is added to a calorimeter (coffee cup) andstirred and kept for 5 minutes. Then 50.0 mL. of hot water at the temperature of 98.0 °C isadded to the calorimeter contained the 50.0 mL. H2O at 23.5 °C. The temperature of themixture is recorded every 30 second, the following data is obtained:Time in secondTemperature in °C3055.06050.09045.512042.015038.018037.521037.524037.027037.030037.033037.0D of H2O = 1.00 g/ mL.Heat capacity of H2O (1) = 4.184 J/ g °Ca. Plot temperature vs. time to get the final temperature.b. Calculate the heat capacity of the calorimeter.Hint: the heat lost by hot water = heat gained by cold water + heat gained by calorimeter2. Determination heat of reaction.50.0 mL. of 0.100 M NaOH(aq) at the temperature of 22.0 °C is added to a calorimeter(coffee cup). Then 50.0 mL. of HCl(aq) at the temperature of 21.0 °C is added to thecalorimeter containing NaOH(aq). The temperature is measured every 30 seconds , thefollowing data is obtained:Time in secondTemperature in °C3025.06026.59028.012029.515029.018029.521029.5a. Plot temperature vs. time to get the final temperature.b. Calculate the heat of reaction in kJ/ mole

"Since you have asked multiple questions, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.1 a) Plot of Temperature vs Time The final temperature can be calculated from the graph. At Tfinal, x=0 Straight line equation from the graph y=-0.0539 x + 50.936 put x = 0 and solve for Tfinal Tfinal = 50.936 oC b) To Calculate the heat capacity of Calorimeter Heat lost by water = Heat gained by cold water + heat gained by calorimeter -qhot water=qcold water+qcalorimeter-mhot water×Chot water×∆Thot water=mcold water×Ccold water×∆Tcold water+ Ccal×∆Tcal (1)mhot water= Volume×density =50 mL×1.00 g/mL = 50 gChot water=4.184 Jg°C∆Thot water=Tf-Thot water=50.936°C-98°C =-47.064°Cmcold water= Volume×density =50 mL×1.00 g/mL = 50 gCcoldwater=4.184 Jg°C∆Tcoldwater=Tf-Tcold water=50.936°C-23.5°C =27.436 °C∆Tcalr=Tf-Tcal=50.936°C-23.5°C =27.436 °CNow, substitute these values in equation(1) and solve for Ccal-50 g×4.184 Jg°C×-47.064°C=50 g×4.184 Jg°C×27.436 °C+ Ccal×27.436 °C9845.7888 = 5739.6112 +27.436 CcalCcal = 149.66 J°CANSWER 1a) Final Temperature Tfinal = 50.936 oC 1b) Heat Capacity of the Calorimeter Ccal = 149.66 J/oC

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Chemistry

1. Determination of heat capacity of calorimeter: 50.0 mL. H2O at the initial temperature of 23.5 °C is added to a calorimeter (coffee cup) and stirred and kept for 5 minutes. Then 50.0 mL. of hot water at the temperature of 98.0 °C is added to the calorimeter contained the 50.0 mL. H2O at 23.5 °C. The temperature of the mixture is recorded every 30 second, the following data is obtained:

1. Determination of heat capacity of calorimeter: 50.0 mL. H2O at the initial temperature of 23.5 °C is added to a calorimeter (coffee cup) and stirred and kept for 5 minutes. Then 50.0 mL. of hot water at the temperature of 98.0 °C is added to the calorimeter contained the 50.0 mL. H2O at 23.5 °C. The temperature of the mixture is recorded every 30 second, the following data is obtained:

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

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Question

Hint: This is the diluted : heat capacity = 3.184 J / g °C and D = 1.00 g/ mole
Heat of reaction = - (heat of solution) - heat of calorimetry
c. Calculate the heat of reaction for the following chemical reaction:
NaOH (s) + HC1(aq) --------→ NaC(aq) + H2O(1)
Using the following reactions:
NaOH(s) -
-→ NaOH(aq).
AH = -102 kj / mole
NaOH(aq) + HCl(aq) --
calculations.
-→ NaCl(aq) + H2O(1). use the heat of reaction from your
- -

Chem. 1A : Thermochemistry
1. Determination of heat capacity of calorimeter:
50.0 mL. H2O at the initial temperature of 23.5 °C is added to a calorimeter (coffee cup) and
stirred and kept for 5 minutes. Then 50.0 mL. of hot water at the temperature of 98.0 °C is
added to the calorimeter contained the 50.0 mL. H2O at 23.5 °C. The temperature of the
mixture is recorded every 30 second, the following data is obtained:
Time in second
Temperature in °C
30
55.0
60
50.0
90
45.5
120
42.0
150
38.0
180
37.5
210
37.5
240
37.0
270
37.0
300
37.0
330
37.0
D of H2O = 1.00 g/ mL.
Heat capacity of H2O (1) = 4.184 J/ g °C
a. Plot temperature vs. time to get the final temperature.
b. Calculate the heat capacity of the calorimeter.
Hint: the heat lost by hot water = heat gained by cold water + heat gained by calorimeter
2. Determination heat of reaction.
50.0 mL. of 0.100 M NaOH(aq) at the temperature of 22.0 °C is added to a calorimeter
(coffee cup). Then 50.0 mL. of HCl(aq) at the temperature of 21.0 °C is added to the
calorimeter containing NaOH(aq). The temperature is measured every 30 seconds , the
following data is obtained:
Time in second
Temperature in °C
30
25.0
60
26.5
90
28.0
120
29.5
150
29.0
180
29.5
210
29.5
a. Plot temperature vs. time to get the final temperature.
b. Calculate the heat of reaction in kJ/ mole

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