My working out is given Heat of capacity is 4.68 J g–1 K–1 and M (mass)= 41.9g-21.2g = 63.1g then temperature, delta T= 43.7 °C- 29.8 °C = 13.9 °C. After you have the delta H or Q = MC(delta)T equation = 63.1 x 4.68 x 13.9 = 4104.7812, but what do you do with the calorimeter constant of 34.7 J K–1 (is that there to trick students or do I use it in conjunction to the Heat of capacity of 4.68 J g–1 K–1 ?I have a second image of a example of a tutor working out but it has two issue of accidently forgetting a minus sign and maybe not including the calorimeter constant. 75.1 g of a basic solution was mixed with 32.5 g of water in a calorimeter with a calorimeter constant of 50.8 J K-1 with all substances initially at 59.4 °C.The resulting solution was observed to be at a temperature of 87.7 °C and have a heat capacity of 4.19 J g-1 K-1.Determine q for the mixing process.From the RelochonQ= MCOTa. 1.28 x 104 Jb. 1.67 x 105 J@= Total Heat RequitedOc. -1.67 x 105 Jm= Total mass =75.19+3R.5g = 107.69Od. -1.28 x 104 JC = Heat capacity 4.19 Jo"k!e. -1.42 x 104 JAT= To-T, = (87-7- 59.4)°CIncorrect answer.change is always same AT= 28.3°c = 28.3 KYou have left out a - sign in your equation.Don't forget to include the heat required to change the temperature of the calorimeter.107.6g x 4.19 k'x28.3 K@= 1.28 X 10%JAnsoption (a) is comect| 41.9 g of a salt solution was mixed with 21.2 g of an acidic solution in a calorimeter with a calorimeter constant of 34.7 J K-1 with all substances initially at 43.7°C.The resulting solution was observed to be at a temperature of 29.8 °C and have a heat capacity of 4.68 J g K-1.Determine g for the mixing process.a. 8.56 x 104 Jb. -3.45 x 104JOc. 4590 Jd. 4100 Je. 3.45 x 104 JIncorrect answer.Don't forget to include the heat required to change the temperature of the calorimeter.

Given data,Mass=63.1g∆T=13.9oCC(heat capacity)=4.68J/g.KCcalo=34.7J/K

Answered: My working out is given Heat of…

Chemistry

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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

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My working out is given Heat of capacity is 4.68 J g^–1 K^–1 and M (mass)= 41.9g-21.2g = 63.1g then temperature, delta T= 43.7 °C- 29.8 °C = 13.9 °C.

After you have the delta H or Q = MC(delta)T equation = 63.1 x 4.68 x 13.9 = 4104.7812, but what do you do with the calorimeter constant of 34.7 J K^–1 (is that there to trick students or do I use it in conjunction to the Heat of capacity of 4.68 J g^–1 K^–1?

I have a second image of a example of a tutor working out but it has two issue of accidently forgetting a minus sign and maybe not including the calorimeter constant.

75.1 g of a basic solution was mixed with 32.5 g of water in a calorimeter with a calorimeter constant of 50.8 J K-1 with all substances initially at 59.4 °C.
The resulting solution was observed to be at a temperature of 87.7 °C and have a heat capacity of 4.19 J g-1 K-1.
Determine q for the mixing process.
From the Relochon
Q= MCOT
a. 1.28 x 104 J
b. 1.67 x 105 J
@= Total Heat Requited
Oc. -1.67 x 105 J
m= Total mass =
75.19+3R.5g = 107.69
Od. -1.28 x 104 J
C = Heat capacity 4.19 Jo"k!
e. -1.42 x 104 J
AT= To-T, = (87-7- 59.4)°C
Incorrect answer.
change is always same AT= 28.3°c = 28.3 K
You have left out a - sign in your equation.
Don't forget to include the heat required to change the temperature of the calorimeter.
107.6g x 4.19 k'x28.3 K
@= 1.28 X 10%J
Ans
option (a) is comect|

41.9 g of a salt solution was mixed with 21.2 g of an acidic solution in a calorimeter with a calorimeter constant of 34.7 J K-1 with all substances initially at 43.7
°C.
The resulting solution was observed to be at a temperature of 29.8 °C and have a heat capacity of 4.68 J g K-1.
Determine g for the mixing process.
a. 8.56 x 104 J
b. -3.45 x 104J
Oc. 4590 J
d. 4100 J
e. 3.45 x 104 J
Incorrect answer.
Don't forget to include the heat required to change the temperature of the calorimeter.

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