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Calorimetry lab 25

Laetitia M Lange DO7 Afternoon11/1119 11/18/19

Calorimetry lab 25 Data

A. Specific heat of metal 1. Mass of metal. (g)

Trial 1

2. Temperature of metal (boiling water) ( OC 3. Mass of calorimeter (g) 4. Mass of calorimeter + water (g) 5. Mass of water (g) 6. Temperature of water in calorimeter ( oC o

7. Maximum temperature of metal and water from graph ( C)

Trial 2 9.14 13.662

100 169.73 188.37 18.64

99.8 1700.24 189.67 19.44

22

22.8

27

25.2

78 6083.22

77 6262.94

78

77

0.5844

0.1915

Table 1. A. Specific heat of metal Results;

Calculations 1. Temperature change of water ( oC) 2. Heat gained by water (J) o

3. Temperature change of metal ( C o

4. Specific heat of metal (J/g x C) o

5. Average specific heat of metal (J/g x C) Table 2.1. Specific heat of metal results

0.388

Data C. Enthalpy of solution for the dissolution of a salt 1. Mass of salt (g) 2. Molar mass of salr (g/mol) 3. Moles of salt (mol) 4. Mass of calorimeter (g) 5. Mass of calorimeter and water (g) 6. Mass of water (g)

5.0109 0.09367 129.12 164.4 35.28

5.0167 53.491 0.09378 171.95 190.81 18.861

21.8

23.7

17.8

17.1

-4

-6.6

325.097 14.184 151.39

4.184 514.24 34.985 373.4 338.1

Table 3. C. Enthalpy of solution for the dissolution of a salt Results; o

7. Initial Temperature of water ( C) o

8. Final temperature of mixture from graph ( C o

9. change in temperature of solution ( C) 10. Specific. Heat of water (J/g x oC) 11. Heat of solution (J) 12. Heat of reaction (J) 13. Hrxn (J/ mol salt) 14. Average Hrxn (J/ mol salt) Table 3.1. Enthalpy of solution for the dissolution of a salt Results description:

Table 1 and 2.1 represent the recorded data for the specific heat of metal and the calculations, and table 3 and 3.1 represent the recorded data for the enthalpy of solution for the dissolution of a salt. In both sections, there is increase for trial 2 maybe because a more accurate (less water) was used and it has a very high specific heat, it absorb higher amount of heat while, the temperature didn’t rise as much.

Calculations:

Conclusions:

In conclusion, for part B, my average ∆H Our hypothesis that …

n

for my NaOH + HCl acidbase reaction was -63.835 kj/mol, my standard deviation was

0.6293, my relative standard error was 0.9858, and my percent error was 14.40%. For my NaOH + HNO 3

acid-base reaction my average ∆H n

was -68.745 kj/mol, my standard deviation was 1.2657, my relative standard deviation was 1.8412, and my percent error was 20.82%. In part

C, my unknown was 21, its heat capacity was 0.903J/g °C, and based on the chart in the lab book my unknown salt was Na 2

SO 4

. The average total ∆H s

per mole of the unknown salt was -7796.12J/mol, my standard deviation was 372.3,

my relative standard deviation was 4.78, and my percent error was 297.6%. In conclusion, for part A the specific heat of the metal was 0.5844 and 0.1915 for trial 1 and 2 respectively. The average specific heat of metal was 0.388 and it was calculated using the specific heat of the metal gathered from both trials. In part C, the heat of the solution was 325.097 (trial 1) and 514.24 (trial 2). The heat of reaction was 14.184 and 34.985 for trials 1 and 2 respectively. The Hrxn was 151.39 for trial 1 and 373.4 for trial 2. These results were then used to determine the average Hrxn, which was 338.1 Post lab questions: 1. The temperature change will be lower because water can absorb higher amount of heat while not rising the temperature too much. 2. This will result in a smaller specific heat of the metal because the temperature of the water will be less, and the temperature of the metal will be higher. 7. The enthalpy will be low. Loss of salt will result in lower amount of salt present in the reaction, meaning there will be less energy...