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Can you help me with the question 5&6? Given data of group 9 only
In this lab, the temperature of chemicals before chemical change and after chemical change was collected and thus, weather bonds were broken or created was determined. To do this a calorimeter was used; a calorimeter can help us measure heat of chemical reaction Data: Grou Tria Volum Volum mol H20 mol 1e H20 eEtOH Ethanolfraction aT H20 22. 23. 0.9 22. 24. 2.2 22. 26. 4 22. 27. 5.1 23 29. 6.2 1 0.9 30.1 0.0499 2 3.05 22.2 0.169 3 7.07 22.2 0.3923 4 12.12 17.5 0.67258 79 1.10 5 19.97.5 27 2 420 18 12 10 12 20. 21. 1.7 2 9 19. 22. 2.7 9 19. 24.5.1 6 4 15 20. 28. 8.2 5 25 22. 26. 41 Grou Tria Volum Volum mol H20 mol mol H20 0.80421 18. 3 Acetonitril fraction a T 6 14.210 0.39444444 0.1919 10620 0.55555555 22. 21-1.7 7 Discussion: Layout View Sec 1 Pages: 1 of 2 Words: 285 of 304 |脹
Can you help me with the question 5&6? Given d
Can you help me with the question 5&6? Given d
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Answer #1

Group

Trial

Volume H2O

Volume ethanol

Moles H2O

Moles ethanol

Mole fraction H2O

Ti

Tf

Delta T

9

1

0.9

30.1

0.0499

0.5155

0.08826

22.5

23.4

0.9

2

3.05

22.2

0.169

0.3802

0.3077

22.7

24.9

2.2

3

7.07

22.2

0.3923

0.3802

0.50783

22.7

26.7

4

4

12.12

17.5

0.67258

0.2958

0.69454

22.8

27.9

5.1

5

19.9

7.5

1.10

0.1285

0.89734

23

29.2

6.2

6

14.2

10

0.394444444

0.1919 (acetonitrile)

0.804

21.3

18.3

-3

1) The mole fraction of trial 5 releases maximum a heat since heat of reaction or enthalpy change is directly proportational to temperature and the temperature change in trial 5 is maximum.

In order for a solute to be soluble in a particular solvent, 3 things must be considered. 1st the intermolecular force holding the solvent molecules together must be broken (requires energy). 2nd the intermolecular forces holding the solute together must be broken (requires energy). Finally the solute & solvent can interact through whatever forces available (bond formation requires energy) called solvation energy. Generally withincrease in temperature the solubility increases.

q = m*C*dT where, q= heat capacity; m= mass; C= specific heat capacity; dT =change in temp.

C for ethanol is 2.460J/gCo

Mass

dT

q

23.749

0.9

52.58

17.5158

2.2

94.795

17.5158

4

172.55

13.8075

5.1

173.23

5.9175

6.2

90.2537

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