what is the Regression equation (displayed on graph):       What units are associated with the slope?  Evaluate Hvap from the slope. Evaluate the normal boiling point of the organic liquid from the regression equation: First, evaluate ln(760) and substitute this result into the regression equation and solve for T1 Next, convert the found value of T1 to T. Report this temperature in both K and °C:

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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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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what is the Regression equation (displayed on graph):     

 What units are associated with the slope? 

Evaluate Hvap from the slope.

Evaluate the normal boiling point of the organic liquid from the regression equation:

  • First, evaluate ln(760) and substitute this result into the regression equation and solve for T1
  • Next, convert the found value of T1 to T. Report this temperature in both K and °C:
Given, Pvapor- Pequil - Pinitial
The first table can be filled as below:
T,°C
Pinitial (mm Hg) Pequil (mm Hg)
Pvapor (mm Hg)
Run
1
11.9
736.8
750.4
750.4 - 736.8 = 13.6
2
19.5
736.6
757.7
757.7 - 736.6 =21.1
28.4
736.7
771.2
771.2 - 736.7 = 34.5
4
36.3
736.7
789.2
789.2 - 736.7 = 52.5
The second table can be filled as below:
Run
T, °C
Т, К
1/T, к1
Pvapor (mm Hg) In(Pvapor)
1
11.9
11.9+273.15 = 285.05
1/285.05 = 0.00350816 13.6
In(13.6) = 2.610
2
19.5
19.5 + 273.15= 292.65 1/292.65 = 0.00341705 21.1
In(21.1) = 3.0493
3
28.4
28.4 + 273.15 = 301.55 1/301.55 = 0.00331620 34.5
In(34.5) = 3.5409
4
36.3
36.3 + 273.15= 309.45 1/309.45 = 0.00323154 52.5
In(52.5) =3.9608
3.
Transcribed Image Text:Given, Pvapor- Pequil - Pinitial The first table can be filled as below: T,°C Pinitial (mm Hg) Pequil (mm Hg) Pvapor (mm Hg) Run 1 11.9 736.8 750.4 750.4 - 736.8 = 13.6 2 19.5 736.6 757.7 757.7 - 736.6 =21.1 28.4 736.7 771.2 771.2 - 736.7 = 34.5 4 36.3 736.7 789.2 789.2 - 736.7 = 52.5 The second table can be filled as below: Run T, °C Т, К 1/T, к1 Pvapor (mm Hg) In(Pvapor) 1 11.9 11.9+273.15 = 285.05 1/285.05 = 0.00350816 13.6 In(13.6) = 2.610 2 19.5 19.5 + 273.15= 292.65 1/292.65 = 0.00341705 21.1 In(21.1) = 3.0493 3 28.4 28.4 + 273.15 = 301.55 1/301.55 = 0.00331620 34.5 In(34.5) = 3.5409 4 36.3 36.3 + 273.15= 309.45 1/309.45 = 0.00323154 52.5 In(52.5) =3.9608 3.
4.5
4
y = -4882.6x + 19.736
R = 1
3.5
2.5
2
1.5
0.5
0.0032
0.00325
0.0033
0.00335
0.0034
0.00345
0.0035
0.00355
1/т, к1
In (Pvapor)
Transcribed Image Text:4.5 4 y = -4882.6x + 19.736 R = 1 3.5 2.5 2 1.5 0.5 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345 0.0035 0.00355 1/т, к1 In (Pvapor)
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