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:
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
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
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
Related questions
Question
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7f9c6ce-7c61-4a7b-90a0-6a641d198bf8%2F83b7be54-fbd6-4e6a-a2cb-84f6d22ad293%2F9fwvqg9_processed.png&w=3840&q=75)
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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7f9c6ce-7c61-4a7b-90a0-6a641d198bf8%2F83b7be54-fbd6-4e6a-a2cb-84f6d22ad293%2Fndq0gmh_processed.png&w=3840&q=75)
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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