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Burka [1]
3 years ago
9

Choose a slideshow or presentation program to organize your research. Your presentation should contain Information collected In

Part 1 and Part 2 Part 1: Include important facts found through your research. Include the answers to questions 1-4 for each of the three occupations you chose. Part 2: Include Information a visual representation of the data and the answers to questions 1-6. Submission Requirements: The student should submit a presentation created with a slideshow or presentation program of their choice. The presentation should include important facts obtained through research, references, Images, data, graphs, and answers to the reflection questions. Upload your presentation in the box below.​
Chemistry
1 answer:
beks73 [17]3 years ago
3 0

Answer:

Do the work

Explanation:

I know

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The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

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Explanation:

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A gas occupies a volume of 2.45 L at a pressure of 1.03 atm. What volume will the
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2.58 L

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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