Part I: 1. Mass of KC103 2. Volume of water displaced 3. Temperature of water 4. Atmospheric pressure 5. Vapor pressure of water (Appendix 3) 6. Pressure of oxygen (4-5) 7. Value of R Calculations: 2.0154 g 615.0 mL 22 °C 760 mmHg 19.83 mmHg 740.17 mmHg 0.615 L 295.15 K 0.973908033 atm 0

Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter10: Solids, Liquids, And Phase Transitions
Section: Chapter Questions
Problem 7P
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Value of R= ?

 

Part I:
+
1. Mass of KC103
2. Volume of water displaced
3. Temperature of water
4. Atmospheric pressure
5. Vapor pressure of water
(Appendix 3)
6. Pressure of oxygen (4-5)
7. Value of R
Calculations:
2.0154 g
615.0 mL
22 °C
760 mmHg
19.83 mmHg
740.17 mmHg
0.615 L
295.15 K
0.973908033 atm
Transcribed Image Text:Part I: + 1. Mass of KC103 2. Volume of water displaced 3. Temperature of water 4. Atmospheric pressure 5. Vapor pressure of water (Appendix 3) 6. Pressure of oxygen (4-5) 7. Value of R Calculations: 2.0154 g 615.0 mL 22 °C 760 mmHg 19.83 mmHg 740.17 mmHg 0.615 L 295.15 K 0.973908033 atm
Calculations
Pressure of oxygen = Atmospheric pressure - water vapor pressure (see appendix)
Moles of oxygen = 3/2 moles of KClO3 (calculate moles of potassium chlorate from its
mass, then convert that to moles of oxygen according to the stoichiometry of the balanced
equation.
R = PV/nT_(pressure should be in atmosphere, volume in liters, and temp. in Kelvins)
Transcribed Image Text:Calculations Pressure of oxygen = Atmospheric pressure - water vapor pressure (see appendix) Moles of oxygen = 3/2 moles of KClO3 (calculate moles of potassium chlorate from its mass, then convert that to moles of oxygen according to the stoichiometry of the balanced equation. R = PV/nT_(pressure should be in atmosphere, volume in liters, and temp. in Kelvins)
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