Assigned Tr Trial 1 DATA $4409 Mass of unknown 3410265 Mass of beaker + water Mass of empty beaker 341, 289 19. Loc= ,9984052gkm? Mass of displaced water Temperature of water K Density of water (lookup @ T) 341,8151 ,3418 ISAL 7341.! Volume of water Liters of gas (mostly CO2) Barometric Pressure mm Hg Vapor Pressure H20 (look up @ T) 719.4mlHtg Partial Pressure of CO2

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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Solve for the Number of moles of the unknown with the given data. I do not no the given substance

Assigned
Tr
Trial 1
DATA
$4409
Mass of unknown
3410265
Mass of beaker + water
Mass of empty beaker
341, 289
19. Loc=
,9984052gkm?
Mass of displaced water
Temperature of water
K
Density of water (lookup @ T)
341,8151
,3418 ISAL
7341.!
Volume of water
Liters of gas (mostly CO2)
Barometric Pressure
mm Hg
Vapor Pressure H20 (look up @ T)
719.4mlHtg
Partial Pressure of CO2
Transcribed Image Text:Assigned Tr Trial 1 DATA $4409 Mass of unknown 3410265 Mass of beaker + water Mass of empty beaker 341, 289 19. Loc= ,9984052gkm? Mass of displaced water Temperature of water K Density of water (lookup @ T) 341,8151 ,3418 ISAL 7341.! Volume of water Liters of gas (mostly CO2) Barometric Pressure mm Hg Vapor Pressure H20 (look up @ T) 719.4mlHtg Partial Pressure of CO2
Expert Solution
Given conditions

Mass of the unknown, m = 0.5940 g
Partial pressure of CO2 , P= 719.4 mm Hg = 719.4 / 760 = 0.9465 atm
1 atm = 760 mm Hg
Volume of the gas, V = 0.3418 L
Temperature of the gas = 19.2oC = 19.2+273.15 = 292.35 K
R = 0.082057 L atm mol-1K-1  
Find n from the following data

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