Pressure, P = 30 MPa 1 MPa = 9.869 Atm 30 MPA = temperature, T = 158°F = 343.15° K 296.077 Alm PV = nRT m 44 296-0770XV = d = m = = X 0.0821 X Nearest Value 44x 296.077 0 0821 X 34315 3 1 litre = d = 462.413 x10³ g/m³ d 462.413 Kg/m3 0.001 m² = 600 kg/m³ 34315 = 462-413 8/litre
Pressure, P = 30 MPa 1 MPa = 9.869 Atm 30 MPA = temperature, T = 158°F = 343.15° K 296.077 Alm PV = nRT m 44 296-0770XV = d = m = = X 0.0821 X Nearest Value 44x 296.077 0 0821 X 34315 3 1 litre = d = 462.413 x10³ g/m³ d 462.413 Kg/m3 0.001 m² = 600 kg/m³ 34315 = 462-413 8/litre
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...
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How did they know n= m/44?
![Pressure, P = 30 MPa
1 MPa = 9.869 Atm
30 MPA =
temperature, T = 158°F
= 343-15° K
296.077 Alm
PV = nRT
m
44
296-0770XV =
d = m =
=
X 0.0821 X
Nearest Value
44x 296.077
0 0821 X 343.15
3
1 litre =
d = 462.413 x10³ g/m³
d
462.413 Kg/m3
0.001 m²
= 600 kg/m³
34315
= 462-413 8/litre](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc012420-4d67-48a3-8e85-bac83b5d2ff3%2F5f763a23-2c91-44e9-84fb-911bc0746bb0%2F13azfia_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Pressure, P = 30 MPa
1 MPa = 9.869 Atm
30 MPA =
temperature, T = 158°F
= 343-15° K
296.077 Alm
PV = nRT
m
44
296-0770XV =
d = m =
=
X 0.0821 X
Nearest Value
44x 296.077
0 0821 X 343.15
3
1 litre =
d = 462.413 x10³ g/m³
d
462.413 Kg/m3
0.001 m²
= 600 kg/m³
34315
= 462-413 8/litre
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