A 5.00 liter sample of gas measured at 27.0?C and 1.25 atm of pressure has a mass of 10.13 g. What is the mole - 1.25 c mount video camera adapter
39.9)
b) 0898
c) 2.57
d) 2.50 x 10
A 5.00 liter sample of gas measured at 27.0?C and 1.25 atm of pressure has a mass of 10.13 g. What is the mole - 1.25 c mount video camera adapter
39.9)
b) 0898
c) 2.57
d) 2.50 x 10
2 comments:
I assume you know the molecular weight of the gas.
PV = nRT
126625 * 5 = n * 8.31 * 300
n = 126.625 * 5/8.31 * 300
n = 0254 mol
The mass is 10.13, so that the molecular weight = 10.13/0.254 = 39.9
The answer is a).
Assuming this is an ideal gas can be used
PV = nRT
P = pressure (atm)
V = volume (L)
T = temperature (° K), R = universal gas constant = 0.08205
n = number of moles of
So the solution for n is
n = PV / RT
n = (1.25) (5) / (0.08205) (27 + 273)
n = 0.25 mol
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