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Answer:
Moles O2 = pV/RT = 2.75 x 5.60 / 0.08206 x 250=0.751
mass = 32 g/mol x 0.751 mol= 24.0 g
Explanation:
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The mass of 5.60 Liter oxygen at 1.75 atm pressure and 250.0 K temperature is 24.0 gram.
Ideal gas equation is a hypothesis which tells about the behavior of gas at standard condition and it is represented as:
PV = nRT, where
P = pressure = 1.75 atm
V = volume = 5.60 L
R = universal gas constant = 0.0821 L.atm /K.mol
T = temperature = 250 K
n is the moles of gas which will further define as:
n = W/M, where
W = required mass
M = molar mass
Now putting all the value in the above ideal gas equation and calculate for moles as:
n = (1.75)(5.60) / (0.0821)(250)
n = 0.751
Molar mass of oxygen is 32 g/mol and we calculate their required mass as:
W = 32 g/mol x 0.751 mol= 24.0 g
Hence, the required mass of oxygen is 24.0 g.
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