What is the volume (in mL) of 0.250 grams of oxygen gas at a pressure of 0.852 atm and a temperature of -196 °C?
R=0.0821
No. of moles (n) = (0.250 g) / (32 g/mol) = 0.0078 mol
Pressure (P) = 0.852 atm
Temperature (T) = -196 °C = 77 K
R = 0.0821
PV = nRT
Volume = nRT / P = (0.0078 mol x 0.0821 x 77 K) / (0.852 atm) = 0.0578 L = 57.8 mL
Volume (V) = 57.8 mL
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