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Q: What is volume of oxygen occupied by 2 mols at 1.3 ATM and 300 k?

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38 L

38 L

38 L

38 L

38 L

V = nRT/p : v = 160*0.08205*300/2 =1.97mL

PV = nRTV = nRT/P = (2moles)(0.0821 Latm/Kmol)(300K)/1.3 atm V = 37.9 liters = 38 liters (2 significant figures)

600 oxygen atoms. there are 300 molecules, each having 2 oxygen atoms. 300 x 2 =

The Carboniferous period, during which oxygen amounted to 35% of the atmosphere by volume (this was around 300 million years ago). Nowadays the amount is 20.95%.

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300. mL

CH3CO2HIn one molecule, there is 2 atoms of oxygen. So, in 300 molecules, there will be 600 oxygen atoms.

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It's the space occupied by 2,241.15 gallons.