What is the mass of oxygen gas in a 16.6 l container at 34.0°c and 6.22 atm?

This is a PV=nRT problem. So to do this, you are looking for grams. The easiest way to get to grams is to convert moles, so set up the equation looking for n

n=PV/RT

To keep everything consistent with the universal gas constant (0.082057 L atm/mol K), one thing needs to be converted.

To convert 34.0 C to kelvin, just add 273.15. That gets 307.2 with sig figs.

Next, plug everything into your equation

n=(6.22atm)(16.6L)/(0.082057Latm/molK)(307.2K)

Cleaning that up gets you 4.10 moles (again, after sig figs)

Now, you need to convert the 4.10 moles of O2 to grams. To do that, you multiply 4.10 by 31.998 (molar mass of o2) to get 131 g (for a third time, sig figs!)

So, your final answer is 131 g O2

Sorry if any math is wrong, but you should be able to walk yourself through this and other problems now. Hope I helped!

yes definetly because you are using energy

astronomers such as edwin hubble.