How many mols of gas are present in a sample of oxygen if it is found to be held at 4.98 atm in a 29.5 L container at 29.0 C ?
Solution:
PV=nRT
n=PVRT=4.98∗29.50.0821(273+29)=5.925molesn= \frac{PV}{RT}=\frac{4.98*29.5}{0.0821(273+29)}=5.925 moles n=RTPV=0.0821(273+29)4.98∗29.5=5.925moles
Answer = 5.925 moles or approximate 6 moles
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