Answer:
V = 0.24 L.
Explanation:
Hello there!
In this case, it is possible to approach this problem by using the ideal gas equation defined by:
[tex]PV=nRT[/tex]
Whereas our unknown is V, volume, and we solve for it as follows:
[tex]V=\frac{nRT}{P}[/tex]
Then since the STP conditions are ate 273.15 K and 1 atm, the volume of 0.011 moles of CO2 gas turns out to be:
[tex]V=\frac{0.011mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=0.24L[/tex]
Regards!
