Question

How many moles of gas are present in 1.13 L of gas at 2.09 atm and 291 K?

Answer 1

n = 0.0989 moles

Step-by-step explanation:

n = PV / RT

P = 2.09atm

V = 1.13L

R = 0.08206

T = 291K

Plug the numbers in the equation.

n = (2.09atm)(1.13L) / (0.08206)(291K)

n = 0.0989 moles