how many moles of O2 are contained in 6.53 L with 297 K and 6.6 atm
PV=nRT
solve for n
To determine the number of moles of O2, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
Rearranging the equation to solve for n, we have:
n = PV / RT
Now, we can plug in the given values:
P = 6.6 atm
V = 6.53 L
R = 0.0821 L·atm/(mol·K)
T = 297 K
n = (6.6 atm * 6.53 L) / (0.0821 L·atm/(mol·K) * 297 K)
n ≈ 1.08 moles
Therefore, approximately 1.08 moles of O2 are contained in the given conditions.