The number of moles of o2 in 45.0L} of 02} gas.
If that's at STP you will have
45.0L x (1 mol/22.4L) = ?
To determine the number of moles of O2 gas in a given volume, we can use the ideal gas law formula:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
Since we don't have the pressure or temperature information, we can assume standard conditions, which are commonly used:
P = 1 atm
T = 273.15 K
So the equation becomes:
PV = nRT
(1 atm) * (45.0 L) = n * (0.0821 L·atm/mol·K) * (273.15 K)
45.0 atm·L = 22.4175 n
Dividing both sides of the equation by 22.4175:
n = 45.0 atm·L / 22.4175
n ≈ 2.01 moles
Therefore, there are approximately 2.01 moles of O2 gas in 45.0 L.
To find the number of moles of a gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature of the gas
However, in this case, we are given the volume of the gas (V = 45.0 L) and we need to find the number of moles of O2. We also need to know the temperature and pressure of the gas to use the ideal gas law equation.
Assuming the temperature and pressure are standard conditions (0 degrees Celsius and 1 atmosphere), we can use the ideal gas law to find the number of moles of O2. The ideal gas constant, R, is 0.0821 L·atm/(mol·K).
Since the pressure and temperature are standard conditions, the equation becomes:
PV = nRT
Rearranging the equation to solve for n:
n = PV / RT
Plugging in the values:
P = 1 atm
V = 45.0 L
R = 0.0821 L·atm/(mol·K)
T = 273 K (0 degrees Celsius is 273 Kelvin)
n = (1 atm * 45.0 L) / (0.0821 L·atm/(mol·K) * 273 K)
Calculate the right side of the equation:
n = 45.0 / (0.0821 * 273)
n ≈ 1.677 moles
Therefore, there are approximately 1.677 moles of O2 in 45.0 L of O2 gas at standard conditions.