How many moles of oxygen gas (O2) are in a 2.50 L container at standard temperature and pressure?
2.24 x 10^1 mol O2
1.12 x 10^–1 mol O2
2.50 x 10^0 mol O2
5.60 x 10^1 mol O2
one mole is in 22.4 L at stp
moles= 2.50/22.4 moles=1.12e-1 moles
To determine the number of moles of oxygen gas (O2) in a 2.50 L container at standard temperature and pressure, we can use the ideal gas law, which states that PV = nRT. In this case, we can assume that the temperature (T) and pressure (P) are constant at standard temperature and pressure (STP), which is defined as 0 degrees Celsius (273.15 K) and 1 atmosphere (1 atm), respectively. The ideal gas constant (R) is 0.0821 L·atm/(mol·K). Rearranging the equation, we get n = (PV) / (RT).
Let's substitute the values into the equation:
P = 1 atm
V = 2.50 L
R = 0.0821 L·atm/(mol·K)
T = 273.15 K
n = (1 atm * 2.50 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n = 2.50 / 0.0821 * (1/(273.15))
n ≈ 2.50 / 0.0821 * 0.00366
n ≈ 91.82
Therefore, there are approximately 91.82 moles of oxygen gas (O2) in a 2.50 L container at standard temperature and pressure.
None of the provided answer choices is approximately equal to 91.82 moles.
To determine the number of moles of oxygen gas (O2) in a 2.50 L container at standard temperature and pressure (STP), we can use the ideal gas law. The ideal gas law equation is:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
At STP, the pressure is 1 atmosphere (atm) and the temperature is 273 Kelvin (K).
Rearranging the ideal gas law equation to solve for n, the number of moles:
n = PV / RT
Plugging in the given values:
P = 1 atm
V = 2.50 L
R = 0.0821 L·atm/(mol·K)
T = 273 K
n = (1 atm)(2.50 L) / (0.0821 L·atm/(mol·K))(273 K)
After performing the calculation, the result is approximately 0.11451 moles of oxygen gas (O2).
Therefore, the correct answer is 1.12 x 10^–1 mol O2, which is the closest approximation to 0.11451 moles.