What is the volume occupied by 0.738g of xenon gas at STP?
Assuming ideal gas, we use ideal gas law formula:
PV = nRT
where
P = pressure (atm)
V = volume (L)
n = moles
T = absolute temperature (K)
R = universal gas constant = 0.0821 L-atm/mol-K
At Standard Temperature & Pressure (STP), T = 273 K and P = 1 atm.
Xenon has molar mass of 131 g/mol and it isn't diatomic. To get its moles, we just divide the given mass by the molar mass.
PV = nRT
1 * V = (0.738 / 131) * 0.0821 * 273
V = ?
Units in L. Hope this helps~ `u`
0.12626729313
To find the volume occupied by a gas at STP (Standard Temperature and Pressure), we need to know the molar mass of the gas and use the ideal gas law equation, PV = nRT.
1. First, we need to determine the number of moles (n) of xenon gas using its molar mass. The molar mass of xenon (Xe) is approximately 131.29 g/mol.
Number of moles (n) = Mass (g) / Molar mass (g/mol)
n = 0.738g / 131.29 g/mol ≈ 0.00562 mol
2. Next, we need to use the ideal gas law equation to find the volume. At STP, the temperature (T) is 273.15 K, and the pressure (P) is 1 atm (or 760 mmHg). The ideal gas constant (R) is 0.0821 L·atm/(mol·K).
PV = nRT
V = (nRT) / P
V = (0.00562 mol)(0.0821 L·atm/(mol·K))(273.15 K) / 1 atm
V ≈ 1.21 L
Therefore, the volume occupied by 0.738g of xenon gas at STP is approximately 1.21 liters.