Respuesta :
Moles of O2 = 5.03/32 = 0.15moles. PV=nRT. 0.998xV = 0.15x301x0.082.(temperature is in Kelvin). V= 0.15x301x0.082/0.998 = 3.71L. Hope it helps.
The 5.03g of Oxygen gas occupies 3.894 L of the volume.
The volume occupied by the oxygen gas can be calculated from the ideal gas equation.
The ideal gas equation has been,
PV = nRT
where, P is the pressure in atm.
V = volume of the gas in liters
n = moles of the gas
T = temperature of the gas in Kelvin.
The moles of oxygen gas in 5.03g of oxygen :
moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
molecular weight of oxygen gas = 32 g/mol
moles of oxygen gas = [tex]\rm \dfrac{5.03}{32}[/tex] moles
moles of oxygen gas (n) = 0.157 moles
The temperature = 28 C
0 C = 273.15 K
28 C = 301.15 K
The volume occupied by oxygen gas:
PV = nRT
[tex]\rm V\;=\;\dfrac{nRT}{V}[/tex]
V = [tex]\rm \dfrac{0.157\;moles\;\times\;0.0821\;atm.L/mol.K\;\times \;301.15\;K}{0.998\;atm}[/tex]
V = 3.894 L.
The 5.03g of Oxygen gas occupies 3.894 L of the volume.
For more information about the volume occupies, refer to the link:
https://brainly.com/question/22622383
