Respuesta :
Answer:
0.56L
Explanation:
This question requires the Ideal Gas Law: [tex]PV=nRT[/tex] where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.
Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:[tex]R=0.0821\frac{L\cdot atm}{mol\cdot K}[/tex]
Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means [tex]T=273.15K[/tex] and [tex]P=1atm[/tex]
Lastly, we must calculate the number of moles of [tex]O_2(g)[/tex] there are. Given 0.80g of [tex]O_2(g)[/tex], we will need to convert with the molar mass of [tex]O_2(g)[/tex]. Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: [tex]32g\text{ }O_2=1mol\text{ }O_2[/tex]
Thus, [tex]\frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n[/tex]
Isolating V in the Ideal Gas Law:
[tex]PV=nRT[/tex]
[tex]V=\frac{nRT}{P}[/tex]
...substituting the known values, and simplifying...
[tex]V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}[/tex]
[tex]V=0.56L \text{ } O_2[/tex]
So, 0.80g of [tex]O_2(g)[/tex] would occupy 0.56L at STP.
