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Answer: The volume of 0.640 grams of [tex]O_{2}[/tex] gas at Standard Temperature and Pressure (STP) is 0.449 L.
Explanation:
Given: Mass of [tex]O_{2}[/tex] gas = 0.640 g
Pressure = 1.0 atm
Temperature = 273 K
As number of moles is the mass of substance divided by its molar mass.
So, moles of [tex]O_{2}[/tex] (molar mass = 32.0 g/mol) is as follows.
[tex]No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol[/tex]
Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L[/tex]
Thus, we can conclude that the volume of 0.640 grams of [tex]O_{2}[/tex] gas at Standard Temperature and Pressure (STP) is 0.449 L.
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