Respuesta :

jacob193

1.31 × 10⁴ grams.

Explanation

Assume that oxygen acts like an ideal gas. In other words, assume that the oxygen here satisfies the ideal gas law:

[tex]P \cdot V = n \cdot R\cdot T[/tex],

where

  • [tex]P[/tex] the pressure on the gas, [tex]\bf P = 10^{5}\;\textbf{kPa}=10^{8}\;\textbf{Pa}[/tex];
  • [tex]V[/tex] the volume of the gas, [tex]V = 10.0 \;\text{L} = 10.0\times 10^{-3}\;\text{m}^{3}=10^{-2}\;\text{m}^{3}[/tex];
  • [tex]n[/tex] the number of moles of the gas, which needs to be found;
  • [tex]T[/tex] the absolute temperature of the gas, [tex]T=20\;\textdegree{}\text{C} = (20 + 273.15)\;\text{K} = 293.15\;\text{K}[/tex].
  • [tex]R[/tex] the ideal gas constant, [tex]R = 8.314[/tex] if P, V, and T are in their corresponding SI units: Pa, m³, and K.

Apply the ideal gas law to find [tex]n[/tex]:

[tex]n = \dfrac{P\cdot V}{R\cdot T} = \dfrac{{\bf 10^{8}\;\textbf{Pa}}\times 10^{-2}\;\text{m}^{3}}{8.314 \;\text{Pa}\cdot\text{m}^{3}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}\times 293.15\;\text{K}} = 410.3\;\text{mol}[/tex].

In other words, there are 410.3 moles of O₂ molecules in that container.

There are two oxygen atoms in each O₂ molecules. The mass of mole of O₂ molecules will be [tex]{\bf 2}\times 16.00 = 32.00\;\text{g}[/tex]. The mass of 410.3 moles of O₂ will be:

[tex]410.3 \times 32.00 = 1.31\times10^{4}\;\text{g}[/tex].

What would be the mass of oxygen in the container if the pressure is approximately the same as STP at [tex]10^{5}\;\textbf{Pa}[/tex] or  [tex]10^{2}\;\text{kPa}[/tex] instead?