The final pressure of the gas mixtures is 72.48 atm.
The given parameters;
- mass of the oxygen = 88 grams
- mass of the argon, = 33 grams
- volume of the gaseous mixture, V = 2400 mL = 2.4 L
The number of moles of the gases is calculated as follows;
[tex]number \ of \ moles \ of \ oxygen = \frac{88}{16} = 5.5 \ moles\\[/tex]
[tex]number \ of \ moles \ of \ argon = \frac{33}{40} = 0.825 \ moles\\[/tex]
The total number of moles of the gas mixture;
= 5.5 + 0.825
= 6.325 moles
The final pressure of the mixture is calculated as follows;
PV = nRT
[tex]P = \frac{nRT}{V} \\\\P = \frac{6.325\ mol \times (0.0821 \ L.atm/mo.K) \times (62+273)}{2.4 \ L} \\\\P = 72.48 \ atm[/tex]
Thus, the final pressure of the gas mixtures is 72.48 atm.
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