• what mass of no2 is contained in a 13.0 l tank at 4.58 atm and 385 k?

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  • No2 has a molar mass of 46.01 g/mol. Mass (g) = Moles x Molar Mass Moles = n = PV/RT Moles = (4.58 atm x 13.0 L) / (8.314 J/K mol x 385 K) = 0.0116 mol Mass (g) = 0.0116 x 46.01 = 0.5386 g

    • Answered:

      Gregory Petty

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  • First, use the ideal gas law to find the number of moles: $n = \frac{PV}{RT}$ $n = \frac{4.58~atm x 13.0~L}{.0821~L ~atm/mol~K x 385~K}$ $n = 4.79$ mol Then, calculate the mass using the molar mass of NO2: $m = nM_{NO2}$ $m = 4.79 mol~x (46.006 g/mol)$ $m = 221.5~g$ Therefore, there is 221.5 g of NO2 in the 13.0 L tank at 4.58 atm and 385 K.

    • Answered:

      Carlo Lewis

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