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A 35. 0-l steel tank at 20. 0?c contains acetylene gas, c2h2, at a pressure of 1. 39 atm. Assuming ideal behavior, how many grams of acetylene are in the tank?.

Sagot :

52.66 grams of acetylene are in the tank.

PV=nRT is the formula for ideal gases, where P is the pressure and R is the constant and T is the temperature (P is 1.39 atm in this case)

Volume = 35.0 L

Gas constant = 0.08206 L.atm/K.mole

Temperature = 20.0 °C  = 293 °C

We also know that we can apply the formula ;

Number of mole(n) = [tex]\frac{MmPV}{RT}[/tex]

m = ( 26.04 × 1.39 × 35 )/ (0.08206 × 293.15)

m = 52.66 g

Thus, the mass is 52.66 grams.

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