What is the volume and L of a 9.21 g sample of oxygen gas at 40.7°C that exerts 886.7 mmHg of pressure? ?

1 Answer

Relevance
  • 4 weeks ago

    1) 760 mm Hg = 1 atm

    886.7 mm Hg = 886.7/760 atm= 1.17 atm

     V = ?

    P = 1.17 atm

    n =  9.21g O2 x (1 mole O2/32 g O2)= .2878

    R = 0.0821 L. atm/mol/K

    Formula: ideal gas equation, PV=nRT

    Temperature in oC = 40.7°C + 273= 313.7°K

    (1.17atm)V=.2878( 0.0821 L. atm/mol/K)( 313.7°K)

    V =.2878( 0.0821 L. atm/mol/K)( 313.7°K)/(1.17atm)

    V= 6.33L

Still have questions? Get answers by asking now.