D dbs756 Guest Feb 23, 2009 #1 a 500. mL flask contains a mixture of 2.56 grams of argon, 1.22 grams of nitrogen and 8.49 grams of chlorine. if the temperature of the gases is 125 degrees C, what is the total pressure in the flask?
a 500. mL flask contains a mixture of 2.56 grams of argon, 1.22 grams of nitrogen and 8.49 grams of chlorine. if the temperature of the gases is 125 degrees C, what is the total pressure in the flask?
J Jacob S Guest Feb 23, 2009 #2 2.56g Ar(1 mole/39.95g)=.064 mole 1.22g(1 mole N/14g)=.087 mole 8.49g (1 mole Cl/35g)=.24g mole .064 +.087 +.24=.391mole PV=nRT P*(.5L)=.391mole(*0.0821L*atm/mol*K)(125+273)K P=25.55 atm
2.56g Ar(1 mole/39.95g)=.064 mole 1.22g(1 mole N/14g)=.087 mole 8.49g (1 mole Cl/35g)=.24g mole .064 +.087 +.24=.391mole PV=nRT P*(.5L)=.391mole(*0.0821L*atm/mol*K)(125+273)K P=25.55 atm
D Dr.A Guest Feb 23, 2009 #3 moles Ar = 2.56 g / 39.948 g/mol=0.0641 moles N2 = 1.22 g / 28.0134 g/mol=0.0436 moles Cl2 = 8.49 g/70.906 g/mol=0.120 total moles =0.278 p = 0.278 x 0.0821 x 398 K / 0.500 L =18.2 atm
moles Ar = 2.56 g / 39.948 g/mol=0.0641 moles N2 = 1.22 g / 28.0134 g/mol=0.0436 moles Cl2 = 8.49 g/70.906 g/mol=0.120 total moles =0.278 p = 0.278 x 0.0821 x 398 K / 0.500 L =18.2 atm
