Question

A 5.00 L tank at 6.27 °C is filled with 3.97 g of dinitrogen difluoride gas and 3.91 g of sulfur tetrafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the partial pressure of each gas, and the total pressure in the tank. Be sure your answers have the correct number of significant digits. dinitrogen difluoride partial pressure: 0. sulfur tetrafluoride partial pressure: X 5 ? Total pressure in tank: atm

Answer 1

The molecular weight of diimogen di florride - 66.01 g/mol The molecular weight of sulfer tehoafluoride = 108.07 1 g/mol 3.97 0.060 3.97 gm of dinitrogen difluoride (n.) = 3.3 66.01 mole -0.036 3.91 sm of 3.31 tetrafluoride (N2) = 108 sulfor 108.07 mole Total mole (n) = ni+n2 = 0.060+ 0.036 1 = 0.096 mole The volume of tanklu= 5.00 L The temperature (T) = 6:27°C =6:27+273) -279.27k .; P= nRT, 0.096 mol x 0.082 2-cam-moi! ' x279.27 R 1 p = 0.4396 atm - Mole Reaction of dinitrogen diflusside - 0.060 0.096 24-0.625 Suufor te hoa fluoride -0.036 " " " 0.006 82=0.375 - The Porticu Poensure of . dinitrogen difluoride = total Proessure & mole Proaction = 0.4396*0.625 - 10.2747 atm sufor tetrafluorside = 0.43968 0.375 = 50 1649 atm The total Proessure - 0.4396 atm]

The molecular weight of chlorine Pentafluoride. 1 = 130.45 g/mal The molecular " " boron trifluorside =67.82 g/mol 7:02 g of chlorine Pentafluoride = 7.02 = 0.0538 moll (nu) 130045 15.9 g of bonon Gospod 15.9 tifluoride (22) = 67.82 1 = 0.2344 mole ..The total mole=natha (W) = 0.0538+0:2344 -0.2882 mole Temperature (T) = 22.7°C = (22+7+273 ) = 295:7K The volume (V) = 8.00L -s P: MRT , 0.2882 male x 295-7 K X0-082 L-am. 8L ทงโ! นี้! = 0.8735 atm The mole fraction of chlorine Pertafluoride -0.0538 o re. 20.1866 0.2882 The Particu Premure " =0.8735X0-1866 = 0.1631 atm The mole hoaction of bonon tirifluoride - 0:33 154 +0.8133 ... The particu Pressure =0.8735x0.8133 20-7104 atm The total Roermore 20.8735 atm.