Due 11/20/17 gRT V = ( in this equation, density = Pressure equalities: 1 atm =

Question

Due 11/20/17 gRT V = ( in this equation, density = Pressure equalities: 1 atm = 760mm Hg 760 torr = 14.7ps-101 .3kPa-10 13mbar 2.54 cm 25.4 mm- 1 in density of Hg is 13.6g/cm3 density of water = 1.0 g/cm3 @ ~ room temp (ht water column /13.62 ht Hg column) 1 Larm = 62.36 Lmmlia uni units in PV-nRT must match units of "R" K mol Complete the following table for dinitrogen tetroxide gas (show work for each problem) # | Pressure 1 1.77 atm 2 673 mm H 30.0899 bar Volume 4.98 L 488 mL Temperature Moles 431 rams 0.783 912 39°F 6.25 1.15 L 0.166

Explanation / Answer

1. PV = nRT
   T = 431+273 = 704K
   R = 0.0821L-atm/mole-K
   V = 4.98L
   P = 1.77atm
   n = PV/RT
      = 1.77*4.98/0.0821*704 = 0.1525 moles
mass of N2O4   = no of moles * gram molar mass
                = 0.1525*92 = 14.03g
2.P = 673/760   = 0.885 atm
V = 0.488L
n = 0.783
T = PV/nR
     = 0.885*0.488/0.783*0.0821 = 6.72K = -266.28C^0
mass of N2O4 = 0.783*92 = 72.036g
3.P = 0.0899bar = 0.0899*0.9869 = 0.08872atm
T = 912+273     = 1185K
n = W/G.M.Wt
    = 6.25/92 = 0.068 moles
V = nRT/P
      = 0.068*0.0821*1185/0.08872   = 74.57L
4.
    mass of N2O4 = no of moles * gram molar mass
                  = 0.166*92 = 15.272g
    T   = 39F = (39+459.67)*5/9 = 55.4K
    n   = 0.166moles
    V   = 1.15L
    P = nRT/V
       = 0.166*0.0821*55.4/1.15 = 0.66atm

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