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CHE111P Exam 2 Key - S6100%wtH2 S2100%wtH...

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N 2 = 0.79(1mol) n p O 2 =1.2(1mol)(0.21) n p N 2 =0.90 P O 2 =0.10P N 2 =(0.79(1mol)- 0.79(1mol)n p ) n p O 2 =1.2((1mol)(0.21)- 1.2(1mol)(0.21)n p ) n p 79% mol N 2 21% mol O 2 MEMBRANE MEMBRANE C    17% wt salt solution SATURATOR A D 4% wt salt solution B 100% wt H 2 O F S2   100% wt H 2 O Evaporated F S7 Dry NaCl (361.7 lb crystals) F S6 100% wt H 2 O Dried F S4   Crystals Adhering Solution S1       10,000 lb/hr         6% wt NaCl S3   X% wt NaCl 837.6 lb F EVAPORATOR DRYER CENTRIFUGE S5 adhering solution 198.7 lb solution (60% of total adhering solution) F CHE111P Exam No. 2 2 nd  Term 2010-2011 Answer Key 1.   Gas separation membranes are used to provide either air with a greater than normal oxygen content (for people   needing supplemental oxygen) or a lower than normal oxygen content (for controlled atmosphere warehouses). To do this,   air at an elevated pressure is fed to one side of the membrane in a separation module. A fraction of the moles of nitrogen  in the air (called this fraction n p ) permeate through the membrane. The fraction of the moles of nitrogen that permeate   through the membrane is 20.0% larger than the fraction of the moles of nitrogen that permeate. This leaves the gas that   does not permeate higher in nitrogen content. This nonpermeating gas is sent to another membrane separation module.   In the second module, the fraction of the moles of nitrogen that enter the second module that permeate through the   membrane is n p ; the fraction of the moles of oxygen that enter the second module that permeate through the membrane is   again 20.0% greater than the fraction of the number of moles of nitrogen that permeate. The gas that does not permeate   through the membrane in the second module is your product. If the mole fraction of nitrogen in this gas is 0.900, what is 
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