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Tarea 4

# Tarea 4 - c PT= P1 P2 0.4 atm 1.2 atm = 1.6 atm N N2 =...

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Red 20% Yellow 30% Blue 50% B) PB=(0.5)(0.90)=0.45 atm PR=(0.2)(0.90)=0.18 atm PY=(0.3)(0.90)=0.27 atm PB+PR+PY= 0.90 atm N 2 V = 1.2 L H 2 V = 3.6 L TN 2 = TH 2 VN 2 = VH 2 = 1.2 + 3.6 = 4.8 L NH 3 a) PMO 2 =2x16=32 V=65 L P = 16500 KPa = 162 atm TO 2 = 23º C = 296º K PV = nRT = n = PV/RT = (162)(65)/(0.082)(296ºK) = 433.83 mol m = (PM)(433.83)=13882 g b) FALTA asi dice en las hojas que mando…

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N 2 molecular weight = 28 Xe atomic weight = 131 There will be less mass in the volume of N 2 so it is less dense. a) VT=V1+V2 El volumen incrementa de 2L a 5L PN 2 = 0.40 atm despues de mezclarse b) Po 2 = 1.2 atm VO 2 aumenta de 3L a 5L

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Unformatted text preview: c) PT= P1 + P2 0.4 atm+ 1.2 atm = 1.6 atm N N2 = PV/RT = (1)(2)/(.082)(298) = 0.0817 moles N o2 = (2)(3)/(.082)(298) = 0.245 moles P = N t + RT/ VT = (.326)(.082)(298)/5L = 1.6 atm M=337.428g M Ar =339.854g M Ar + M – Ne= 1.648g .0539 mol = M Ne /MM Ne + M Ar /MM Ar M Ar = 1.648 - M Ne 0.0539 mol = M Ne /20 + M Ar /40 0.0539 = M Ne /20 + (1648 – M Ne) /40 2.156g = 2M Ne +1648 - M Ne M Ne = 0.508 g R Ne = 0.0254 moles R Ar = 1.14g/40 = 0.0285 moles R T = 0.0539 moles % R Ne = (0.0254/0.0534) x 100% = 47.2%...
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