Question

Incorrect Question 1 0/2.5 pts A container has 704 torr of gas in 666mL container. If after heating to 61C the pressure is 1.32 atm find the moles of gas that were in the container. 25.7 Incorrect Question 2 0/2.5 pts 2 NO + O2 -> 2 NO2 The NO in the reaction is 28C, 786torr, 984mL. What volume in Liter of Oz is required at 47C, 790 torr? 98.6

Answer 1

760tour -0.6660 L 10oomL lao Question 1 : Tinitial) Pressure of gas in container (P) = 704 torr = 704 torrxlatm = 019263at, volume of the gas in container (v) = 666 m2 = 666 m2x IL Initial temperature of the gas (T.) in unknown Final pressure of the gas (P2) = 1:32 afm Final temperature of the gas (12) =61'c=[61+273'15)K=334'15K volume is constant. Therefore according to Gay Lussack's Pi P2 ㅠ Tz → T- PITZ 0.9263 afm x 334.15k 234,5 K P2 1.32 atm Therefore initial temperature of the gas (T.) = 234.5k Initial pressure of the gas (Pi) = 0.9263 atm. And volume (V)=0-66602 Now at STP : Temperature (T3): = 273.15K Volum Pressure (P₂)= 10 atom volume (V3) =? P3 V3 Ti T3 P. V T3 0.9263 storax 0:6660L x 273:15K > V3 PIN 이 Ti P3 234.5x 1.0 atom ²/3=0.71862 1 mole .: At a stp volume of the gas = 0-71862 Now at STP 22:42 of gas is equivalent to Thencore at STP 0.718620f gas is equivalent to Imolex 0.71862 = 0.0321 mol 22.4L ... The moles of the gas in container = 0.0321 mol

Question 2: The given reaction is ZNO + O2 2N02 from the balanced reaction we see that I mole of Oz reacts with 2 moles of NO and produces 2 mole of No2. Now for no Temperature (T.) = 28°C = {28+273:15) K = 301.15% Pressure (P.) =786 forr = 786 forry laton = 1.0342 atom Volume (v.) = 984 m2 =984mL X 12 =0.9842 760 torr 1000m2 AT STP Temperature (72) = 273.15K, Pressure (P2) = loafn volume (V2) =? Now PVL = P2 V2 PIVA Tz 1.0342 af mx 0.9842*273.15k Povitz => V₂ Ti P2 30115K X1-0 atm → V2 = 0.9230 L - O•Ouizmo Now 0.92302 Noat at stp = Imol x0.92302 22:42 .:. According to balanced reaction motes of O2 should react = moles of No reacted - 00412 mol = 0·0206 mol 2 2 Now at STP, 0.0206 mol O: 22:42 X 0.0206mol I mol = 0.461444 ATSIP: Volume of Oz (VB) =0.461442 Temperahne (73)=273,15K, Pressure (P) 2.0am at 47%, and 790 ton: Temperature (T4)=47°2= (47+273.15)K Pressure (P4) = 790 forr= 790 torx latim =110395 atm volume (v (V4) =? - 320.15k 760 four

Now PzV3 TB P3 V3 o Puvu Tu P303 Tu T3 Pu → V4= 1.0 afm x 0.461442320.15* 273.15% x 1.0395 afon => vu=0.5202 vy- 520ML .: Volume of O2 required at 47°C, 1790 tor is 0.5202