Question

If a gaseous mixture is made by combining 3.97 g Ar and 3.90 g Kr in an evacuated 2.50 L container at 25.0 ∘C, what are the partial pressures of each gas, ?Ar and ?Kr, and what is the total pressure, ?total, exerted by the gaseous mixture?

?Ar=______atm

?Kr=_____atm

?total=____atm

Answer 1

Ari mass= 3.9288 3.97 Ar molar mals = 40 glmol Mass of Ar moles of Arnar " Molar mass it 47 0.01925 40 ka- масс = 34 976ws Kor: Molar mass = 84 gimol mass of Ka moles ob kr = n = kr Molar mass of 39 -0.04843 4 total molesa na nutno= 0.099 25+ 0.04643=0.14566 mol n=0-14568 mol total v= 2.54 n=0-14568 mol R=0.0821 z.atm. metky T= 25+273= 298 K pua nRT PX 2.5=0-14568 x 0.08 21 X 898 . Protol=1.426 atm n .. partral pressure= mole fraction x total pressure moles of one component mole fraction total number of moles of all components 1.426 am (6.09925 - 0.97 15 atm xPtotal - WA 0.09925+0-04643 КИ Ath 1.426 ка 10.04643 -10.04643+0.09925 -0.4545 atry Vyti Rate please

Ari mass= 3.92 gr 3.97 Ari nolan mass= 40 glmol. 1 mass of Ar moles of Assa Molar mass of 41 0 .0 2925 40 kr. mass= 39 grams kor: Molar mass = 84 glmol O mass of Ka moles ob kran - . kr molar mass of 3-9 - E004643 84 - kr ni 0-099 25+ 0.04643=0.1456p mol fohal molesans nu n20.14568 mol Probaja ? v=2. 5t n=0-14568 mol R=0.0821 z.atm. metlky T=25+273= 298 K puanRT PX 2.5=0-14568 X 0.0P21 x 298 • Pobla 1426 atm 1 partial press 491€ = male Boraction x total pressu91e moles of one component mole fraction - total number of moles of all components primar xPtotal o 1426 = 0.97 15 atm 10-09925+0.04643 (0.02925 Mast n. 10.04643 71-426 0.04643+0.09925 0.4545atry ka 2 Rate please .