Gasni zakoni

Submitted by Marija on Thu, 11/21/2013 - 20:13
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Koliko molova molekula vodonika zauzima zapreminu od 15.dm(3) na 49 stepeni i 720mmhg?koliko atoma H sadrzi ovaj uzorak?

milicaradenkovic's picture

milicaradenkovic - Thu, 11/21/2013 - 20:22

1 mm Hg----------------------------133.322 Pa

720 mm Hg--------------------------x Pa

x=133.322*720/1=95991.8 Pa=95.9918 kPa

p*V=n*R*T

n=p*V/R*T

n=(95.9918*10(3)*15*10(-3))/(8.314*322.15)

n=0.537 mol molekula H2

n(H)=2*n(H2)=2.0537=1.074 mol

1 mol---------------------6*10(23)

1.074 mol-------------------x

x=1.074*6*10(23)/1=6.44*10(23) atoma H

No love's picture

No love - Sun, 12/01/2013 - 21:05

U sudu zapremine 2,0dm3 nalazi se 16,0g CO2 na temperaturi 200C.Izracunati masu N2 koju treba uneti u sud da bi se pritisak povecao tri puta.

milicaradenkovic's picture

milicaradenkovic - Sun, 12/01/2013 - 21:26

p*V=n*R*T

n(CO2)=m/Mr=16/44=0.364 mol

p=n*R*T/V=0.364*8.314*293.15/2*10(-3)

p=443.58 kPa

p1=3*p=3*443.58=1330.74 kPa

n(N2)=V*p/R*T

n(N2)=2*10(-3)*1330.74*10(3)/8.314*293.15

n(N2)=1.092 mol

m(N2)=n(N2)*Mr=1.092*28=30.58 g

aksabg's picture

aksabg - Sun, 12/01/2013 - 22:18

Jedan zadatak po temi. Tema zakljucana.