Azot

Submitted by Hemicarka5 on Sun, 12/15/2013 - 19:29
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U 1 zapremini vode moze se rastvoriti 700 zapremina amonijaka.Izracunajte koliko se molova NH3 moze rastvoriti u 0.5 dm3 vode.

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milicaradenkovic - Sun, 12/15/2013 - 20:57

1:700=0.5:x

x=0.5*700/1=350  dm3 NH3

V(NH3)=n*Vm

n=V(NH3)/Vm=350/22.4=15.625 mol