Molim pomoc za zadatak koji glasi:
Na sobnoj temperaturi gasoviti ugljovodonik ima sledeci sastav:
82.76% C, 17,24% H. Pri normalnim uslovima 1 dm3 ovog jedinjenja ima masu 2,59 g. Odredite molekulsku formulu jedinjenja.
n(C):n(H)=(82.76/12):(17.24/1)
n(C):n(H)=2:5
n(CxHy)=1dm^3/22.4dm^3/mol=0.04464mol
M(CxHy)=m/n(CxHy)=2.59g/0.04464mol=58g/mol
(2*12+5*1)*a=58
a=2. Odatle je formula C4H10.
The goal of life is living in agreement with nature.
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n(C):n(H)=(82.76/12):(17.24/1
n(C):n(H)=(82.76/12):(17.24/1)
n(C):n(H)=2:5
n(CxHy)=1dm^3/22.4dm^3/mol=0.04464mol
M(CxHy)=m/n(CxHy)=2.59g/0.04464mol=58g/mol
(2*12+5*1)*a=58
a=2.
Odatle je formula C4H10.
The goal of life is living in agreement with nature.