zapremina

Submitted by laky1234 on Sat, 05/31/2014 - 15:35
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Kolika je ukupna zapremina gasova (svedena na normalne uslove) dobijenih razlaganjem 33,6cm3  vodene pare?

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milicaradenkovic - Sat, 05/31/2014 - 16:34

2H2O-------------------->2H2+O2

n(H2O)=V(H2O)/Vm=33.6*10(-3)/22.4=0.0015 mol

n(H2)=n(H2O)=0.0015 mol

n(O2)=n(H2O)/2=0.0015/2=0.00075 mol

V(O2)=n(O2)*Vm=22.4*0.00075=0.0168 l=16.8 ml

V(H2)=n(H2)*Vm=22.4*0.0015=0.0336 l=33.6 ml

Vu=V(O2)+V(H2)=33.6+16.8=50.4 ml

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laky1234 - Sat, 05/31/2014 - 16:47

hvala puno :)