Analiticka hemija

Submitted by Milance 41 on Sun, 01/11/2015 - 15:33
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Odrediti molaritet 17.5% rastvora hlorovodonicne kiseline,cija je gustina q=1.18g/cm3
V=m/q=17.5g/1.18g/cm3=14.8cm3
c=n/V=m(HCl)/M(HCl)xm(H20)=17.5g/36.5g/molx0.0815=5.811mol/kg
mrr=m(H2o)-m(HCl)
m(H2o)=mrr-m(HCl)=100-17.5=82.5g=0.825
Da li neko zna pravilno da uradi ovaj zadatak
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Milance 41 - Sun, 01/11/2015 - 15:39

Izracunati procentni sastav rastvora nastalog rastvaranjem 30g bakar (II) sulfata pentahidrata u 150g vode.

30g CuSO4x5H2o

mrr(H2o)=150g:30g CuSO4=100g rr : xg CuSO4

x=20                    w=20%

Da li neko zna pravilno da uradi ovaj zadatak

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milicaradenkovic - Sun, 01/11/2015 - 18:19

Jedan zadatak jedna tema.

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Milance 41 - Sun, 01/11/2015 - 22:16
da
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Milance 41 - Sun, 01/11/2015 - 22:41

oba zadatka su analiticka hemija