Pri sagorjevanju uzorka celika mase 5g u struji kiseonika dobijen je o,1g CO2.Koliki je maseni udeo ugljenika u uzorku celika?
n(CO2)=0.1g/44g/mol=0.00227mol n(C)=n(CO2)=0.00227mol m(C)=n(C)*12g/mol=0.00227mol*12g/mol=0.02724g w=0.02724g/5g=0.00545=0.545%
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n(CO2)=0.1g/44g/mol=0.00227mo
n(CO2)=0.1g/44g/mol=0.00227mol
n(C)=n(CO2)=0.00227mol
m(C)=n(C)*12g/mol=0.00227mol*12g/mol=0.02724g
w=0.02724g/5g=0.00545=0.545%