Koliko je grama natrijum-karbonata i vode potrebno za pripremanje 100 g rastvora molalnosti 1mol/kg?
b=n/m(H2O) m(H2O)=mR-m(Na2CO3)=mR-n*106g/mol 1mol/kg=n/0.1kg-0.106kg/mol*n 0.1mol-0.106n=n 0.1mol=1.106n n=0.090416mol m(Na2CO3)=0.090416mol*106g/mol=9.584g m(H2O)=100g-9.584g=90.416g
Je l` moze neko da mi objasni odakle ovde 1,106?
b=n/m(H2O)m(H2O)=mR-m(Na2CO3)=mR-n*106g/mol1mol/kg=n/0.1kg-0.106kg/mol*n0.1mol-0.106n=n0.1mol=1.106nn=0.090416molm(Na2CO3)=0.090416mol*106g/mol=9.584gm(H2O)=100g-9.584g=90.416g
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b=n/m(H2O) m(H2O)=mR-m(Na2CO3
b=n/m(H2O)
m(H2O)=mR-m(Na2CO3)=mR-n*106g/mol
1mol/kg=n/0.1kg-0.106kg/mol*n
0.1mol-0.106n=n
0.1mol=1.106n
n=0.090416mol
m(Na2CO3)=0.090416mol*106g/mol=9.584g
m(H2O)=100g-9.584g=90.416g
?
Je l` moze neko da mi objasni odakle ovde 1,106?
b=n/m(H2O)
m(H2O)=mR-m(Na2CO3)=mR-n*106g/mol
1mol/kg=n/0.1kg-0.106kg/mol*n
0.1mol-0.106n=n
0.1mol=1.106n
n=0.090416mol
m(Na2CO3)=0.090416mol*106g/mol=9.584g
m(H2O)=100g-9.584g=90.416g