Koliko mg NaOH treba dodati u 200 ml vode da bi se u toj zapremini nalazilo 1,2x1018 jona OH- ?
n(OH-)=1.2*1018/(6*1023/mol)=2*10-6mol
n(NaOH)= n(OH-)= 2*10-6mol
m(NaOH)= 2*10-6 mol*40 g/mol=0.08mgThe goal of life is living in agreement with nature.
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n(OH-)=1.2*1018/(6*1023/mol)=
n(OH-)=1.2*1018/(6*1023/mol)=2*10-6mol
n(NaOH)= n(OH-)= 2*10-6mol
m(NaOH)= 2*10-6 mol*40 g/mol=0.08mg
The goal of life is living in agreement with nature.