6.Oblicz stężenie molowe 96% kwasu siarkowego(VI) o gęstości 1,84g/cm3.
Cp=96%
d=1,84g/cm3=1840g/dm3
M H2SO4=98g/mol
Cm=?
Cm=Cp*d/100%*M
Cm=96*1840/100*98
Cm = 18mol/dm3
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Cp=96%
d=1,84g/cm3=1840g/dm3
M H2SO4=98g/mol
Cm=?
Cm=Cp*d/100%*M
Cm=96*1840/100*98
Cm = 18mol/dm3