how to calculate weight percent of oxygen in BaNi0.5Nb0.5O3
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Ethanol=C2H5OH
C2= 2*12
H5=5*1
O=16
H=1
24+5+16+1= 46g/mole
Carbon= (24/46)*100= 52.2%
Hydrogen= 5+1=6 (6/46)*100= 13.0%
Oxygen= (16/46)*100= 34.8%
C2= 2*12
H5=5*1
O=16
H=1
24+5+16+1= 46g/mole
Carbon= (24/46)*100= 52.2%
Hydrogen= 5+1=6 (6/46)*100= 13.0%
Oxygen= (16/46)*100= 34.8%
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