masaNa2O = 23*2 + 16 = 62
masa Na2 = 23*2= 46
62 - 100%
46 - x
x=46*100%/62
x= 74,19% - zawartosc % Na w Na2O
masaNaOH=23+16+1 = 40
masaNa = 23
40 - 100%
23 - x
x=23*100%/40
x=57,5% - zawartosc % Na w NaOH
Odp. Większą zawartosc % Na ma Na2O
Na2O = 23*2+16=62u%Na=46*100%/62=74,19%NaOH = 23+16+1=40u%Na=23/40*100%=57,5%74,19>57,5 Wiekszą zawartość % sodu ma Na2O
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masaNa2O = 23*2 + 16 = 62
masa Na2 = 23*2= 46
62 - 100%
46 - x
x=46*100%/62
x= 74,19% - zawartosc % Na w Na2O
masaNaOH=23+16+1 = 40
masaNa = 23
40 - 100%
23 - x
x=23*100%/40
x=57,5% - zawartosc % Na w NaOH
Odp. Większą zawartosc % Na ma Na2O
Na2O = 23*2+16=62u
%Na=46*100%/62=74,19%
NaOH = 23+16+1=40u
%Na=23/40*100%=57,5%
74,19>57,5
Wiekszą zawartość % sodu ma Na2O