mNa2O=46u+16u=62u%Na = (46/62)*100% = 74,2% NamNaOH=23u+1u+16u=40u%Na = (23/40)*100% = 57,7% Na
Więcej sodu jest zawarte w Na2O
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mNa2O=46u+16u=62u
%Na = (46/62)*100% = 74,2% Na
mNaOH=23u+1u+16u=40u
%Na = (23/40)*100% = 57,7% Na
Więcej sodu jest zawarte w Na2O