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%N = 14u/47u * 100%=1400/47%= 29,8%
HNO3 = 1u + 14u + 3*16u= 15u +48u = 63u
%N = 14u/63u*100% = 1400/63% = 22,2%
większa zawartość procentowa jest w kw. azotowym(III) - HNO2