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mr=ms+m alkoholu
ms=30g
m alkoholu=V*d=60*0,85=51g
mr=30g+51g=81g
Cp=ms*100%/mr
Cp=30g*100%/81=37,04%
51g + 30g = 81g
81g - 100%
30g - x %
x = 3000 / 81 ~ 37.04%
Zawartość procentowa azotanu srebra w etanolu wynosi ~ 37.04%