pewien tlenek azotu o masie czasteczkowej 92u zawiera 30,43% azotu,... jaki jest wzór tego tlenku
92u-----100%
xu-------30,43%
x = 28u azotu
Masa N=14u
n=m/M
n=28/14
n=2 atomy azotu
92u - 28u = 64u tlenu
Masa O=16u
n=64/16
n=4 atomy tlenu
N₂O₄
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92u-----100%
xu-------30,43%
x = 28u azotu
Masa N=14u
n=m/M
n=28/14
n=2 atomy azotu
92u - 28u = 64u tlenu
Masa O=16u
n=m/M
n=64/16
n=4 atomy tlenu
N₂O₄