Oblicz jaką objętość w warunkach normalnych zajmie:4,4 tlenku azotu(III).
1 mol tlenku azotu (III) - 76g/mol
x moli - 4,4g
x = 0.0579 moli
1mol - 22.4 dm3
0.0579 moli - x dm3
x = 12.97 dm3
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1 mol tlenku azotu (III) - 76g/mol
x moli - 4,4g
x = 0.0579 moli
1mol - 22.4 dm3
0.0579 moli - x dm3
x = 12.97 dm3