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Oblicz stosunki masowe
a) ZnS
b) Cu₂O
c) Na₂SO


Odpowiedź :

a) ZnS

[tex]\frac{mZn}{mS} = \frac{65u}{32u} = \frac{65}{32} = 65 : 32[/tex]

b) Cu2O

[tex]\frac{mCu}{mO} = \frac{64u * 2}{16 u} = \frac{128}{16} = \frac{8}{1} = 8 : 1[/tex]

c) Na2SO

mNa2 = 23u * 2 = 46 u

mS = 32 u

mO = 16 u

46 : 32 : 16 = 23 : 16 : 8