Ответ:
534г
n(Al) =n(AlCl3) =24,08×10^23:(6, 02×10^23)=4 моль
m(AlCl3) =4моль*(27+35.5*3)г/моль=534г
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Answers & Comments
Ответ:
534г
n(Al) =n(AlCl3) =24,08×10^23:(6, 02×10^23)=4 моль
m(AlCl3) =4моль*(27+35.5*3)г/моль=534г