M(AlBr3) = 27 + 3×80 = 267 г/моль
w(Al) = 27/267 = 0,1011 (10,11%)
Ответ: 10,11%
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M(AlBr3) = 27 + 3×80 = 267 г/моль
w(Al) = 27/267 = 0,1011 (10,11%)
Ответ: 10,11%