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Biserka1999
@Biserka1999
June 2022
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Найти массовую долю каждого элемента
Al(HCO₃)₃
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Senpoliya
Verified answer
Mr(Al(HCO₃)₃) = 27 + (1 + 12 + 16 * 3) * 3 = 210
w%(Al) = 27 : 210 * 100 % = 12,857 %
w%(H) = (1 * 3) : 210 * 100 % = 1,429%
w%(C) = (12 * 3) : 210 * 100 % = 17,143 %
w%(O) = (16 * 3 * 3) : 210 * 100 % = 68,571 %
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Answers & Comments
Verified answer
Mr(Al(HCO₃)₃) = 27 + (1 + 12 + 16 * 3) * 3 = 210w%(Al) = 27 : 210 * 100 % = 12,857 %
w%(H) = (1 * 3) : 210 * 100 % = 1,429%
w%(C) = (12 * 3) : 210 * 100 % = 17,143 %
w%(O) = (16 * 3 * 3) : 210 * 100 % = 68,571 %