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Вычислите массу осадка который образуется при действии раствора избытка хлорида меди(II) на 80г 10 % раствора гидроксида натрия
Пожалуйста решите задачу Приведите подробное решение
Ответы 1
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Решение:CuCl2 + 2NaOH = Cu(OH)2 + 2NaClm(NaOH в-ва) = 80г*10%/100% = 8гm(Cu(OH)2) = 8г*98г/80г = 9,8гОтвет: 9,8г
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Автор:
munchkin - 5 лет назад
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