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Ответ

Ответ дан
denisgr98

Найти массовые доли элементов в соединении оксида хлора 7 (Cl2O7)

Дано:

Cl2O7

Решение:

M(Cl2O7) = 2*35,5 + 7*16 = 183

w(Cl) = (2*M(Cl))/M(Cl2O7) = (2*35,5)/183 = 0,388 или 38,8%

w(O) = (7*M(O))/M(Cl2O7) = (7*16)/183 = 0,612 или 61,2%

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