Предмет: Химия, автор: safaryanz

вычислите процентное содержание кислорода в молекуле органического вещества C12H22O11

Ответы

Автор ответа: 270719
1
M(C12H22O11)=342г\моль.
342 - 100%
176 - x%
x=176*100/342=51,462%
Ответ: 51.5%
safaryanz: спасибо большое
270719: пожалуйста
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