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lola134
@lola134
October 2021
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массовая доля водорода в амиловом спирте равна(%)
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Аксолотльььь
12,5%
M(C5H11OH) = 5*12 + 1*11 + 16 + 1
w(H) = 11/88 * 100 = 0.125 * 100 = 12.5%
Могу ошибаться , но по сути всё по формуле
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Answers & Comments
M(C5H11OH) = 5*12 + 1*11 + 16 + 1
w(H) = 11/88 * 100 = 0.125 * 100 = 12.5%
Могу ошибаться , но по сути всё по формуле