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aruka13
@aruka13
August 2022
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сколько процентов кислорода в сахарозе (C_12H_12O_11)
a) 3.6
b) 43
c) 63
d) 75
e) 53
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Verified answer
w(э) =
n (э)
∙
Ar(э) \
Mr(в-ва) ∙ 100% = 11
∙ 16\332 =0,53 = 53%
ОТВЕТ: е
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Verified answer
w(э) = n (э) ∙ Ar(э) \Mr(в-ва) ∙ 100% = 11 ∙ 16\332 =0,53 = 53%ОТВЕТ: е