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dapenko24
@dapenko24
August 2022
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Какой объём хлора вступит в реакцию с 6 моль пропана.
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Platinum15
Verified answer
С3H8 + Cl2 = C3H7Cl + HCl
n(C3H8) = n(Cl2) = 6 моль
V(Cl2) = 22.4 *6 =134.4 л
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Answers & Comments
Verified answer
С3H8 + Cl2 = C3H7Cl + HCln(C3H8) = n(Cl2) = 6 моль
V(Cl2) = 22.4 *6 =134.4 л