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shamsutdinovav
@shamsutdinovav
July 2022
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ПОМОГИТЕ РЕШИТЬ ЗАДАЧУ!!!!! ПОЖАЛУЙСТА!!
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shakha2017
Verified answer
Решение
ω(CuO)=100%-20%=80%
m(чистый CuO)=40г*80%/100%=32г
M(CuO)=64+16=80г/моль
Vm=22.4л (при н.у и длч кождого ГАЗА)
32г Хл
CuO+H₂=Cu+H₂O
80г 22,4л
V(H₂)=X=32*22.4/80=8.96л
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shakha2017
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Answers & Comments
Verified answer
Решениеω(CuO)=100%-20%=80%
m(чистый CuO)=40г*80%/100%=32г
M(CuO)=64+16=80г/моль
Vm=22.4л (при н.у и длч кождого ГАЗА)
32г Хл
CuO+H₂=Cu+H₂O
80г 22,4л
V(H₂)=X=32*22.4/80=8.96л