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sempaiuu
@sempaiuu
August 2022
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ПОМОГИТЕЕ
Сколько граммов Br получится при взаимодействии 10-ти литров Cl c KBr ?
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svent007
Cl2 + 2 KBr = 2 KCL + Br2
n(Cl2)=10/22.4=0,446
n(Br2)= n(Cl2)=0,446
M(Br2)=160 г/моль
m(Br2)=0,446*160 = 71,36 г
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Answers & Comments
n(Cl2)=10/22.4=0,446
n(Br2)= n(Cl2)=0,446
M(Br2)=160 г/моль
m(Br2)=0,446*160 = 71,36 г