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Irisha20115
@Irisha20115
July 2022
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Смешали раствор массой 200 г с массовой долей натрий сульфата 14,2% и раствор массой 200 г с массовой долей барий хлорида 5,2%. Вычислите массу образовавшегося осадка.
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Irina108
Na2SO4 + BaCl2= 2 NaCl+BaSO4
m(Na2SO4)=200*0,142=28,4
n(Na2SO4)=28,4/142=0,2 моль-изб.
m(BaCl2)=200*0,052=10,4
n(BaCl2)=10,4*208=0,05 моль
m(BaSO$)=0,05*233=64,05 г
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Answers & Comments
m(Na2SO4)=200*0,142=28,4
n(Na2SO4)=28,4/142=0,2 моль-изб.
m(BaCl2)=200*0,052=10,4
n(BaCl2)=10,4*208=0,05 моль
m(BaSO$)=0,05*233=64,05 г