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Anna290100
@Anna290100
July 2022
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Сколько граммов растворенного вещества содержат 3 л 0,001н раствора Н3РО4 (f экв=1/3)
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mrvladimir2
Verified answer
N=3 л*0,001 л*экв/моль/3 = 0,001 моль
М(Н3РО4) = 98 г/моль
m(H3PO4) = 0,001 моль*98 г/моль = 0,098 г
Ответ: 0,098 г
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Answers & Comments
Verified answer
N=3 л*0,001 л*экв/моль/3 = 0,001 мольМ(Н3РО4) = 98 г/моль
m(H3PO4) = 0,001 моль*98 г/моль = 0,098 г
Ответ: 0,098 г