m(BaCl2)=20×0,1+15×0,13=3,95г
m(ppa)=20+15=35г
w(BaCl2)=3,95:35≈0,1128=11,28≈11,3%
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m(BaCl2)=20×0,1+15×0,13=3,95г
m(ppa)=20+15=35г
w(BaCl2)=3,95:35≈0,1128=11,28≈11,3%