1)даноW(C) = 8.7%W(K) = 56.52%W(O) = 34.78%-----------------CxKyOz-?CxKyOz = 8.7 / 12 : 56.52/39 : 34.78/16CxKyOz = 0.7 : 1.4 : 2.1 = 1:2:3K2CO3ответ K2CO37)даноV(H2O) = 80 mLW(K2CO3) = 21%---------------------------m(K2CO3)-?m(H2O) = V(H2O)*p(H2O) = 80*1 = 80 gОбозначим карбонат калия через - ХW(X) = X / X+m(H2O) * 100%21% = X / X+80 * 100%21X+1680 = 100XX = 21.3 gответ 21.3 г