Ответ:
Элементтің массалық үлесі: \(w(E)=\frac{A_r(E)\cdot N(E)}{M_r(зат)}\cdot100\%\).
- а) \(\mathrm{Al_2O_3}\): \(M_r=2\cdot27+3\cdot16=102\). \(w(Al)=54/102\cdot100\%=52,94\%\), \(w(O)=48/102\cdot100\%=47,06\%\).
- ә) \(\mathrm{Al(OH)_3}\): \(M_r=27+3(16+1)=78\). \(w(Al)=27/78\cdot100\%=34,62\%\), \(w(O)=48/78\cdot100\%=61,54\%\), \(w(H)=3/78\cdot100\%=3,85\%\).
- б) \(\mathrm{K_2O}\): \(M_r=2\cdot39+16=94\). \(w(K)=78/94\cdot100\%=82,98\%\), \(w(O)=16/94\cdot100\%=17,02\%\).
- в) \(\mathrm{KCl}\): \(M_r=39+35,5=74,5\). \(w(K)=39/74,5\cdot100\%=52,35\%\), \(w(Cl)=35,5/74,5\cdot100\%=47,65\%\).
