1-45
Вычислите массовые доли элементов в следующих соединениях:
а) Mg(NO3)2; б) Al2(SO4)3; в) Ca3(PO4)2.
а)
Ответ: `ω(Mg) = 16.2%`, `ω(N) = 18.9%`, `ω(O) = 64.9%`. |
|
Дано: | Решение |
---|---|
`Mg(NO_3)_2` |
`ω(Mg) = (100*k*A_r(Mg))/(M_r(Mg(NO_3)_2)) = (100*1*24)/148 = 16.2%` `ω(N) = (100*k*A_r(N))/(M_r(Mg(NO_3)_2)) = (100*2*14)/148 = 18.9%` `ω(O) = (100*k*A_r(O))/(M_r(Mg(NO_3)_2)) = (100*6*16)/148 = 64.9%` |
`ω(Mg) = ?` `ω(N) = ?` `ω(O) = ?` |
б)
Ответ: `ω(Al) = 15.8%`, `ω(S) = 28.1%`, `ω(O) = 56.1%`. |
|
Дано: | Решение |
---|---|
`Al_2(SO_4)_3` |
`ω(Al) = (100*k*A_r(Al))/(M_r(Al_2(SO_4)_3)) = (100*2*27)/342 = 15.8%` `ω(S) = (100*k*A_r(S))/(M_r(Al_2(SO_4)_3)) = (100*3*32)/342 = 28.1%` `ω(O) = (100*k*A_r(O))/(M_r(Al_2(SO_4)_3)) = (100*12*16)/342 = 56.1%` |
`ω(Al) = ?` `ω(S) = ?` `ω(O) = ?` |
в)
Ответ: `ω(Ca) = 38.7%`, `ω(P) = 20%`, `ω(O) = 41.3%`. |
|
Дано: | Решение |
---|---|
`Ca_3(PO_4)_2` |
`ω(Ca) = (100*k*A_r(Ca))/(M_r(Ca_3(PO_4)_2)) = (100*3*40)/310 = 38.7%` `ω(P) = (100*k*A_r(P))/(M_r(Ca_3(PO_4)_2)) = (100*2*31)/310 = 20%` `ω(O) = (100*k*A_r(O))/(M_r(Ca_3(PO_4)_2)) = (100*8*16)/310 = 41.3%` |
`ω(Ca) = ?` `ω(P) = ?` `ω(O) = ?` |