Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]
Cations forming precipitates with \( K_4[{Fe(CN)}_6] \) are:
Cations that do not form characteristic precipitates with \( K_4[{Fe(CN)}_6] \) are:
The cations yielding characteristic precipitates with \( K_4[{Fe(CN)}_6] \) are \( Cu^{2+} \), \( Fe^{3+} \), and \( Zn^{2+} \).
Total precipitating cations: 3. Thus, the answer is \( \boxed{(3)} \).
Kjeldahl's method cannot be used for the estimation of nitrogen in which compound? 
In the group analysis of cations, Ba$^{2+}$ & Ca$^{2+}$ are precipitated respectively as