C, to make sure the design works as expected.
A prototype is first, typical model of the said product. Hope this helps!
Answer:
Explanation:
first write the equilibrium equaion ,
⇄ 
assuming degree of dissociation 
=1/10;
and initial concentraion of =c;
At equlibrium ;
concentration of 
![[C_3H_5O_3^{-} ]= c\alpha](https://tex.z-dn.net/?f=%5BC_3H_5O_3%5E%7B-%7D%20%20%5D%3D%20c%5Calpha)
![[H^{+}] = c\alpha](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20c%5Calpha)
is very small so 
can be neglected
and equation is;
= 
![P_H =- log[H^{+} ]](https://tex.z-dn.net/?f=P_H%20%3D-%20log%5BH%5E%7B%2B%7D%20%5D)
composiion ;
![c=\frac{1}{\alpha} \times [H^{+}]](https://tex.z-dn.net/?f=c%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%20%5Ctimes%20%5BH%5E%7B%2B%7D%5D)
![[H^{+}] =antilog(-P_H)](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3Dantilog%28-P_H%29)
![[H^{+} ] =0.0014](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%20%3D0.0014)
13.5 Moles Kcl x 6.022 x 10^23 /1
= 8.12 x 10^24
whenever you need to find molecules just multiply moles by avogadro number
Cu2S is the one that is going to form a precipitate and Cu(NO3)2 will not form a precipitate because it is soluble. It is a fact that those that are insoluble will form a precipitate. I hope that this is the answer that you were looking for and the answer has actually come to your desired help.
