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)
Answer:
The SI base unit for amount of substance is the mole. 1 grams NaOH is equal to 0.025001806380511 mole.
Answer:
1.06 V
Explanation:
The standard reduction potentials are:
Ag^+/Ag E° = 0.7996 V
Ni^2+/Ni E° = -0.257 V
The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are
Ni → Ni^2+ + 2e- E° = 0.257 V
<u>2Ag^+ 2e- → 2Ag </u> <u>E° = 0.7996 V
</u>
Ni + 2Ag^+ → Ni^2+ + 2Ag E° = 1.0566 V
To three significant figures, the standard potential for the cell is 1.06 V
.
Hope you understand how to work out those types of questions now xD ;)
