Explanation:
For what I can see, is missing the concentration of [Ag+] in the half-cell. To calculate it:
Niquel half-cell
Oxidation reaction: 
![E=E^0 - \frac{R*T}{n*F}*ln(1/[Ni^{2+}])](https://tex.z-dn.net/?f=E%3DE%5E0%20-%20%5Cfrac%7BR%2AT%7D%7Bn%2AF%7D%2Aln%281%2F%5BNi%5E%7B2%2B%7D%5D%29)
Assuming T=298 K / R=8.314 J/mol K / F=96500 C
Silver half-cell
Reduction reaction: 
![E=E^0 - \frac{R*T}{n*F}*ln(1/[Ag+])](https://tex.z-dn.net/?f=E%3DE%5E0%20-%20%5Cfrac%7BR%2AT%7D%7Bn%2AF%7D%2Aln%281%2F%5BAg%2B%5D%29)
Assuming T=298 K / R=8.314 J/mol K / F=96500 C
![0.835V=0.8V - \frac{8.314*298}{1*96500}*ln(1/[Ag+])](https://tex.z-dn.net/?f=0.835V%3D0.8V%20-%20%5Cfrac%7B8.314%2A298%7D%7B1%2A96500%7D%2Aln%281%2F%5BAg%2B%5D%29)
![[Ag+]=0.26 M](https://tex.z-dn.net/?f=%5BAg%2B%5D%3D0.26%20M)
The answer is Single Replacement.
<em>C = 0,75 mol/dm³</em>
<em>V = 500mL = 500cm³ = 0,5dm³</em>
C = n/V
n = 0,75×0,5dm³
<u>n = 0,375 moles</u>
<em>M NaCl: 23+35,5 = 58,5g</em>
1 mole ---------- 58,5g
0,375 ----------- X
X = 0,375×58,5
<u>X = 21,9375g NaCl</u>
:)
Answer:
. The products are lithium sulfate, Li2SO4 , and water, H2O
Explanation:
hoped i helped!!
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