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)
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
1.D 2.C 3.B 4.A is the answer i believe you're looking for.
Answer:
<em>lithosphere</em><em> </em><em>(land)</em>
<em>hydrosphere (water)</em>
<em>biosphere</em><em> </em><em>(living things)</em>
<em>atmosphere (air).</em>
