Answer : The concentration of silver ion is, 
Explanation :
Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.
The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.
As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.
The given equilibrium reaction is,
The expression of 
will be,
![K_{eq}=[Ag^+]^2[S^{2-}]](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5BAg%5E%2B%5D%5E2%5BS%5E%7B2-%7D%5D)
![2.4\times 10^{-4}=(2.5\times 110^{-1})^2[S^{2-}]](https://tex.z-dn.net/?f=2.4%5Ctimes%2010%5E%7B-4%7D%3D%282.5%5Ctimes%20110%5E%7B-1%7D%29%5E2%5BS%5E%7B2-%7D%5D)
![[S^{2-}]=3.8\times 10^{-3}M](https://tex.z-dn.net/?f=%5BS%5E%7B2-%7D%5D%3D3.8%5Ctimes%2010%5E%7B-3%7DM)
Therefore, the concentration of silver ion is,
I would say Na. Oxygen has 2 valence electrons and when reacting with other molecules, the ones with the fullest or emptiest shells will react the least. Both H2 and Na are in the Alkali Metals in the first row, but since H2 has 2 molecules, it would use more oxygen than Ana
Answer:
At the cathode in an electrolytic cell, ions in the surrounding solution are reduced into atoms, which precipitate or plate out on to the solid cathode. The anode is where oxidation takes place, and the cathode is where reduction takes place. The anode is defined as the electrode where oxidation occurs. The cathode is the electrode where reduction takes place. ... At the cathode, the metal ion in the solution will accept one or more electrons from the cathode, and the ion's oxidation state will reduce to 0. This forms a solid metal that deposits on the cathode.
examining the structure of plant cells
