Answer :
AgI should precipitate first.
The concentration of 
when CuI just begins to precipitate is, 
Percent of remains is, 0.0076 %
Explanation :
for CuI is 
for AgI is 
As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.
Now we have to calculate the concentration of iodide ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Cu^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BI%5E-%5D)
![1\times 10^{-12}=0.0079\times [I^-]](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-12%7D%3D0.0079%5Ctimes%20%5BI%5E-%5D)
![[I^-]=1.25\times 10^{-10}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D1.25%5Ctimes%2010%5E%7B-10%7DM)
Now we have to calculate the concentration of silver ion.
The solubility equilibrium reaction will be:
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ag^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BI%5E-%5D)
![8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M](https://tex.z-dn.net/?f=8.3%5Ctimes%2010%5E%7B-17%7D%3D%5BAg%5E%2B%5D%5Ctimes%201.25%5Ctimes%2010%5E%7B-10%7DM)
![[Ag^+]=6.64\times 10^{-7}M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D6.64%5Ctimes%2010%5E%7B-7%7DM)
Now we have to calculate the percent of remains in solution at this point.
Percent of remains = 
Percent of remains = 0.0076 %
The mass of a substance per unit volume is the substance's density.
D = m/v.
1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).
2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.
3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.
4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
Answer:
0.15M
Explanation:
The equation for molarity is M= n/L. Where "M" is Molarity, "n" is the number of moles of solute, and "L" is the total liters in solution.
You need to calculate the number of moles from the given grams. The molar mass of KOH is (39.098+ 16 +1.008)= 56.106g. To calculate the mols of KOH, 
× 
= 0.44558... mol, you see that the grams unit cancel out leaving you with mol as the unit.
The volume is given in L already so no need to do any conversion. M= 
= 0.1485M ≈ 0.15M
