According to Dalton's law of partial pressures, when two or more gases are in one container without chemical interaction, each behaves independently of the others. The partial pressure of oxygen can, therefore, be found by multiplying the molar fraction of oxygen in the container by the total pressure of the three gases.

\(\displaystyle X_{oxygen}= \frac{2mol\ O_2}{1mol\ H_2+2.5mol\ He+2mol\ O_2}= 0.37\)

\(\displaystyle X_{oxygen}*P=P_{oxygen}\)

\(\displaystyle 0.37 * 4 atm\approx 1.45 atm\)

This is an equation which requires us to find the partial pressures of each gas in the container. The equation for partial pressure is \(\displaystyle P_{a} = X_{a}P_{total}\) where \(\displaystyle X\) is the molar fraction of one of the gases. Since there are twice as many moles of gas A as there are gas B, we know that gas A accounts for 2/3 of the moles in the container.

\(\displaystyle X_a=\frac{mol\ A}{total\ mol}=\frac{2mol\ A}{(2mol A)+(1molB)}=\frac{2}{3}\)

\(\displaystyle 5 = \frac{2}{3}P_{total}\)

\(\displaystyle P_{total} = 7.5atm\)

The molar quantity of a gas is directly proportional to its partial pressure, so if we have a total of 12 moles of gas, and \(\displaystyle \frac{1}{12}\) of the gas sample is water vapor, then water vapor makes up \(\displaystyle \frac{1}{12}\) of the total pressure. Thus to get the partial vapor pressure of any of the substances, we multiply the total vapor pressure by the proportion of that gas.

\(\displaystyle \frac{1}{12}(780) = 65mmHg\)

Iron (II) has a positive two charge: \(\displaystyle Fe^{2+}\).

Phosphate has a negative three charge: \(\displaystyle PO_4^{3-}\).

The initial compound must be constructed to cancel these charges. The dissociation is: \(\displaystyle Fe_3(PO_4)_2\rightleftharpoons 3Fe^{2+}+2PO_4^{3-}\).

Lead (II) hydroxide dissociates according to this reaction: \(\displaystyle Pb(OH)_2\rightleftharpoons Pb^{2+}+2OH^-\). Solubility is equal to the number of moles that will dissociate at equilibrium, and can be found using this reaction and the value of K_sp.

K_sp = [Pb²⁺][OH^-]²

If [Pb²⁺] = x, then [OH^-] = 2x. They exist in a 1:2 ratio.

K_sp = 1.43 * 10^-20 = x(2x)² = 4x³

x = 1.53 x 10^-7M

Increasing the temperature of a solution will generally increase its solubility, while decreasing temperature will have the opposite effect. The common ion effect refers to the phenomenon when two compunds in a solution dissociate to produce at least one similar ion, and will make it harder for additional amounts of that ion in the solution to dissociate, decreasing solubility.

First, set up a balanced reaction and enter X for the amount of each solute that is going to be dissolved. As NaCl dissociates, each ion amount will increase by some amount, X.

\(\displaystyle NaCl (s) \rightarrow Na^{+}(aq) + Cl^{-}(aq)\)

    \(\displaystyle \wr\)                 \(\displaystyle +X\)                 \(\displaystyle +X\)

    \(\displaystyle \wr\)                   \(\displaystyle X\)                    \(\displaystyle X\)

Then set up the equation for the value of K_sp and enter in the value of X for each solute. Use this equation to solve for X, the solubility (in mol/L). Remember that NaCl si not included in the equation, as it is a pure solid.

\(\displaystyle K_{sp}= [Na^{+}][Cl^{-}] = (X)(X) = X^{2}\)

\(\displaystyle 2.70 * 10^{-4} = X^{2}\)

\(\displaystyle X= 1.64 * 10^{-2}M\)

Use the balanced reaction of the dissociation to assign values of X for the solubility of each ion. As CaF₂ dissociates, each ion amount will increase by some amount; calcium will increase by X and flouride will increase by 2X.

\(\displaystyle CaF_{2}(s) \rightarrow Ca^{2+}(aq) + 2F^{-}(aq)\)

       \(\displaystyle \wr\)               \(\displaystyle +X\)                \(\displaystyle +2X\)

       \(\displaystyle \wr\)                 \(\displaystyle X\)                   \(\displaystyle 2X\)

Then plug in the equation to solve K_sp, using the X values assigned for each solute. Use the given value for solubility to find K_sp. Remember that NaCl si not included in the equation, as it is a pure solid.

\(\displaystyle K_{sp}= [Ca^{2+}][F^-]^2=(X)(2X)^{2} = 4X^{3}\)

\(\displaystyle K_{sp}= 4(1.9 * 10^{-2})^{3} = 1.46 * 10^{-8}\)

In order to solve for the solubility of barium hydroxide, we need to establish an ICE table for the reaction. 

I. Initially, there are no barium ions or hydroxide ions in the solution, so we can call their concentrations zero at the beginning of the reaction.

C. For every molecule of dissolved barium hydroxide, there will be one ion of barium and two ions of hydroxide. As a result, the increases in each ion can be designated \(\displaystyle \small +x\) and \(\displaystyle \small +2x\), respectively.

E. Finally, we plug these values into the equilibrium expression and set it equal to the solubility product constant.

\(\displaystyle \begin{matrix} Ba(OH)_2 & Ba^{2+} &OH^- \\ /& 0 & 0\\ / & x & 2x \end{matrix}\)

\(\displaystyle K_{sp}=[Ba^{2+}][OH^-]^2\)

\(\displaystyle \small 5*10^{-3} = [x][2x]^{2}\)

\(\displaystyle \small x = 0.108M\)

The solubility product constants help us envision which ions will have the largest concentrations after dissolving to equilibrium. \(\displaystyle \small CaCrO_{4}\) has the largest solubility product constant, meaning it will dissolve the most out of the four options.

Upon the dissolving of \(\displaystyle \small MgF_{2}\), twice the number of fluoride ions will be released as magnesium ions; however, the solubility constant for magnesium fluoride is so much smaller, compared to the solubility constant of \(\displaystyle \small CaCrO_{4}\), that it will not become the ion with the largest concentration.

\(\displaystyle 2(3.7*10^{-8})< 7.1*10^{-4}\)

\(\displaystyle [F^-]< [CrO_4^{2-}]\)