Use the dilution formula:

$\displaystyle M_{1}V_{1}=M_{2}V_{2}$

Rearranging this equation gives:

$\displaystyle \frac{M_{1}V_{1}}{V_{2}}=M_{2}$

Plugging in the values gives:

$\displaystyle M_{2}=\frac{0.35\ M * 30\ mL}{50\ mL}= 0.21 M$

Therefore, after diluting the solution to 50mL, the solution concentration would be lowered from 0.35M to 0.21M.

By using the concentration as a conversion factor, the number of moles can calculated by multiplying the concentration by the number of liters.

$\displaystyle \frac{5*10^{-3}moles}{L} * 0.010L = 5*10^{-5}moles$

Therefore, there are $\displaystyle 5*10^{-5}moles$ in $\displaystyle 0.010 L$ of a $\displaystyle 5*10^{-3}M$ solution.

In order to calculate the concentration, we must use molarity formula:

$\displaystyle Molarity = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}$

We must use the molecular weight of sodium chloride to calculate the moles of solute:

$\displaystyle MW=(atomic\ weight\ of\ Na)+(atomic\ weight\ of\ Cl)$

$\displaystyle MW_{NaCl}=(23 \frac{g}{mole})+(35 \frac{g}{mole})= 58 \frac{g}{mole}$

$\displaystyle 0.082g*\frac{mole}{58g}=0.0014\ moles\ NaCl$

$\displaystyle \frac{0.0014\ moles\ CH_{3}CH_{2}OH}{0.035\L} = 0.040\ M$

Therefore, the concentration in molarity of this solution is 0.040M.

By simply using the concentration as a conversion factor, the number of moles can be calculated by multiplying the concentration by the number of liters. Before calculating the number of moles, the number of milliliters must be converted to liters using the fact that $\displaystyle 1\ liter = 1000\ mL$.

$\displaystyle \frac{3*10^{-5}\ moles}{L} * 1\ L = 3*10^{-5}\ moles$

Therefore, there are $\displaystyle 3*10^{-5}\ moles$ in $\displaystyle 0.010\ L$ of a $\displaystyle 5*10^{-3}M$ solution.

First, let us write out an ion exchance reaction for the reactants:

$\displaystyle Na_2SO_{4\hspace{1 mm}(aq)}+BaCl_{2\hspace{1 mm}(aq)}\rightarrow 2NaCl\hspace{1 mm}+BaSO_4$

By solubility rules, $\displaystyle NaCl$ is soluble in water and $\displaystyle BaSO_4$ is not. Our new reaction is:

$\displaystyle Na_2SO_{4\hspace{1 mm}(aq)}+BaCl_{2\hspace{1 mm}(aq)}\rightarrow 2NaCl_{(aq)}\hspace{1 mm}+BaSO_{4\hspace{1 mm}(s)}$

Now we will calculate the theoretical yield of each reactant.

$\displaystyle 300\hspace{1 mm}mL\times\frac{1 L}{1000 mL}\times \frac{0.3\hspace{1 mm}moles\hspace{1 mm}Na_2SO_4}{1 L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}Na_2SO_4}\times \frac{233.4\hspace{1 mm}g\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}=21.0\hspace{1 mm}g\hspace{1 mm}BaSO_4$

Now we perform the same calculation beginning with $\displaystyle BaCl_2$:

$\displaystyle 200\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.4\hspace{1 mm}moles\hspace{1 mm}BaCl_2}{1\hspace{1 mm}L}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}BaCl_2}\times\frac{233.4\hspace{1 mm}g\hspace{1 mm}BaSO_4}{1\hspace{1 mm}mole\hspace{1 mm}BaSO_4}=18.7\hspace{1 mm}g\hspace{1 mm}BaSO_4$

The limiting reagent is $\displaystyle BaCl_2$ and this reaction produces 18.7 g precipitate.

$\displaystyle 234\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.4\hspace{1 mm}moles\hspace{1 mm}KCl}{1\hspace{1 mm}L}\times\frac{74.55\hspace{1 mm}g\hspace{1 mm}KCl}{1\hspace{1 mm}mole\hspace{1 mm}KCl}=6.98\hspace{1 mm}g\hspace{1 mm}KCl$

First, you must recognize that the chemical formula for sodium hydroxide is $\displaystyle NaOH$. The mass of the boiled solution is

$\displaystyle 344\hspace{1 mm}mL\times\frac{0.35\hspace{1 mm}mol\hspace{1 mm}NaOH}{1000\hspace{1 mm}mL}\times\frac{39.997\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mol\hspace{1 mm}NaOH}= 4.82\hspace{1 mm}g\hspace{1 mm}NaOH$

The remaining mass with be equal to the mass of the sodium chloride in the solution. Once the solvent (water) evaporates, the solute will remain.

Atomic mass of sodium is ~23. Atomic mass of chlorine is ~35.5. Molecular mass of NaCl is ~58.5.

$\displaystyle 0.100L*\frac{0.4mol}{1L}=0.040mol\ NaCl$

$\displaystyle 0.040mol\ NaCl*\frac{58.5g\ NaCl}{1mol\ NaCl}=2.34g\ NaCl$

First we will figure out the number of moles of $\displaystyle Li_3 PO_4$ that we have. We have 5.2 L of a 0.3 M solution, so:

$\displaystyle 0.3\hspace{1mm} \frac{mol}{L} \times 5.2\hspace{1 mm} L = 1.56 \hspace{1 mm}moles$

Now the problem is to find the mass of 1.56 moles of $\displaystyle Li_3 PO_4$. First, we need to know the molecular weight of $\displaystyle Li_3 PO_4$. We will go to the periodic table and add up the mass of each element present.

$\displaystyle (3\hspace{1 mm}\times6.94\hspace{1 mm}\frac{grams\hspace{1 mm} Li}{mol\hspace{1 mm}Li})+30.97\hspace{1 mm}\frac{grams\hspace{1 mm}P}{mol\hspace{1 mm}P}+(4\hspace{1 mm}\times 16\hspace{1 mm}\frac{grams\hspace{1 mm}O}{mol\hspace{1mm}O})= 115.79\hspace{1mm}\frac{grams\hspace{1 mm} Li_3 PO_4}{mol\hspace{1 mm}Li_3 PO_4}$

Now the problem is simple as we have the molar mass and the number of desired moles.

$\displaystyle 115.79\hspace{1 mm}\frac{grams\hspace{1 mm}Li_3 PO_4}{mol\hspace{1 mm}Li_3 PO_4}\times1.56\hspace{1 mm}moles\hspace{1 mm}Li_3 PO_4 = 181\hspace{1 mm}grams\hspace{1 mm}Li_3 PO_4$

Use the Henderson-Hasselbalch equation to determine the ratio of bicarbonate to carbonic acid in solution:

$\displaystyle pH = pK_a + \log\frac{\left [A^- \right ]}{\left [HA \right ]}$

$\displaystyle pH = pK_a + \log\frac{[HCO_3^-]}{[H_2CO_3]}$

$\displaystyle 5.60-6.37 = -0.77 = \log\frac{\left [HCO_3^- \right ]}{\left [H_2CO_3 \right ]}$

Solve for the ratio we need to answer the question:

$\displaystyle 10^{-0.77} = \frac{\left [ HCO_3^-\right ]}{[H_2CO_3]} = 0.170$