11. Solubility is the maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature to form a saturated solution.
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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11. Solubility is the maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature to form a saturated solution.
12. Step 1: Obtain sodium metal and chlorine gas.
Step 2: Heat a small piece of sodium metal in a combustion spoon until it starts to burn.
Step 3: Carefully lower the burning sodium into a gas jar filled with chlorine gas.
Step 4: The sodium will react vigorously with the chlorine gas, producing a bright yellow flame and forming white solid sodium chloride.
$$2\text{Na(s)} + \text{Cl}_2\text{(g)} \to 2\text{NaCl(s)}$$
Step 5: Allow the reaction to complete and cool. Collect the solid sodium chloride formed.
13. To prepare a solid sample of $\text{PbSO}_4$ from $\text{PbO}$, follow these steps:
Step 1: Prepare a soluble lead(II) salt solution from $\text{PbO}$.
Since $\text{PbO}$ is an insoluble base, it must first be reacted with a dilute acid to form a soluble lead(II) salt. Nitric acid is commonly used as it forms soluble nitrates.
Add dilute nitric acid to $\text{PbO}$ and warm gently until all the $\text{PbO}$ dissolves, forming lead(II) nitrate solution.
$$\text{PbO(s)} + 2\text{HNO}_3\text{(aq)} \to \text{Pb(NO}_3)_2\text{(aq)} + \text{H}_2\text{O(l)}$$
Step 2: Precipitate lead(II) sulfate.
Add a solution of a soluble sulfate, such as sodium sulfate ($\text{Na}_2\text{SO}_4$) or dilute sulfuric acid ($\text{H}_2\text{SO}_4$), to the lead(II) nitrate solution. Lead(II) sulfate is an insoluble salt and will precipitate out as a white solid.
$$\text{Pb(NO}_3)_2\text{(aq)} + \text{Na}_2\text{SO}_4\text{(aq)} \to \text{PbSO}_4\text{(s)} + 2\text{NaNO}_3\text{(aq)}$$
or
$$\text{Pb(NO}_3)_2\text{(aq)} + \text{H}_2\text{SO}_4\text{(aq)} \to \text{PbSO}_4\text{(s)} + 2\text{HNO}_3\text{(aq)}$$
Step 3: Isolate and purify the lead(II) sulfate.
Filter the mixture to separate the solid $\text{PbSO}_4$ precipitate from the solution.
Wash the precipitate thoroughly with distilled water to remove any soluble impurities.
Dry the solid $\text{PbSO}_4$ in an oven or desiccator.
14. Graham's Law of Diffusion/Effusion:
Graham's Law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass (or density), assuming constant temperature and pressure.
Expressed mathematically:
$$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}$$
where $\text{Rate}_1$ and $\text{Rate}_2$ are the rates of diffusion/effusion of gas 1 and gas 2, respectively, and $M_1$ and $M_2$ are their respective molar masses.
