You're on a roll — Step 1: Calculate the concentration of HCl. 3) Step 1: Write the balanced chemical equation for the neutralization reaction. $$ \text{NaOH(aq)} + \text{HCl(aq)} \to \text{NaCl(aq)} + \text{H}_2\text{O(l)} $$ From the equation, the mole ratio of $\text{NaOH}$ to $\text{HCl}$ is $1:1$. Step 2: Calculate the moles of $\text{NaOH}$ used. Given: Volume of $\text{NaOH} = 26 \text{ cm}^3 = 0.026 \text{ dm}^3$ Concentration of $\text{NaOH} = 0.1 \text{ M}$ $$ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} $$ $$ \text{Moles of NaOH} = 0.1 \text{ mol/dm}^3 \times 0.026 \text{ dm}^3 $$ $$ = 0.0026 \text{ mol} $$ Step 3: Determine the moles of $\text{HCl}$ reacted. Since the mole ratio of $\text{NaOH}$ to $\text{HCl}$ is $1:1$, $$ \text{Moles of HCl} = \text{Moles of NaOH} = 0.0026 \text{ mol} $$ Step 4: Calculate the concentration of $\text{HCl}$. Given: Volume of $\text{HCl} = 12.4 \text{ cm}^3 = 0.0124 \text{ dm}^3$ $$ \text{Concentration of HCl} = \frac{\text{Moles of HCl}}{\text{Volume of HCl}} $$ $$ \text{Concentration of HCl} = \frac{0.0026 \text{ mol}}{0.0124 \text{ dm}^3} $$ $$ \approx 0.209677 \text{ mol/dm}^3 $$ Rounding to three significant figures: $$ \text{Concentration of HCl} \approx 0.210 \text{ mol/dm}^3 $$ The concentration of HCl is $\boxed{\text{0.210 mol/dm}^3}$. 1) Step 1: Write the balanced chemical equation for the combustion of carbon(II) oxide. $$ \text{2CO(g)} + \text{O}_2\text{(g)} \to \text{2CO}_2\text{(g)} $$ Step 2: Determine the limiting reactant using the given volumes and the mole ratio from the balanced equation. According to the equation, $2 \text{ volumes of CO}$ react with $1 \text{ volume of O}_2$. Given: Volume of $\text{CO} = 36 \text{ cm}^3$ Volume of $\text{O}_2 = 72 \text{ cm}^3$ For $36 \text{ cm}^3$ of $\text{CO}$, the required volume of $\text{O}_2$ is: $$ 36 \text{ cm}^3 \text{ CO} \times \frac{1 \text{ volume O}_2}{2 \text{ volumes CO}} = 18 \text{ cm}^3 \text{ O}_2 $$ Since $18 \text{ cm}^3$ of $\text{O}_2$ is required and $72 \text{ cm}^3$ of $\text{O}_2$ is available, $\text{O}_2$ is in excess. $\text{CO}$ is the limiting reactant. Step 3: Calculate the volume of $\text{CO}_2$ produced. From the equation, $2 \text{ volumes of CO}$ produce $2 \text{ volumes of CO}_2$. $$ \text{Volume of CO}_2 = 36 \text{ cm}^3 \text{ CO} \times \frac{2 \text{ volumes CO}_2}{2 \text{ volumes CO}} = 36 \text{ cm}^3 \text{ CO}_2 $$ Step 4: Calculate the volume of excess $\text{O}_2$ remaining. $$ \text{Excess O}_2 = \text{Initial O}_2 - \text{O}_2 \text{ reacted} $$ $$ \text{Excess O}_2 = 72 \text{ cm}^3 - 18 \text{ cm}^3 = 54 \text{ cm}^3 $$ Step 5: Determine the total volume of the resulting gaseous mixture. The gaseous mixture consists of the product ($\text{CO}_2$) and the unreacted excess reactant ($\text{O}_2$). $$ \text{Total volume} = \text{Volume of CO}_2 + \text{Volume of excess O}_2 $$ $$ \text{Total volume} = 36 \text{ cm}^3 + 54 \text{ cm}^3 = 90 \text{ cm}^3 $$ The total volume of the resulting gaseous mixture is $\boxed{\text{90 cm}^3}$. 4) To prepare crystals of copper(II) chloride from copper metal: 1. Heat the copper metal in air to form copper(II) oxide. This is because copper does not react with dilute hydrochloric acid. $$ \text{2Cu(s)} + \text{O}_2\text{(g)} \xrightarrow{\text{heat}} \text{2CuO(s)} $$ 2. Add the black copper(II) oxide to a beaker containing dilute hydrochloric acid. Stir the mixture gently. The copper(II) oxide will react with the acid to form a blue solution of copper(II) chloride. $$ \text{CuO(s)} + \text{2HCl(aq)} \to \text{CuCl}_2\text{(aq)} + \text{H}_2\text{O(l)} $$ 3. Heat the solution gently to evaporate some of the water, concentrating the solution until it is saturated. This can be tested by dipping a clean glass rod into the solution and observing if crystals form on cooling. 4. Allow the concentrated solution to cool slowly at room temperature. Copper(II) chloride crystals will form. 5. Filter the crystals from the remaining solution (mother liquor) and wash them with a small amount of cold distilled water to remove impurities. 6. Dry the crystals between filter papers or in a desiccator. 5) An isotope refers to atoms of the same element that have the same number of protons (atomic number) but different numbers of neutrons, resulting in different mass numbers. They have identical chemical properties but differ in their physical properties. Got more? Send 'em.