1. How many grams of NaNO_(3) must be taken to prepare 300 ml of a 0.2M solution? 2. For a complex compound [Ni(NH_(3))_(6)]SO_(4) indicate the outer and inner spheres, central ion , its charge and coordinate number , ligands and there dentation. Name the compound. 3. Finish and balance the redox-reaction CrCl_(3)+H_(2)O_(2)+NaOH= 4. 12 ml of CuCl_(2) solution was react with 7,5 ml of a 0.1245 N solution of NaOH Calculate the mass of Cu(OH)_(2) wich was formed. 5. Calculate the mass of Cd^2+ in 0.1 M saturated solution of [Cd(NH_(3))_(4)]Cl_(2) (K_(inst)=7.6ast 10^-8) 6. Suggest a scheme for analyzing a solution containing the following ions: Pb^2+;Hg_(2)^2+;NO_(3)^- What is the color of the solution?Write the equations of analytical reactions.

Let's address each question step by step:<br /><br />1. **Grams of \( \text{NaNO}_3 \) for a 0.2M solution:**<br /><br /> To find the mass of \( \text{NaNO}_3 \) needed, use the formula:<br /> \[<br /> \text{mass} = \text{molarity} \times \text{volume (L)} \times \text{molar mass}<br /> \]<br /> - Molarity (\( M \)) = 0.2 M<br /> - Volume = 300 mL = 0.3 L<br /> - Molar mass of \( \text{NaNO}_3 \) = 85 g/mol<br /><br /> \[<br /> \text{mass} = 0.2 \, \text{mol/L} \times 0.3 \, \text{L} \times 85 \, \text{g/mol} = 5.1 \, \text{g}<br /> \]<br /><br />2. **Complex compound \([Ni(NH_3)_6]SO_4\):**<br /><br /> - Outer sphere: \( \text{SO}_4^{2-} \)<br /> - Inner sphere: \([Ni(NH_3)_6]^{2+}\)<br /> - Central ion: \( \text{Ni}^{2+} \)<br /> - Charge of central ion: +2<br /> - Coordination number: 6 (from six \( \text{NH}_3 \) ligands)<br /> - Ligands: \( \text{NH}_3 \), monodentate<br /> - Name: Hexaammine nickel(II) sulfate<br /><br />3. **Balancing the redox reaction \( \text{CrCl}_3 + \text{H}_2\text{O}_2 + \text{NaOH} \):**<br /><br /> The balanced equation is:<br /> \[<br /> 2\text{CrCl}_3 + 3\text{H}_2\text{O}_2 + 10\text{NaOH} \rightarrow 2\text{Na}_2\text{CrO}_4 + 6\text{NaCl} + 8\text{H}_2\text{O}<br /> \]<br /><br />4. **Mass of \( \text{Cu(OH)}_2 \) formed:**<br /><br /> First, calculate the moles of \( \text{NaOH} \):<br /> \[<br /> \text{Normality} = 0.1245 \, \text{N}, \quad \text{Volume} = 7.5 \, \text{mL} = 0.0075 \, \text{L}<br /> \]<br /> \[<br /> \text{moles of NaOH} = 0.1245 \, \text{mol/L} \times 0.0075 \, \text{L} = 0.00093375 \, \text{mol}<br /> \]<br /><br /> Reaction: <br /> \[<br /> \text{CuCl}_2 + 2\text{NaOH} \rightarrow \text{Cu(OH)}_2 + 2\text{NaCl}<br /> \]<br /><br /> Moles of \( \text{CuCl}_2 \) = half of moles of \( \text{NaOH} \) = 0.000466875 mol<br /><br /> Molar mass of \( \text{Cu(OH)}_2 \) = 97.56 g/mol<br /><br /> Mass of \( \text{Cu(OH)}_2 \):<br /> \[<br /> \text{mass} = 0.000466875 \, \text{mol} \times 97.56 \, \text{g/mol} = 0.0455 \, \text{g}<br /> \]<br /><br />5. **Mass of \( \text{Cd}^{2+} \) in a saturated solution:**<br /><br /> For the complex \([Cd(NH_3)_4]Cl_2\), the dissociation is:<br /> \[<br /> [Cd(NH_3)_4]^{2+} \rightleftharpoons Cd^{2+} + 4NH_3<br /> \]<br /><br /> Given \( K_{inst} = 7.6 \times 10^{-8} \), assume \( x \) is the concentration of \( \text{Cd}^{2+} \).<br /><br /> \[<br /> K_{inst} = x \cdot (4x)^4 = 256x^5<br /> \]<br /><br /> Solving for \( x \):<br /> \[<br /> 256x^5 = 7.6 \times 10^{-8}<br /> \]<br /> \[<br /> x^5 = \frac{7.6 \times 10^{-8}}{256}<br /> \]<br /> \[<br /> x = \left(\frac{7.6 \times 10^{-8}}{256}\right)^{1/5}<br /> \]<br /><br /> Calculate \( x \) and then find the mass using the molar mass of \( \text{Cd}^{2+} \) (112.41 g/mol).<br /><br />6. **Scheme for analyzing ions \( Pb^{2+}; Hg_2^{2+}; NO_3^{-} \):**<br /><br /> - **Color:** Typically colorless.<br /> - **Reactions:**<br /> - \( \text{Pb}^{2+} + 2\text{I}^- \rightarrow \text{PbI}_2 \) (yellow precipitate)<br /> - \( \text{Hg}_2^{2+} + 2\text{Cl}^- \rightarrow \text{Hg}_2\text{Cl}_2 \) (white precipitate)<br /> - \( \text{NO}_3^- \) can be tested with brown ring test.<br /><br />These steps should help you solve each part of the problem accurately.