is for each) . How many grams of NaNO_(3) must be taken to prepare 300 ml of a 0.2M solution? For a complex compound [Ni(NH_(3))_(6)]SO_(4) indicate the outer and in spheres, central ion, its charge and coordina dentation. Name the

To prepare a 0.2M solution of $NaNO_{3}$, we need to calculate the amount of $NaNO_{3}$ required in grams.<br /><br />Given information:<br />- Volume of the solution: 300 ml (0.3 L)<br />- Molarity of the solution: 0.2M<br /><br />Step 1: Calculate the number of moles of $NaNO_{3}$ required.<br />Molarity (M) = Moles of solute / Volume of solution (in liters)<br />0.2M = Moles of $NaNO_{3}$ / 0.3L<br />Moles of $NaNO_{3}$ = 0.2M × 0.3L = 0.06 moles<br /><br />Step 2: Calculate the mass of $NaNO_{3}$ required.<br />Molar mass of $NaNO_{3}$ = 85.0 g/mol<br />Mass of $NaNO_{3}$ = Moles of $NaNO_{3}$ × Molar mass of $NaNO_{3}$<br />Mass of $NaNO_{3}$ = 0.06 moles × 85.0 g/mol = 5.1 grams<br /><br />Therefore, to prepare 300 ml of a 0.2M solution of $NaNO_{3}$, you need to take 5.1 grams of $NaNO_{3}$.<br /><br />For the complex compound $[Ni(NH_{3})_{6}]SO_{4}$:<br /><br />- Outer sphere: $SO_{4}^{2-}$<br />- Inner sphere: $[Ni(NH_{3})_{6}]^{2+}$<br />- Central ion: $Ni^{2+}$<br />- Charge: +2<br />- Coordination: 6 (hexaammine)