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.
ScanSolve AI Answer
Welcome back Noura — missed you this week. Here's the solution for question 7: Step 1: Calculate the moles of silver nitrate (AgNO_3) used. The volume of silver nitrate solution is 200 cm^3 = 0.200 L and its concentration is 0.20 M. Moles of AgNO_3 = Concentration × Volume Moles of AgNO_3 = 0.20 mol/L × 0.200 L = 0.040 mol Step 2: Determine the moles of chloride ions (Cl^-) that reacted. The precipitation reaction is Ag^+(aq) + Cl^-(aq) → AgCl(s). From the stoichiometry, 1 mole of Ag^+ reacts with 1 mole of Cl^-. Since AgNO_3 provides Ag^+ ions in a 1:1 ratio, the moles of Ag^+ are equal to the moles of AgNO_3. Moles of Cl^- = Moles of Ag^+ = Moles of AgNO_3 = 0.040 mol Step 3: Calculate the moles of the metal chloride in the initial solution. The volume of the metal chloride solution is 100 cm^3 = 0.100 L and its concentration is 0.10 M. Moles of metal chloride = Concentration × Volume Moles of metal chloride = 0.10 mol/L × 0.100 L = 0.010 mol Step 4: Determine the formula of the metal chloride. Let the formula of the metal chloride be MCl_x. This means that 1 mole of MCl_x contains x moles of Cl^-. We can find x by taking the ratio of moles of chloride ions to moles of metal chloride. x = Moles of Cl^-Moles of metal chloride = 0.040 mol0.010 mol = 4 Therefore, the formula of the metal chloride is MCl_4. The correct option is A. The right formula for the metal chloride is A. MCl_4. Send me the next one 📸