I add 2.00 g NaCl and 6.42 g BaCl₂ to a a container. You can assume that both of these salts are strong electrolytes. I add water so that the final concentration of chloride ions (CI) is 0.034 M. What is the volume of the solution? 2.8 L solution

The stock solution in the stock beaker contains 10.00 mL of 0.1000 M iron(III) nitrate solution, 10.00 mL of 0.1000 M potassium thiocyanate solution, and 10.00 mL of 0.5000 M nitric acid solution.   What is the total volume of the stock solution?  How many moles of iron (III) nitrate are present in the stock solution?      How many moles of potassium thiocyanate are present in the stock solution?        Based on the moles of iron (III) nitrate and moles of potassium thiocyanate, how many moles of iron (III) thiocyanate do you expect to be formed    Once you determine the moles of iron (III) thiocyanate formed, determine the concentration of iron (III) thiocyanate in the final solution.

In the laboratory you dilute 2.50 mL of a concentrated 12.0 M hydroiodic acid solution to a total volume of 50.0 mL. What is the concentration of the dilute solution?

Consider the following reaction. MgCl2(aq) + 2 NaOH(aq) → Mg(OH)2(s) + 2 NaCl(aq) A 185.0 mL solution of 0.395 M MgCl2 reacts with a 37.62 mL solution of 0.811 M NaOH to produce Mg(OH)2 and NaCl. Identify the limiting reactant. ○ Mg(OH)2 NaOH MgCl2 NaCl Caclulate the mass of Mg(OH)2 that can be produced. mass of Mg(OH)2: The actual mass of Mg(OH)2 isolated was 0.742 g. Calculate the percent yield of Mg(OH)2. percent yield: 6.0 g %