You used 80 mL of deionized water instead of the necessary 50 mL of water to make a KHP solution for standardizing NaOH. What effect would this have on the calculated molarity of the NaOH? Select the single best answer. O The calculated molarity will be unaffected. O The calculated molarity will be too high. O The calculated molarity will be too low. O It is not possible to tell without knowing the volume of NaOH used.

Which of the following solutions will have the highest concentration of chloride ions? (hint: write down the dissociation of the ionic compounds in water) OA. 0.10 mol/L MgCl2 O B. All of these solutions have the same concentration of chloride ions. O c. 0.10 mol/L LICI OD.0.05 mol/L CaCl2 O E. 0.10 mol/L AICI3

1125 mL of a 0.75 M All solution was mixed with 920 mL of a 0.4 M NaCl solution. What is the [I] after mixing? 1.4 M 1.2 M O 0.55 M 0.65 M 0.92 M

Use only 1 decimal point (X.X) for all atomic weights. What will be the final molarity of 10.0 mL of 12.0 M HNO2 dilluted to 500.0 mL?

In a solution the substance which is being disolved is the

14. What volume of 0.10 M NAOH is required to neutralize 45.0 ml of 0.8M hydroiodic acid, HI? А. 36 mL в. О 178 mL 5.6 mL D. 360 mL

Mix together 150.0 ml of a 1.50 mol / dm³ NaCl solution and 250.0 ml of a 0.750 mol / dm³ MgCl₂ solution. What is the chloride ion concentration of the resulting solution?

In your first task, you are required to prepare a 100 mL solution of sodium bicarbonate at a concentration of 0.50 M. Complete the following table and calculate the mass of sodium bicarbonate needed to prepare this solution. Molecular weight of sodium bicarbonate Required final concentration Required final volume Amount of sodium bicarbonate needed to make 1 L at 0.5 M Amount of sodium bicarbonate needed to make 100 mL at 0.5 M Mass (g) = Amount (mol) × Molecular Weight (g/mol) Value 84.01 Now that you know the amount of sodium bicarbonate in mol needed to make your solution, you can convert this value into the mass by using the molecular weight of the compound. The following equation may be helpful: Mass of sodium bicarbonate needed to make 100 mL at 0.5 M Units g/mol M (mol/L) L mol mol g In the 100 mL 0.5 M solution, how many molecules of sodium bicarbonate have been added. Hint - In 1 mole there are 6.022 × 10²³ molecules.

Calculate the [CI] for a solution prepared by dissolving 1.525 g BaCl2.2H2O (244 g/mol) and 1.17 g NaCl (58.5 g/mol) in 1 L solution. Select one: a. 12.5 mM b. 42.5 mM с. 32.5 mM d. 2.5 mM е. 22.5 mM

A student is asked to standardize a solution of barium hydroxide. He weighs out 0.929 g potassium hydrogen phthalate ( KHC8H404, treat this as a monoprotic acid). It requires 36.8 mL of barium hydroxide to reach the endpoint. A. What is the molarity of the barium hydroxide solution? M This barium hydroxide solution is then used to titrate an unknown solution of hydrobromic acid. B. If 28.5 mL of the barium hydroxide solution is required to neutralize 13.3 mL of hydrobromic acid, what is the molarity of the hydrobromic acid solution? M