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Unit 10B Review

Unit 10B Review. Reg Chem 2012-13. 11. When a solution sits out over a long period of time and water evaporates the concentration of the solution __________. Increases. As the volume of solution decreases, the molarity (concentration) increases. 12.

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Unit 10B Review

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  1. Unit 10B Review Reg Chem 2012-13

  2. 11 • When a solution sits out over a long period of time and water evaporates the concentration of the solution __________. Increases. As the volume of solution decreases, the molarity (concentration) increases

  3. 12 • List an example of a solution that would naturally dilute. Ice in pop

  4. 13 • What does solution concentration describe? Molarity: amount of solute dissolved in a given amount of solvent

  5. 14 • What do we use to describe solution concentration in chemistry? Molarity

  6. 15 • If a solution is “strong” it is • If a solution is “weak” it is concentrated dilute

  7. 16 • What does it mean to dilute a solution? What equation do we use for dilutions? To lower its concentration “water it down” Yet keep the number of moles the same M1V1 = M2V2

  8. 17 • What is the molarity of a sodium chloride solution that contains 1.73 moles in 3.94 L of solution?

  9. 18 • What is the molarity of sodium hydroxide solution that contains 23.5 g NaOH in 500.0 mL of solution?

  10. 19 • How many grams of potassium nitrate are in 275 mL of 1.25 M solution?

  11. 20 • How many mL of 3.25 M hydrochloric acid would contain 16.0 grams of solute?

  12. 21 • You have 12.0 M HCl in your stock room, how would you prepare 600.0 mL of 2.50 M HCl solution? Measure out 125 mL of stock solution. Add 475 mL of distilled water. (12.0 M)(x) = (2.50 M)(600.0 mL) X=125 mL

  13. 22 • How would you correctly prepare 500.0 mL of a 3.0 M solution of NaOH from solid solute? NaOH Measure out 60 g of NaOH (s) Add water to the 500.0 mL line

  14. 23 • How would you prepare 500 mL of 3.0 M NaOH from 12.0 M concentrated stock solution? Measure out 125 mL of stock solution. Add it to 375 mL of distilled water (12.0 M)(X) = (3.0 M)(500 mL) X=125 mL

  15. 26 • An excess of zinc is added to 125 mL of 0.100 M HCl solution. What mass of zinc chloride is formed? Zn + 2HCl  ZnCl2 + H2

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