Solutions and Gases Calculations (2)

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Solutions and Gases Calculations presented by Diana Mason, Deborah Koeck and Joyce Kulhanek, ACT2 2010
The TEA has issued and implemented major revisions to the chemistry TEKS. Chemistry is usually the first course students take that requires them to understand how to solve problem that can't just simply be memorized. Even though these new standards have been aligned with algebra I, the actual use of this knowledge is in the middle to higher levels of Bloom's Taxonomy (application, analysis and synthesis). There is also a new emphasis on mathematics that will be incorporated into the Chemistry STAAR exams. This workshop will take 12 (or more) calculations from the revised TEKS that are also documented in the Texas College and Career Readiness Standards and will share teaching suggestions and activities supporting calculations of pH, redox processes, Dalton's law of partial pressure and the ideal gas law.

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Solutions and Gases Calculations (2)

  1. 1. A Dozen or More Calculations Gases Mason, Koeck, and Kulhanek
  2. 2. BCCE August 1-5 University of North Texas Denton, Texas
  3. 3. TEKS Development and Correlation to CCRS Lone Star Chemistry Solutions https://callevents.unt.edu/ei/getdemo.ei?id=3&s=_3000VYEX3
  4. 4. Engagement Gases <ul><li>Dynamic Diana’s Daring Demo </li></ul>
  5. 5. Exploration Boyle’s Law <ul><li>The pressure of a gas at constant temperature is directly proportional to its volume. </li></ul><ul><li>PV= k </li></ul>
  6. 6. Exploration Boyle’s Law <ul><li>A gas that is at a constant temperature has a volume of 5.0 L at a pressure of 2 atm. What is the volume of the same gas if the pressure is increased to 3 atm.? </li></ul>
  7. 7. Exploration Boyle’s Law <ul><li>A gas that is at a constant temperature has a volume of 9.0 L at a pressure of 2 atm. What is the pressure of the same gas if the volume is decreased to 3.0 L? </li></ul>
  8. 8. Explanation Conceptual Understanding Boyle’s Law 5.0 L X 2 atm = 3.3 L 3 atm P 1 V 1 = P 2 V 2 V 2 = P 1 V 1 P 2 = 5.0 L X 2 atm = 3.3 L 3 atm
  9. 9. Explanation Conceptual Understanding Boyle’s Law #2 2 atm X 9.0 L = 6 atm 3.0 L P 1 V 1 = P 2 V 2 P 2 = P 1 V 1 V 2 = 2 atm X 9.0 L = 6 atm 3.0 L
  10. 10. Elaboration Derivation of Graham’s Law of Diffusion Materials: Glass Tubes 50-80 cm in length 6 M HCl 6 M NH 4 OH Meter Stick Cotton Balls
  11. 11. Evaluation What is the Concentration? <ul><ul><li>Using the expression for kinetic energy, </li></ul></ul><ul><ul><li>KE = ½ m v 2 </li></ul></ul><ul><ul><li>Derive Graham’s Law of Diffusion. </li></ul></ul>
  12. 12. A Dozen or More Calculations Additional Concepts Charles’ Law, Dalton’s Law, Ideal Gas Law, Gas Stoichiometry

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