1. CH4: Gases and their Properties NOTES #5

2. Diffusion  Caused by random motion!!! And collision of molecules  Random motion of molecules and their collisions produce PRESSURE!!! PRESSURE!!!  Same kinetic E at a given temp

3. 38) When someone standing at one end of a large room opens a bottle of vinegar, it may take several minutes for a person at the other end to smell it. Gas molecules at room temperature move at very high velocities, so what is responsible for the delay in detection of the vinegar? A the increase in the airspace occupied by vinegar molecules B the chemical reaction with nerves, which is slower than other sensory processes C attractive forces between the air and vinegar molecules D random collisions between the air and vinegar molecules

4. 39) Methane (CH 4 ) gas diffuses through air because the molecules are A moving randomly. B dissolving quickly. C traveling slowly. D expanding steadily.

5. Temperature  Temperature = average kinetic energy  Temperature must be changed to Kelvin for all gas laws  K = ˚C + 273  Why use Kelvin over Celsius? Never negative K Never negative K Causes V-T relationship to be proportional Causes V-T relationship to be proportional  0 K = absolute zero Lowest temperature possible Lowest temperature possible No temperatures below 0K No temperatures below 0K Because no more heat to release Because no more heat to release

6. 45) What is the equivalent of 423 kelvin in degrees Celsius? A –223 ºC B –23 ºC C 150 ºC D 696 ºC

7. 46) Theoretically, when an ideal gas in a closed container cools, the pressure will drop steadily until the pressure inside is essentially that of a vacuum. At what temperature should this occur? A 0ºC B −460 ºC C −273 K D 0 K

8. 47) The temperature at which all molecular motion stops is A −460 ºC. B −273 K. C 0 K. D 0C.

9. Gas Laws  Boyle’s Law  P 1 V 1 =P 2 V 2  Indirect proportion P = V P = V V P

10. 40 The volume of 400 mL of chlorine gas at 400 mm Hg is decreased to 200 mL at constant temperature. What is the new gas pressure? A 400 mm Hg B 300 mm Hg C 800 mm Hg D 650 mm Hg

11. 41) Under what circumstance might a gas decrease in volume when heated? A The gas is held constant at STP. B The gas remains under uniform temperature. C The gas is placed under increasing pressure. D The gas undergoes a decrease in pressure.

12.  Charles’ Law  V 1 = V 2 T 1 T 2 T 1 T 2  Temperature MUST be in Kelvin  Direct proportion V = T V = T V T

13.  Combined Gas Law  P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2  Temperature MUST be in Kelvin!  Dalton’s Law of Partial Pressures Addition of the pressures of all gases collected together (including water vapor) Addition of the pressures of all gases collected together (including water vapor) P total = P 1 + P 2 + P 3 + P 4... P total = P 1 + P 2 + P 3 + P 4...

14. 37) When a cold tire is inflated to a certain pressure and then is warmed up due to friction with the road, the pressure increases. This happens because the A air molecules hit the walls of the tire less frequently. B rubber in the tire reacts with oxygen in the atmosphere. C air molecules speed up and collide with the tire walls more often. D air molecules diffuse rapidly through the walls of the tire.

15.  Ideal Gas Law Use when moles or grams are given Use when moles or grams are given PV = nRT (R = 0.0821) PV = nRT (R = 0.0821) P pressure in atm P pressure in atm V volume in L V volume in L n moles n moles T temp in Kelvin T temp in Kelvin

16. STP  Standard Temperature and Pressure Numbers set for comparison Numbers set for comparison  Pressures 101325 Pa 101325 Pa 101.325 kPa 101.325 kPa 1.00 atm 1.00 atm 760 torr, mmHg 760 torr, mmHg  These ALL equal 1 MOLE!! at STP  Look for simple ratios 0.5 mole = ? Torr 0.5 mole = ? Torr 2 mole = ?kPa 2 mole = ?kPa

17. 44) Under which of the following sets of conditions will a 0.50 mole sample of helium occupy a volume of 11.2 liters? A 298 K and 0.90 atm B 273 K and 1.10 atm C 373 K and 0.50 atm D 273 K and 1.00 atm

18. STP  Temperature 273K 273K  1 ˚C = 1 K  Water freezes at 0 ˚C, 273 K

19. 43 Standard temperature and pressure (STP) are defined as A 0 ºC and 1.0 atm pressure. B 0 ºC and 273 mm Hg pressure. C 0 K and 1.0 atm pressure. D 0 K and 760 mm Hg pressure.

20. 42) A sample of carbon dioxide gas occupies a volume of 20 L at standard temperature and pressure (STP). What will be the volume of a sample of argon gas that has the same number of moles and pressure but twice the absolute temperature? A 10 L B 20 L C 40 L D 80 L

21. Homework #5  Pg. 22 – 26; #1 – 43

22. CH5: Acids and Bases Notes #6

23. Arrhenius  Acids Donate H + ions Donate H + ions  Bases Donate OH - Donate OH -  Acid + Base  H 2 O + salt  Salt Ionic compound formed when an acid and base react Ionic compound formed when an acid and base react

25. Bronsted-Lowry  Acids Proton (H + ) donor Proton (H + ) donor  Bases Proton (H + ) acceptor Proton (H + ) acceptor

26. Strong Acids/Bases  Strong acids and bases completely dissociate in water  Completely break apart  Are strong electrolytes  Conduct electricity well  HBr (aq)  H + (aq) + Br - (aq)

28. Weak Acids/Bases  Partially dissociate in water  Partly break up  Weak electrolytes  Weakly conduct electricity  NH 3 + H 2 O  NH 4 + + OH - Only some break up

29. Non electrolytes  Do not conduct electricity  Covalent compounds  Compounds without metals

30. pH  Scale 0-14 to represent acidity/basicity  0-6 acids  7 neutral  8-14 bases  Is log based Therefore each difference in pH is a factor of 10 Therefore each difference in pH is a factor of 10 Example pH 3 to pH of 8, difference is 10x10x10x10x10 = 100000 Example pH 3 to pH of 8, difference is 10x10x10x10x10 = 100000

31. Homework #6  Pg. 27 – 29; #1 – 30

32. CH6: Solutions NOTES #7  Solute Substance being dissolved Substance being dissolved Have a smaller amount Have a smaller amount Causes the boiling point to rise and freezing point to lower Causes the boiling point to rise and freezing point to lower  Solvent Does the dissolving Does the dissolving Have the larger amount Have the larger amount  Solution Solute + Solvent Solute + Solvent Solute Solvent Solution

34. Solubility  Increase temp, surface area Increase solubility of solid Increase solubility of solid  Increase pressure, decrease temp Increase solubility of gas Increase solubility of gas

36. Concentration  Molarity Moles (solute) Moles (solute) Liters (solution Liters (solution  Solubility Grams/Liter Grams/Liter Use proportions Use proportions  Molality Moles (solute) Moles (solute) kg (solvent) kg (solvent)

37. % composition  % of one substance in a compound  Mass substance x 100 Total mass Total mass Ex: what is the percent composition of C in C 12 H 22 O 11 ? Ex: what is the percent composition of C in C 12 H 22 O 11 ?

38. Units  Grams/liter Density of a substance Density of a substance  Parts per million (ppm)

39. Homework #7  Pg. 31 – 33; #1 – 23