Wednesday, May 11, 2010 Weston Preparatory Academy Chemistry Classes

Wednesday, May 12, Warmup: Daily class diary/ journal OR: What happens when water and electricity mix? Guess or tell a story. 2. HW due today: p.311 #1-7 all. 3. HW due tomorrow: p. 319 #1-14 odd; due Friday: p. 327 #1-9 all; due Mon. May 17: p. 329 #1-21 odd. 4. Notes, discussion, group tutoring!

Chapter 9 Reading Notes: Page 302. Page 302. Balanced equations show proportions. Balanced equations show proportions. 2H 2 + O 2  2H 2 O 2H 2 + O 2  2H 2 O

Chapter 9 Reading Notes: Balanced equations show proportions. 2H 2 + O 2  2H 2 O 2H 2 + O 2  2H 2 O For every 2 moles of Hydrogen gas, There is 1 mole of oxygen gas and there are 2 moles of water produced from the reaction. For every 2 moles of Hydrogen gas, There is 1 mole of oxygen gas and there are 2 moles of water produced from the reaction.

Chapter 9 Reading Notes: 2H 2 + O 2  2H 2 O 2H 2 + O 2  2H 2 O 12 hydrogen atoms- 6 oxygen atoms- 12 water molecules 12 hydrogen atoms- 6 oxygen atoms- 12 water molecules 500 hydrogen- 250 oxygen – 500 water 500 hydrogen- 250 oxygen – 500 water ½ mole hydrogen- ¼ mole oxygen – ½ mole water ½ mole hydrogen- ¼ mole oxygen – ½ mole water The coefficients show the RATIO of molecules and atoms of reactants and products to each other. The coefficients show the RATIO of molecules and atoms of reactants and products to each other.

Chapter 9 Reading Notes: Page 303. stoichiometry: the proportional relationship between two or more substances during a chemical reaction. Page 303. stoichiometry: the proportional relationship between two or more substances during a chemical reaction. See the diagram… use the coefficients from the balanced equation to predict moles on one type of atom or molecule, given the number of moles of limiting reactant… See the diagram… use the coefficients from the balanced equation to predict moles on one type of atom or molecule, given the number of moles of limiting reactant…

Chapter 9 Reading Notes: Page 304. Getting into Moles and Getting Out of Moles Page 304. Getting into Moles and Getting Out of Moles Problem – solving method given in detail- pages 304 and 305. Problem – solving method given in detail- pages 304 and 305. Convert given information to moles. Convert given information to moles. Based on mole ratio, calculate number of moles of desired atom or compound Based on mole ratio, calculate number of moles of desired atom or compound Convert to desired unit (grams, number of particles) Convert to desired unit (grams, number of particles)

Chapter 9 Reading Notes: Page 306: Mass, Volume, Particles. Page 306: Mass, Volume, Particles. I may assign mass and particle problems. I may assign mass and particle problems. Volume- requires an understanding of gas pressure laws and STP- Coming soon!- Next week. Volume- requires an understanding of gas pressure laws and STP- Coming soon!- Next week. Page 307: sample problem b, practice Page 307: sample problem b, practice

Chapter 9 Reading Notes: Page 308. more volume info. Read for a preview before next week! (required) Page 308. more volume info. Read for a preview before next week! (required) Pages – more sample problems. Pages – more sample problems. Read Chapter 9: required! Was assigned on MONDAY MAY 10. Read Chapter 9: required! Was assigned on MONDAY MAY 10.

PAGE 311 #1-7 ; SKIP #5; due today. 1. What conversion factor is present in almost all stoichiometry calculations? (pg. 303) 2. For a given substance, what information links mass to moles? (see pg. 230) number of particles to moles? (it’s Avogadro’s number) 3. What conversion factor will change moles CO 2 to grams CO 2 ? (see P.Table) Moles H 2 O to molecules H 2 O? (see P. Table)

PAGE 311 #1-7 ; SKIP #5; due today. 4. Br 2 + Cl 2  2BrCl 4. Br 2 + Cl 2  2BrCl –A. Look at the mole ratios! The number of moles of Cl 2 is the HALF as many as the moles of BrCl –B. Convert grams to moles of Cl 2, then multiply the number of moles by 2 (due to the mole ratio). Multiply this number of moles by the molar mass of BrCl found in the periodic table. –C. Convert molecules to moles. That is the same number of moles needed of Br 2 (mole ratio is 1:1). Multiply by molar mass of bromine.

PAGE 311 #1-7 ; SKIP #5; due today. 6. Why do you need to use amount in moles to solve stoichiometry problems? Why can’t you just convert from mass to mass? (Using coefficients from balanced equation shows relationships of numbers of particles, not masses!) 6. Why do you need to use amount in moles to solve stoichiometry problems? Why can’t you just convert from mass to mass? (Using coefficients from balanced equation shows relationships of numbers of particles, not masses!)

PAGE 311 #1-7 ; SKIP #5; due today. 7. TWO reactions are described. 7. TWO reactions are described. A. LiOH + CO 2  Li 2 CO 3 + H 2 O (NOW BALANCE THIS BEFORE WRITING IT DOWN!) and NaOH + CO 2  Na 2 CO 3 + H 2 O A. LiOH + CO 2  Li 2 CO 3 + H 2 O (NOW BALANCE THIS BEFORE WRITING IT DOWN!) and NaOH + CO 2  Na 2 CO 3 + H 2 O B. Convert 288 g CO 2 to moles CO 2 using the molar mass from the periodic table. Then… multiply or divide according to the mole ration (look at the coefficients from your balanced chemical equations, part A). Then… multiply number of moles by the molar mass (of LiOH, then NaOH) of the substance to calculate the number of grams. B. Convert 288 g CO 2 to moles CO 2 using the molar mass from the periodic table. Then… multiply or divide according to the mole ration (look at the coefficients from your balanced chemical equations, part A). Then… multiply number of moles by the molar mass (of LiOH, then NaOH) of the substance to calculate the number of grams.

PAGE 311 #1-7 ; SKIP #5; due today. 7. TWO reactions are described. 7. TWO reactions are described. A. LiOH + CO 2  LiCO 3 + H 2 O (NOW BALANCE THIS BEFORE WRITING IT DOWN!) and NaOH + CO 2  NaCO 3 + H 2 O A. LiOH + CO 2  LiCO 3 + H 2 O (NOW BALANCE THIS BEFORE WRITING IT DOWN!) and NaOH + CO 2  NaCO 3 + H 2 O B. Convert 288 g CO 2 to moles CO 2 using the molar mass from the periodic table. Then… multiply or divide according to the mole ration (look at the coefficients from your balanced chemical equations, part A). Then… multiply number of moles by the molar mass (of LiOH, then NaOH) of the substance to calculate the number of grams. B. Convert 288 g CO 2 to moles CO 2 using the molar mass from the periodic table. Then… multiply or divide according to the mole ration (look at the coefficients from your balanced chemical equations, part A). Then… multiply number of moles by the molar mass (of LiOH, then NaOH) of the substance to calculate the number of grams. C. Comment on the number of grams of each that are needed, the cost per mole of each, and the cost to transport large masses of material into space. C. Comment on the number of grams of each that are needed, the cost per mole of each, and the cost to transport large masses of material into space.

Thursday, May 13, 2010 Weston Preparatory Academy Chemistry Classes

Thursday, May 13, Warmup: Daily class diary/ journal OR: Why is mass as important as cost when choosing which materials to put on an airplane or space shuttle? Guess or tell a story. 2. CHARACTER COUNTS!!! PLEDGE… 3. HW due yesterday: p.311 #1-7, skip #5 (accepted late today, -20%: 14 of 70 points) 4. HW : p. 319 #1-14 odd; HW p. 327 #1-9 all; due Mon. May 17: p. 329 #1-21 odd.-all accepted by Monday, full credit. 5. Notes, discussion, group tutoring!