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Wednesday, October 24th •Review •Quiz •Finish Chapter 6, •Begin Chapter 7 Frequency: CD Remaining Scedule Month October November December Holidays: Day Topic 22 24 26 Finish Chapter 6 Quiz #7, Review Chapter 6, Lecture on Chapter 7.1-7.4 Lecture on Chapter 7.4-7.5, In Class Group Work 29 31 2 Lecture on Chapter 7..6-7.8 Quiz #8, Lecture on Chapter 7.9 In Class Group Work 5 7 9 Test #2 Lecture on Chapter 8.1-8.3 Lecture on Chapter 8.4, 8.6 In Class Group Work 12 14 16 No Class Quiz #9, Lecture on Chapter 10.1 In Class Group Work 19 No Class 26 28 30 Lecture on Chapter 10.1-10.3 Quiz #10 Lecture on Chapter 10.4-10.5 In class Group Work 3 5 7 Lecture on Chapter 10.6 In Class Group Work Test #3 Final Exam Review 10 12 Final Exam- 3pm class, 11:30-1:00 Final Exam- 8am class, 1:15- 2:45 Monday, November 12- Veteran’s Day November 21-25- Thanksgiving Break 2 Review 3 Practice Balance the equations for the combustion of propane (C3H8). 4 Avogadro’s Number as an Equality Avogadro’s number (6.02 x 1023) can be written as an equality and two conversion factors. Equality: Conversion Factors: Subscripts State Atoms and Moles Molar Mass Molar mass is the mass of one mole of a substance. the number of grams that equals the atomic mass of that element. Molar mass is rounded to the tenths (0.1 g) place for use in this text. Quiz • 7 questions, 2 pts each 1. What are the molar coefficients needed to balance the following equation? Al (s) + HCl (aq) AlCl3 (aq) + H2 (g) a) b) c) d) © 2013 Pearson Education, Inc. 1:4:1:2 2:3:2:3 2:6:2:3 4:6:4:3 Copyright © 2011 Pearson Education Inc. 2. How many moles of oxygen can be produced from one mole of KClO3 in the reaction KClO3 KCl + O2? a) b) c) d) © 2013 Pearson Education, Inc. 1 mol 1.5 mol 2 mol 3 mol Copyright © 2011 Pearson Education Inc. 3. The number of C atoms in 0.500 mole C is_______. a) b) c) d) 2.50 × 1023 C atoms 3.01 × 1023 C atoms 5.01 × 1023 C atoms 6.00 × 1023 C atoms © 2013 Pearson Education, Inc. Copyright © 2011 Pearson Education Inc. 4. How many moles of O are there in 2 moles of glucose (C6H12O6)? a) 2 moles b) 24 moles c) 6 moles d) 12 moles © 2013 Pearson Education, Inc. Copyright © 2011 Pearson Education Inc. 5. What is the molar mass of S3 if the molar mass of sulfur is 32.0? a) 64 g b) 32 g c) 94 g d) 10.7 g © 2013 Pearson Education, Inc. Copyright © 2011 Pearson Education Inc. 6. What is the percent yield if 50 g of CO2 were collected and 200g were expected? a) b) c) d) © 2013 Pearson Education, Inc. 50% 25% 75% 100% Copyright © 2011 Pearson Education Inc. 7. Which of the following reactions can be classified as decomposition? a) b) c) d) CuCO3 (s) CuO (s) + CO2 (g) 4Fe (s) + 3O2 (g) 2Fe2O3 (s) Mg (s) + 2AgNO3 (aq) Mg(NO3)2 (aq) + 2Ag (s) NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l) © 2013 Pearson Education, Inc. Copyright © 2011 Pearson Education Inc. Continuing on in Chapter 6 16 Learning Check If 48.2 grams of Ca are mixed with 31.0 grams of N2, how many grams of Ca3N2 can be produced? Solution If 48.2 grams of Ca are mixed with 31.0 grams of N2, how many grams of Ca3N2 can be produced? Step 1 State the given and needed quantities. Analyze the Problem. Solution If 48.2 grams of Ca are mixed with 31.0 grams of N2, how many grams of Ca3N2 can be produced? Step 2 Use coefficients to write mole–mole factors; write molar mass factors, if needed. 1 mole of Ca = 40.1 g of Ca 1 mole of N2 = 28.0 g of N2 Solution If 48.2 grams of Ca are mixed with 31.0 grams of N2, how many grams of Ca3N2 can be produced? Step 2 Use coefficients to write mole–mole factors; write molar mass factors, if needed. Solution If 48.2 grams of Ca are mixed with 31.0 grams of N2, how many grams of Ca3N2 can be produced? Step 3 Calculate the number of moles of product from each reactant and determine the limiting reactant. The limiting reactant is Ca, which produces 0.402 mole Ca3N2. Solution If 48.2 grams of Ca are mixed with 31.0 grams of N2, how many grams of Ca3N2 can be produced? Use the molar mass to convert the smaller number of moles of product to grams. Reaction Conditions There are three conditions required for a chemical reaction to occur. 1. Collision: The reactants must collide. 2. Orientation: The reactants must align properly to . break and form bonds. 3. Energy: The collision must provide the energy of . activation. Activation Energy Activation energy is the amount of energy required to break the bonds between atoms of the reactants. If the energy of a collision is less than the activation energy, the molecules bounce apart without reacting. Many collisions occur, but only a few actually lead to the formation of product. Activation Energy The activation energy is the energy needed to convert reacting molecules into products. Heat of Reaction The heat of reaction is the amount of heat absorbed or released during a reaction. is the difference between the energy of breaking bonds in the reactants and forming bonds in the products. ΔH = has the symbol ΔH. ΔHproducts − ΔHreactants Exothermic Reactions In an exothermic reaction, the energy of the reactants is greater than that of the products and heat is released along with the products. the heat of reaction (ΔH) value is written with a negative sign (–) indicating heat is released. Endothermic Reactions In an endothermic reaction, the energy of the reactants is lower than that of the energy of the products. heat is absorbed and used to convert the reactants to products and written with a (+) sign. Heat of Reaction Summary Reaction Type Endothermic Exothermic Energy Heat Change in Reaction ΔH Heat absorbed Reactant side Heat released Product side Sign of + – Common Gases Of the elements on the periodic table, some exist as a gas at room temperature, these include the Noble Gases, Group 8A (18), H2, N2, O2, F2, Cl2, and many oxides of nonmetals such as CO, CO2, NO, NO2, SO2, and SO3. 30 Gases and Environmental Concerns Some gases are responsible for environmental and health concerns including methane, CH4, chlorofluorocarbons (CFCs), nitrogen oxides found in smog, and volatile organic compounds (VOCs), such as compounds found in paint thinners. 31 Kinetic Theory of Gases A gas consists of small particles that • move randomly with high velocities. • have essentially no attractive (or repulsive) forces toward each other. • have a very small volume compared to the volume of the containers they occupy. • are in constant motion. • have kinetic energies that increase with an increase in temperature. 32 Properties of Gases Gases are described in terms of four properties: 1. pressure (P), 2. volume (V), 3. temperature (T) 4. amount (n). 33 Gas Pressure Gas pressure • is the force acting on a specific area. • has units of atm, mmHg, torr, lb/in.2, and kilopascals(kPa). 1 atm = 760 mmHg (exact) 1 atm = 760 torr (exact) 1 atm = 14.7 lb/in.2 1 atm = 101.325 kPa 34 Practice 1. What is 475 mmHg expressed in atmospheres (atm)? 35 2. The pressure in a tire is 2.00 atm. What is this pressure in mmHg? Solution 1. What is 475 mmHg expressed in atmospheres (atm)? B. 0.625 atm 2. The pressure of a tire is measured as 2.00 atm. What is this pressure in mmHg? B. 1520 mmHg 37 Atmospheric Pressure Atmospheric pressure is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth. • is about 1 atmosphere at sea level. • depends on the altitude and the weather. • is lower at high altitudes where the density of air is less. • is higher on a rainy day than on a sunny day. 38 Boyle’s Law Boyle’s law states that • the pressure of a gas is inversely related to its volume when temperature (T) and amount of gas (n) are constant. • if the pressure (P) increases, then the volume (V) decreases. 39 Solving for a Gas Law Factor A sample of nitrogen gas (N2) has a volume of 4.2 L at 1.0 atm. If the pressure is decreased to 0.75 atm with no change in the temperature or amount of gas, what will be the new volume? 40 Solving for a Gas Law Factor A sample of nitrogen gas (N2) has a volume of 4.2 L at 1.0 atm. If the pressure is decreased to 0.75 atm with no change in the temperature or amount of gas, what will be the new volume? Step 2 Rearrange the gas law equation to solve for the unknown quantity. P1V1 = P2V2 Boyle’s Law To solve for V2 , divide both sides by P2. 41 Charles’s Law Charles’s law states that, • the Kelvin temperature of a gas is directly related to the volume of the gas, • P and n are constant, and • as the temperature of a gas increases, the molecules move faster and its volume increases to maintain constant P. 42 Practice A balloon has a volume of 785 mL at 21 ˚C. If the temperature drops to 0 ˚C, what is the new volume of the balloon (P constant)? 43 Solution A balloon has a volume of 785 mL at 21 ˚C. If the temperature drops to 0 ˚C, what is the new volume of the balloon (P constant)? -Rearrange the gas law equation to solve for the unknown quantity. -Substitute values into the gas law equation and the table. 44 Gay-Lussac’s Law: P and T Gay-Lussac’s law states that • the pressure exerted by a gas is directly related to the Kelvin temperature. • V and n are constant. • an increase in temperature increases the pressure of a gas. 45 Practice Solve for the final pressure of a gas with an initial pressure of 1.20 atm at 75 ˚C when cooled to −22 ˚C. (V and n constant) 46 Solution Solve for the final pressure of a gas with an initial pressure of 1.20 atm at 75 ˚C when cooled to −22 ˚C. Step 3 Substitute values into the gas law equation and calculate. 47 Combined Gas Law The combined gas law uses Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law (n is constant). 48 Learning Check A gas has a volume of 675 mL at 35 °C and 0.850 atm pressure. What is the volume (mL) of the gas at −95 °C and a pressure of 802 mmHg (n constant)? 49 Solution A gas has a volume of 675 mL at 35 °C and 0.850 atm pressure. What is the volume (mL) of the gas at −95 °C and a pressure of 802 mmHg (n constant)? 50 Next Week Quiz - Written Chapter 7 Finish Chapter 7 51