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FTCE Chemistry SAE Preparation Course. Session 4. Lisa Baig Instructor. Course Outline. Session 1 Review Pre Test Competencies 1 & 2 Session 2 Competency 5 Session 3 Competency 3 Session 4 Competency 4 Session 5 Competencies 6, 7 and 8 Post Test. Session Norms. Respect
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FTCE Chemistry SAEPreparation Course Session 4 Lisa Baig Instructor
Course Outline Session 1 Review Pre Test Competencies 1 & 2 Session 2 Competency 5 Session 3 Competency 3 Session 4 Competency 4 Session 5 Competencies 6, 7 and 8 Post Test
Session Norms • Respect • No side bars • Work on assigned materials only • Keep phones on vibrate • If a call must be taken, please leave the room to do so
Homework Review Any questions from last night?
Chemistry Competencies • Knowledge of the nature of matter (11%) • Knowledge of energy and its interaction with matter (14%) • Knowledge of bonding and molecular structure (20%) • Knowledge of chemical reactions and stoichiometry (24%) • Knowledge of atomic theory and structure (9%) • Knowledge of the nature of science (13%) • Knowledge of measurement (5%) • Knowledge of appropriate laboratory use and procedure (4%)
Determining Empirical Formulas • Say you have 65.0g of compound containing Na and Cl. • Determine the Empirical Formula if the compound is 39.3% Na and 60.7%Cl
Higher Level Practice • 1st Step: Convert your percentages to mass of each element present • Na: (.393)(65.0g)= 25.545g Na • Cl: (.607)(65.0g) = 39.455g Cl
Higher Level Practice • 2nd Step: Determine number of moles of each element in the sample 25.545g Na 1 mole = 1.11 mol Na 22.989 g/mol 39.455g Cl1 mole = 1.11 mol Cl 35.453 g/mol
Higher Level Practice • 3rd Step: Use these moles to determine the smallest whole number ratio of elements to each other. That is your empirical formula! 1.11 mol Na : 1.11 mol Cl 1 mol Na : 1 mol Cl Empirical Formula = NaCl
Balancing Equations • __ C3H8 + __ O2 __ CO2 + __ H2O • __ Ca2Si + __ Cl2 __ CaCl2 + __ SiCl4 • __ C7H5N3O6 __ N2 + __ CO + __ H2O + __ C • __ C2H2 + __ O2 __ CO2 + __ H2O • __ Fe(OH)2 + __ H2O2 __ Fe(OH)3 • __ FeS2 + __ Cl2 __ FeCl3 + __ S2Cl2 • __ Al + __ Hg(CH3COO)2 __ Al(CH3COO)3 + __ Hg • __ Fe2O3 + __ H2 __ Fe + __ H2O • __ NH3 + __ O2 __ NO + __ H2O
Types of Chemical Reactions • Synthesis • A+B AB • Decomposition • AB A + B • Combustion • Burn in the presence of O2, to form dioxide gas, and other products **(CO2 + H2O) • Single Displacement • ACTIVITY SERIES • AB + C AC + B • Double Displacement • AB + CD AD + CB
Predict the Product CaO + H2O H2SO3 + O2 CaCO3 KClO3 C6H10 + O2 C6H12O6 + O2 Al + CuCl2 Ca + KCl Na2SO4 + CaCl2 KCl + NaOH Ca(OH)2 H2SO4 CaO + CO2 KCl + O2 CO2 + H2O CO2 + H2O AlCl3 + Cu No Reaction NaCl + CaSO4 KOH + NaCl
Identifying Redox Reactions Redox Redox Not Redox Redox Not Redox 2 KNO3(s) 2 KNO2(s) + O2(g) +1 -1 +1 -1 0 H2(g) + CuO(s) Cu(s) + H2O(l) 0 -2 +2 0 2(+1) -2 NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) +1 -1 +1 -1 +1 -1 2(+1) -2 H2(g) + Cl2(g) 2HCl(g) 0 0 +1 -1 SO3(g) + H2O(l) H2SO4(aq) +6 3(-2) 2(+1) -2 2(+1) -2
Balancing Redox Reactions • The following unbalanced equation represents a redox reaction that takes place in a basic solution containing KOH. Balance the redox reaction. Br2(l) + KOH(aq) KBr(aq) + KBrO3(aq)
Br2(l) + KOH(aq) KBr(aq) + KBrO3(aq) Ionic Reaction: Br2 Br- + BrO3- 0 -1 +5 3(-2)- Reduction ½ Rxn: Br2 Br- Br2 + 2e- 2Br- 5(Br2 + 2e- 2Br-) Oxidation ½ Rxn: Br2 BrO3- 12OH- + Br2 2BrO3- + 6H2O + 10e- Combined Rxn: 5Br2 + 12OH- + Br2 + 10e- 10Br- + 2BrO3- + 6H2O + 10e- 6Br2 + 12KOH 10KBr + 2KBrO3 + 6H2O 3Br2 + 6KOH 5KBr + KBrO3 + 3H2O
Standard Reduction Potentials in Voltaic Cells Write the overall cell reaction and calculate the cell potential for a voltaic cell consisting of the following half-cells: an Iron electrode in an Iron (III) Nitrate solution, and a Silver electrode in a Silver(I) Nitrate solution. • Fe3+(aq)+3e-Fe(s) E0=-0.04V • Ag+(aq)+e-Ag(s) E0=+0.80V • E0cell=E0cathode- E0anode • E0cell= (+0.80 V)- (-0.04 V)= +0.84 V • E0cell= positive = spontaneous
Acid/Base Properties • Strong Acids and Bases • Will ionize completely in a solvent • Weak Acids and Bases • Will ionize partially in a solvent • Buffer Systems • Solution containing a weak acid, and a salt of the weak acid • Acetic Acid and Sodium Acetate • Carbonic Acid and Bicarbonate
Break Time Take a 10 minute break
Mass-Mass Stoichiometry 3 Cu + 8 HNO3 3 Cu(NO3)2 + 4 H2O + 2NO • Copper Nitrate is used in creation of some light sensitive papers • Specialty photographic film • Your company needs 150 grams of Copper nitrate to fill an order. How many grams of Nitric Acid are needed to undergo reaction?
Step 3: Compute 150g Cu(NO3)2 1 mole8 mol HNO3 63.012 g = 187.554g 3 molCu(NO3)2 1 mole 134 g HNO3
Gas Stoichiometry Xenon gas reacts with fluorine gas according to the shown reaction. If a researcher needs 3.14L of XeF6 for an experiment, what volumes of Xenon and Fluorine should be reacted? Assume all volumes are measured under the same temperatures and pressures. Xe(g) + 3 F2(g) XeF6 (g)
Gas Stoichiometry • Xenon 3.14L XeF6 1mole1Xe22.4L = 22.4L 1XeF6 1 mole 3.14L Xe • Fluorine 3.14L XeF61 mole3 F222.4L = 22.4L 1 XeF6 1 mole 9.42L F2
Solution Stoichiometry • How many milliliters of 18.0M Sulfuric Acid are required to react with 250mL of 2.50M Aluminum Hydroxide? • H2SO4 + Al(OH)3 H2O + Al2(SO4)3 • 3 H2SO4 + 2 Al(OH)3 6 H2O + Al2(SO4)3
250mL Al(OH)3 1L 2.5 mol 3 H2SO4 1L 1000mL 1000mL 1 L 2 Al(OH)3 18.0 mol 1L 52.1 mL H2SO4
Titrations • In a titration, 27.4mL of 0.0154M Ba(OH)2 is added to a 20.0mL sample of HCl solution with unknown concentration until the equivalence point is reached. What is the molarity of the acid solution? 0.0154M Ba(OH)2 x 27.4mL Ba(OH)2 x 2 mol HCl x 1 = 1 1 mol Ba(OH)2 20.0mL 4.22 x 10-2 M HCl
Limiting Reactant • The reaction of Ozone with Nitrogen Monoxide to form Oxygen and Nitrogen Dioxide in the atmosphere is responsible for the Ozone hole over Antarctica. • If 0.960g of Ozone reacts with 0.900g of Nitrogen Monoxide, how many grams of Nitrogen Dioxide are produced?
Limiting Reactant 0.960g O3 1 mole 1 NO244.0g NO2 48g O3 1 O3 1 mole 0.880g NO2 0.900gNO 1 mole 1 NO244.0g NO2 30g O3 1 O3 1 mole 1.32g NO2
Break Time Take a 10 minute break
Chemical Equilibrium • Chemical Equilibrium • Point in a reversible chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. • The concentrations of its products and reactants remain unchanged • Le Chatelier’s Principle • If a system at equilibrium is stressed, the equilibrium is shifted in the direction that relieves the stress
How to Affect Equilibrium • Change in Pressure • Only affects reactions with gases • Increased pressure increases concentration • Decreased pressure decreases concentration • Change in Concentration • Of reactants or products. • Increase one- it moves to the other • Decrease one- it moves towards the one you lowered • Change in Temperature • Exothermic • Increase temperature will direct in reverse • Decrease temperature will direct forward • Endothermic • Increase temperature will direct forward • Decrease temperature will direct in reverse
Equilibrium Constant nA + mB ↔ xC + yD K= [C]x[D]y [A]n[B]m
Rate Laws A chemical reaction is expressed by the balanced chemical equation A + 2B C Three reaction rate experiments yield the following data. What is the Rate Law for the Reaction? What is the Order of the reaction with respect to B?
Rate Law for the Reaction A + 2B C R = k[A][B]2 Order of the Reaction with respect to B B is of a 2nd order reaction A is of a 1st order reaction
Calculating pH and pOH pH + pOH = 14 pH = -log[H+] pOH= -log[OH-] • What is the pH of a 2.5x10-6M HNO3solution? • pH = -log [2.5x10-6] • pH = 5.6
Homework • Diagnostic Exam in your AP Chem Prep book- Page 17-26 • Only answer the questions for these Chapters & Questions • Chapter 6 #6-7, 11 • Chapter 7 #14, 16 • Chapter 8 #20 • Chapter 13 #59 • Chapter 14 #63 • Chapter 15 #66