1. FTCE Chemistry SAE Preparation Course Session 4 Lisa Baig Instructor

2. 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

3. 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

4. Homework Review Any questions from last night?

5. Chemistry Competencies 1.Knowledge of the nature of matter (11%) 2.Knowledge of energy and its interaction with matter (14%) 3.Knowledge of bonding and molecular structure (20%) 4.Knowledge of chemical reactions and stoichiometry (24%) 5.Knowledge of atomic theory and structure (9%) 6.Knowledge of the nature of science (13%) 7.Knowledge of measurement (5%) 8.Knowledge of appropriate laboratory use and procedure (4%)

6. 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

7. Higher Level Practice 1 st Step: Convert your percentages to mass of each element present Na: (.393)(65.0g)= 25.545g Na Cl: (.607)(65.0g) = 39.455g Cl

8. Higher Level Practice 2 nd 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 Cl 1 mole = 1.11 mol Cl 35.453 g / mol

9. Higher Level Practice 3 rd 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

10. Balancing Equations __ C 3 H 8 + __ O 2  __ CO 2 + __ H 2 O __ Ca 2 Si + __ Cl 2  __ CaCl 2 + __ SiCl 4 __ C 7 H 5 N 3 O 6  __ N 2 + __ CO + __ H 2 O + __ C __ C 2 H 2 + __ O 2  __ CO 2 + __ H 2 O __ Fe(OH) 2 + __ H 2 O 2  __ Fe(OH) 3 __ FeS 2 + __ Cl 2  __ FeCl 3 + __ S 2 Cl 2 __ Al + __ Hg(CH 3 COO) 2  __ Al(CH 3 COO) 3 + __ Hg __ Fe 2 O 3 + __ H 2  __ Fe + __ H 2 O __ NH 3 + __ O 2  __ NO + __ H 2 O

11. 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 **(CO 2 + H 2 O) Single Displacement – ACTIVITY SERIES – AB + C  AC + B Double Displacement – AB + CD  AD + CB

12. Predict the Product CaO + H 2 O  H 2 SO 3 + O 2  CaCO 3  KClO 3  C 6 H 10 + O 2  C 6 H 12 O 6 + O 2  Al + CuCl 2  Ca + KCl  Na 2 SO 4 + CaCl 2  KCl + NaOH  Ca(OH) 2 H 2 SO 4 CaO + CO 2 KCl + O 2 CO 2 + H 2 O AlCl 3 + Cu No Reaction NaCl + CaSO 4 KOH + NaCl

13. Identifying Redox Reactions 2 KNO 3 (s)  2 KNO 2 (s) + O 2 (g) +1 -1 +1 -1 0 H 2 (g) + CuO(s)  Cu(s) + H 2 O(l) 0 -2 +2 0 2(+1) -2 NaOH(aq) + HCl(aq)  NaCl(aq) + H 2 O(l) +1 -1+1 -1 +1 -1 2(+1) -2 H 2 (g) + Cl 2 (g)  2HCl(g) 0 0 +1 -1 SO 3 (g) + H 2 O(l)  H 2 SO 4 (aq) +6 3(-2) 2(+1) -2 2(+1) -2 Redox Not Redox Redox Not Redox

14. Balancing Redox Reactions The following unbalanced equation represents a redox reaction that takes place in a basic solution containing KOH. Balance the redox reaction. Br 2 (l) + KOH (aq)  KBr (aq) + KBrO 3 (aq)

15. Ionic Reaction: Br 2  Br - + BrO 3 - 0-1 +5 3(-2) - Reduction ½ Rxn: Br 2  Br - Br 2 + 2e -  2Br - 5(Br 2 + 2e -  2Br - ) Oxidation ½ Rxn: Br 2  BrO 3 - 12OH - + Br 2  2BrO 3 - + 6H 2 O + 10e - Combined Rxn: 5Br 2 + 12OH - + Br 2 + 10e -  10Br - + 2BrO 3 - + 6H 2 O + 10e - 6Br 2 + 12KOH  10KBr + 2KBrO 3 + 6H 2 O 3Br 2 + 6KOH  5KBr + KBrO 3 + 3H 2 O

16. 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. Fe 3+ (aq) +3e -  Fe (s) E 0 =-0.04V Ag + (aq) +e -  Ag (s) E 0 =+0.80V E 0 cell = E 0 cathode - E 0 anode E 0 cell = (+0.80 V)- (-0.04 V)= +0.84 V E 0 cell = positive = spontaneous

17. 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

18. Break Time Take a 10 minute break

19. Mass-Mass Stoichiometry 3 Cu + 8 HNO 3  3 Cu(NO 3 ) 2 + 4 H 2 O + 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?

20. Step 3: Compute 150g Cu(NO 3 ) 2 1 mole8 mol HNO 3 63.012 g = 187.554g3 mol Cu(NO 3 ) 2 1 mole 134 g HNO 3

21. Gas Stoichiometry Xenon gas reacts with fluorine gas according to the shown reaction. If a researcher needs 3.14L of XeF 6 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 F 2 (g)  XeF 6 (g)

22. Gas Stoichiometry Xenon 3.14L XeF 6 1mole 1Xe 22.4L = 22.4L 1XeF 6 1 mole 3.14L Xe Fluorine 3.14L XeF 6 1 mole 3 F 2 22.4L = 22.4L 1 XeF 6 1 mole 9.42L F 2

23. Solution Stoichiometry How many milliliters of 18.0M Sulfuric Acid are required to react with 250mL of 2.50M Aluminum Hydroxide? H 2 SO 4 + Al(OH) 3  H 2 O + Al 2 (SO 4 ) 3 3 H 2 SO 4 + 2 Al(OH) 3  6 H 2 O + Al 2 (SO 4 ) 3

24. 250mL Al(OH) 3 1L 2.5 mol 3 H2SO4 1L 1000mL 1000mL1 L 2 Al(OH)3 18.0 mol 1L 52.1 mL H2SO4

25. 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

26. 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?

27. Limiting Reactant 0.960g O 3 1 mole 1 NO 2 44.0g NO 2 48g O 3 1 O 3 1 mole 0.880g NO 2 0.900gNO 1 mole 1 NO 2 44.0g NO 2 30g O 3 1 O 3 1 mole 1.32g NO 2

28. Break Time Take a 10 minute break

29. 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

30. 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

31. Equilibrium Constant nA + mB ↔ xC + yD K= [C] x [D] y [A] n [B] m

32. Factors affecting Reaction Rates

33. 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? Experiment Number Initial [A] Initial [B] Initial Rate of Formation of C 10.20 M 2.0 x 10 -4 M/min 20.20 M0.40 M8.0 x 10 -4 M/min 30.40 M 1.6 x 10 -3 M/min

34. 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 2 nd order reaction A is of a 1 st order reaction

35. Calculating pH and pOH pH + pOH = 14 pH = -log[H + ] pOH = -log[OH - ] What is the pH of a 2.5x10 -6 M HNO 3 solution? pH = -log [2.5x10 -6 ] pH = 5.6

36. 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