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FTCE Chemistry SAE Preparation Course

FTCE Chemistry SAE Preparation Course

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FTCE Chemistry SAE Preparation Course

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  1. FTCE Chemistry SAEPreparation Course Session 2 Lisa Baig Instructor

  2. 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

  3. Course Outline Session 1 Review Pre Test Competencies 6, 7 and 8 Competencies 1 & 2 Competency 5 Session 2 Competency 3 Competency 4 Post Test

  4. Required Materials • Scientific Calculator • 5 Steps to a 5: AP Chemistry • Langley, Richard, & Moore, John. (2010). 5 steps to a 5: AP chemistry, 2010-2011 edition. New York, NY: McGraw Hill Professional. • Paper for notes • State Study Guide

  5. 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%)

  6. Electronegativity • Fluorine is the most electronegative element. • Pattern of increasing electronegativity moves from bottom to top, and from left to right across the periodic table.

  7. Chemical Bond Mutual electrical attraction between the nuclei and valence electrons of different atoms that bind the atoms together Atoms would like to have 8 Valence electrons. These bonds help the atoms to achieve their full valence shells Three Types Ionic Covalent Metallic

  8. Force of attraction between oppositely charged ions Occurs between Metal and Non-Metal elements The Non-metal “steals” the valence electron(s) from the Metal Forms a crystalline structure of these positive and negative charges Typically solids at room temperature Ionic Bond

  9. Ionic Character • Ionic Bonds are bonds with > 50% ionic character • Difference in Electronegativity of involved atoms is >1.7

  10. Covalent Bond • Sharing of valence electron pairs by 2 atoms • Occurs between 2 Non-metal elements • Or the SAME non-metal element • Can share one, two or three pairs of electrons • Single Bond = 1 pair (1 sigma) • Double Bond = 2 pairs (1 sigma, 1 pi) • Triple Bond = 3 pairs (1 sigma, 2 pi) • Sharing can also be “unequal” • Called a POLAR covalent bond • Typically liquids or gases at room temperature

  11. Character • Ionic Character: • Polar Covalent Bonds have between 5% and 50% ionic Character • Non-Polar Covalent Bonds have less than 5% ionic character • Difference in Electronegativities • Polar Covalent Bonds have between 0.3 and 1.7 as a difference in electronegativities • Non-Polar Covalent bonds have less than 0.3 difference in electronegativities

  12. Break Time Take a 10 minute break

  13. Ionic Compounds • Ion names are used in combination • Cation- same as the element • Transitional Metals use Roman Numerals to represent Charge • Anion- replace the ending syllable of the element name with –ide • Polyatomic Ions- use the name of that ion- do not try to rename. Use “criss-cross” to determine charges

  14. CuCl2 Copper (II) Chloride CuO Copper (II) Oxide NaCl Sodium Chloride KI Potassium Iodide Mg3N2 Magnesium Nitride PbO2 Lead (IV) Oxide

  15. Lewis Structures • A way to show the octet rule in molecules

  16. Practice • Draw the lewis structures for • Ammonia (NH3) • Water (H2O) • Phosphorus Trifluoride (PF3) • Hydrogen Cyanide (HCN) • Ozone (O3) • Formaldehyde (CH2O)

  17. VSEPR • AB5 • Trigonal bipyramidal • AB6 • Octahedral

  18. VSEPR • AB4 • Tetrahedral • 109.50 Bond Angles • AB3 • Trigonal Planar • 1200 Bond Angles • AB2 • Linear • 1800 Bond Angles

  19. VSEPR • AB2E • Bent or Angular • AB2E2 • Bent or Angular • AB3E • Trigonal Pyramidal

  20. Polarity? • The potential for opposite charges at different areas of a molecule

  21. Shape and Polarity? • What is the shape and polarity of the following molecules? • Water • Ammonia • Carbon Tetrachloride • Carbon Dioxide • Hydrogen Chloride

  22. Hybrids • Atoms “don’t like” to have empty orbitals • Hybridization • Mixing of 2 or more atomic orbitals of similar energies to produce new hybrid orbitals of equal energies • It works like this • Methane: CH4 Normally: 1s22s22p2 • Through hybridization- it forms an “sp” orbital, with 4 electrons total • New arrangement: 1s22(sp3) 4

  23. Hybrid Orbitals

  24. What type of hybrid? • Beryllium fluoride • BeF2 • sp • Ammonia • NH3 • sp2 • Methane • CH4 • sp3

  25. Break Time Take a 10 minute break

  26. Spectroscopy • Devices that measure the interaction between matter and energy • Absorption • Measures the wavelengths of electromagnetic waves absorbed by a substance • X-Ray spectroscopy • Used to determine elemental composition and types of bonding

  27. Spectroscopy • UV • Used to quantify DNA and Protein concentration • Infrared • Used to determine bond type • Bonds resonate when exposed to the radiation • Nuclear Magnetic Imaging (NMR) • Used to determine bond structure

  28. Simple Organics • Alkanes (end in –ane) • Containing only single bonds • CnH2n+2 • Alkenes (end in –ene) • Containing at least one double bond • CnH2n • Alkynes (end in –yne) • Containing at lease one triple bond • CnH2n-2

  29. Simple Organics

  30. Functional Groups

  31. Naming and Formulas • Numbers are used in the name to designate locations of the following • Types of bonds • Branches • Attached functional groups • For Example • 2,2,4- trimethylpentane • 1-pentyne • 2,3,4- trimethylnonane • 2-methyl 3-hexene • 2- propanol

  32. Macromolecules • Carbohydrates • Chains of carbon, hydrogen and oxygen. • Isomers • Lipids • Fatty acids- Chains of Carbon and Hydrogen • Proteins • Chains of Amino acids • Differ in their R group • Nucleic Acids • Chains of Nucleic Acids

  33. Organic Compound Naming • Numbers are used in the name to designate locations of the following • Types of bonds • Branches • Attached functional groups • For Example • 2,2,4- trimethylpentane • 1-pentyne • 2,3,4- trimethylnonane • 2-methyl 3-hexene • 2- propanol

  34. Lunch Time We start Again In ONE HOUR

  35. 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

  36. 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

  37. 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

  38. 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

  39. 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

  40. 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

  41. 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

  42. 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

  43. 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)

  44. 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

  45. 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

  46. 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

  47. Break Time Take a 10 minute break

  48. 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?

  49. 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

  50. 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)