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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
Course OutlineSession 1
Review Pre TestCompetencies 6, 7 and 8
Competencies 1 & 2 Competency 5
Session 2Competency 3Competency 4
Post Test
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
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%)
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.
Chemical BondMutual 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 TypesIonic
CovalentMetallic
Ionic Bond• 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 Character
• Ionic Bonds are bonds with > 50% ionic character
• Difference in Electronegativity of involved atoms is >1.7
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
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
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
CuCl2
Copper (II) ChlorideCuO
Copper (II) OxideNaCl
Sodium ChlorideKI
Potassium IodideMg3N2
Magnesium NitridePbO2
Lead (IV) Oxide
Practice
• Draw the lewis structures for– Ammonia (NH3)
– Water (H2O)
– Phosphorus Trifluoride (PF3)– Hydrogen Cyanide (HCN)– Ozone (O3)
– Formaldehyde (CH2O)
VSEPR• AB4
– Tetrahedral– 109.50 Bond Angles
• AB3– Trigonal Planar– 1200 Bond Angles
• AB2– Linear– 1800 Bond Angles
Shape and Polarity?
• What is the shape and polarity of the following molecules?– Water– Ammonia– Carbon Tetrachloride– Carbon Dioxide– Hydrogen Chloride
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
Hybrid OrbitalsAtomic Orbitals
Type of Hybrid
Number of
Orbitals
MolecularGeomet
rys p sp 2 1800
Linear
s p p sp2 3 1200
Trigonal Planar
s p p p sp3 4 109.50
Tetrahedral
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
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
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
Simple OrganicsName Number of Carbons
Meth- 1
Eth- 2
Prop- 3
But- 4
Pent- 5
Hex- 6
Hept- 7
Oct- 8
Non- 9
Dec- 10
Functional GroupsCompound
Type Image Suffix or Prefix
Alcohol - -OH -ol
Haloalkane -X Halo-
Amine -CN- -amine
Aldehyde -COH -al
Ketone -COC- -one
Carboxylic Acid -COOH -oic acid
Ester -COOC- -oate
Amide -CON- -amide
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
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
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
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 Cl 1 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 Cl1 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 ReactionNaCl + CaSO4
KOH + NaCl
Identifying Redox Reactions2 KNO3(s) 2 KNO2(s) + O2(g)
+1 -1 +1 -1 0H2(g) + CuO(s) Cu(s) + H2O(l)
0 -2 +2 0 2(+1) -2NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
+1 -1 +1 -1 +1 -1 2(+1) -2H2(g) + Cl2(g) 2HCl(g)
0 0 +1 -1SO3(g) + H2O(l) H2SO4(aq)+6 3(-2) 2(+1) -2 2(+1) -2
Redox
Redox
Not Redox
Redox
Not Redox
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• E0
cell= E0cathode- E0
anode
• 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
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 mole 8 mol HNO3 63.012 g =
187.554g 3 mol Cu(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
• Xenon3.14L XeF6 1mole 1Xe 22.4L =
22.4L 1XeF6 1 mole
3.14L Xe • Fluorine3.14L XeF6 1 mole 3 F2 22.4L =
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
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 NO2 44.0g NO2
48g O3 1 O3 1 mole
0.880g NO2
0.900gNO 1 mole 1 NO2 44.0g NO2
30g O3 1 O3 1 mole
1.32g NO2
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
Rate LawsA 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 ofFormation of C
1 0.20 M 0.20 M 2.0 x 10-4 M/min
2 0.20 M 0.40 M 8.0 x 10-4 M/min
3 0.40 M 0.40 M 1.6 x 10-3 M/min
Rate Law for the ReactionA + 2B CR = k[A][B]2
Order of the Reaction with respect to BB is of a 2nd order reactionA is of a 1st order reaction
Calculating pH and pOHpH + pOH = 14 pH = -log[H+] pOH = -log[OH-]
• What is the pH of a 2.5x10-6M HNO3 solution?
• pH = -log [2.5x10-6]• pH = 5.6
Post-Test
• You will have one and a half hours to complete the post-test
• This test will include examples from all the competencies.
• Scores will be posted on the Quia Website tomorrow as a class file.
• Also to be posted- a reference key of the correct answers AND which competency and skill were covered for each question.