FTCE Chemistry SAE Preparation Course
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Transcript of FTCE Chemistry SAE Preparation Course
FTCE Chemistry SAEPreparation Course
Session 4
Lisa BaigInstructor
Course OutlineSession 1
Review Pre TestCompetencies 1 & 2
Session 2Competency 5
Session 3Competency 3
Session 4Competency 4
Session 5Competencies 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 Competencies1. 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%)
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 + H2OCO2 + H2OAlCl3 + CuNo 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
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 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 mole3.14L Xe
• Fluorine3.14L XeF6 1 mole 3 F2 22.4L =
22.4L 1 XeF6 1 mole9.42L 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 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
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
Factors affecting Reaction Rates
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
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