Pre-Test
• The answer key for the pre-test is now on the wiki
• You will receive a listing of competencies covered by each question, to better review the information you need further assistance in
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
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%)
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
Required Materials
• Scientific Calculator• Paper for notes• State Study Guide• SUGGESTED Book
– 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.
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%)
Knowledge of theNature of Matter
Differentiate between pure substances, homogeneous mixtures and heterogeneous mixtures
Knowledge of theNature of Matter
Determine the effects of changes in temperature, volume, pressure or quantity
on an ideal gas
(Work with the various gas laws and their constants.)
P1V1=P2V2 P1 = P2 V1 = V2
T1 T2 T1 T2
P1V1 = P2V2 PV= nRT Values for R are given
T1 T2 on your reference sheet
Apply units of mass, volume and moles to determine concentrations and
dilutions of solutions.
Molarity (M) = moles/LiterMolality (m) = moles/kilogram
How many liters of solution are needed to make a 0.200M solution with 36.7g of
Calcium chloride?
Knowledge of theNature of Matter
How many liters of solution are needed to make a 0.200M solution with 36.7g of Calcium chloride?
Molarity = moles/Liter
36.7g CaCl2 =
110.984 g/mol=0.331 moles CaCl2
0.331 moles CaCl2 =
0.200 M solution= 1.65 L of solution
Analyze the effects of physical conditions on solubility and the
dissolving process
How do changes in the following affect solubility?
pressureheat
agitation
Knowledge of theNature of Matter
Evaluate problems relating colligative properties, molar mass and solution
processPactual = POXsolvent
If 18g of Sucrose (C12H22O11) are used in a 250mL cup of coffee. (80oC), What is the vapor pressure of the sugared coffee?
Knowledge of theNature of Matter
• How many moles of Sucrose? (C12H22O11) – Molar mass = 342 g/mol– Moles = 0.0526 mol
• 1 mL = 1g of water, so 250g of water – 13.89 mol H2O
13.89 mol H2O = X
13.89 mol H2O+ 0.0526 mol C12H22O11
X =.996• Vapor pressure of water at 80oC = 355.1
(reference sheet)• P = (355.1)(0.996)• P = 354 mmHg
• Analyze the effects of forces between chemical species on properties (eg, melting point, boiling point, vapor pressure, solubility, conductivity of matter)– ie- boiling point elevation, freezing point
depression
DT =kbm
DTt = -kf moles solute kg solvent
Knowledge of theNature of Matter
Practice problemWhat is the Freezing Point Depression if
2.84 moles of a solute are added to 0.687 kg of benzene?
Normal F.P = 5.48oCKf = 5.12 DTt = -kf moles solute
kg solventDTt = -5.12(2.84/.687)
DTt = -21.16
5.48oC -21.16oC=-15.68oC
• Solve problems involving an intensive property of matter– Density– Specific Heat
D = m/V Cp= . Q . m*DT
Knowledge of theNature of Matter
Practice problem
What is the energy absorbed by an 8.32g sample of Gold that goes from 37oC to 100oC? (Specific Heat of Gold = 0.129)
Cp= . Q . m*DT
0.129 = Q/(8.32•63)0.129•8.32•63=Q67.6J=Q
• Differentiate physical methods for separating the components of mixtures– Chromatography
• Combined liquids
– Extraction• Combined liquids
– Filtration• Solids within liquids
Knowledge of theNature of Matter
Knowledge of Energy and its Interaction with Matter
• Distinguish between different forms of energy– Thermal– Electrical– Nuclear– Mechanical– Potential– Kinetic
The Kinetic Molecular Theory of Matter1) Gases consist of large numbers of tiny particles that
are far apart relative to their size2) Collisions between gas particles and between
particles and container walls are elastic collisions3) Gas particles are in continuous, rapid random
motion. They therefore possess kinetic energy, which is energy of motion
4) There are no forces of attraction between gas particles
5) The temperature of a gas depends on the average kinetic energy of the particles of the gas EK= ½ mv2
Knowledge of Energy and its Interaction with Matter
Points on Diagram
A = Triple PointB = Normal Melting Point
C = Normal Vaporization PointD = Critical Pressure Boiling Point
E = Critical Point
Wood, A. (2006, May). CO2 info. Retrieved from http://www.teamonslaught.fsnet.co.uk/co2_info.htm
Knowledge of Energy and its Interaction with Matter
As substance is heated, temperatures do NOT rise when it reaches a
melting/boiling point. Temperatures remain constant until all matter
reaches next state!
Calculate the enthalpy change for:C (s) + 2H2 (g) CH4 (g)
Given the following equations:Equation DHC + O2 CO2 -393.5
H2 + 1/2 O2 H2O -285.8
CH4 + 2 O2 CO2 + 2 H2O -890.3
Knowledge of Energy and its Interaction with Matter
We want C (s) + 2H2 (g) CH4 (g), so:
C + O2 CO2 -393.5
CO2 + 2 H2O CH4 + 2 O2 +890.3
2(H2 + ½ O2 H2O) 2(-285.8)
-74.8
• Predicting Entropy changes• Look at States of Matter
– Solids- LOW entropy– Liquids- Medium entropy– Gases- HIGH entropy
• Look at compounds-vs-elements– The more items in combination, the
more entropy
Knowledge of Energy and its Interaction with Matter
Knowledge of Energy and its Interaction with Matter
DH DS DG Spontaneous?
- + - Yes
- --
@ low temps
Yes@ low temps
+ +-
@ high temps
Yes@ high temps
+ - + No
Knowledge of Energy and its Interaction with Matter
DGo=DHo-TDSo
Temperature must be in KELVINS!!!DHo- • + = endothermic• - = exothermic
Knowledge of Energy and its Interaction with Matter
• Relate regions of the electromagnetic spectrum to the energy, wavelength and frequency of photons
E = h x vE = Energy of Quantum
h = 6.626 x 10-34 J•s (Planck’s Constant)v = frequency of the wave
C = l x vC = Speed of Light
3 x 108 m/s
l = wavelengthv= frequency
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