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FTCE Chemistry SAE Preparation Course

FTCE Chemistry SAE Preparation Course. Session 1. Lisa Baig Instructor. 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 Outline. Session 1 Review Pre Test Competencies 1 & 2

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FTCE Chemistry SAE Preparation Course

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  1. FTCE Chemistry SAEPreparation Course Session 1 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 1 & 2 Session 2 Competency 5 Session 3 Competency 3 Session 4 Competency 4 Session 5 Competencies 6, 7 and 8 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. Pre-Test • Your homework coming into this session was to complete the pre-test and bring in to this session. • We will now go over your test answers. • You will receive a listing of competencies covered by each question, to better review the information you need further assistance in

  7. Pre-Test Review

  8. Break Time Take a 10 minute break!

  9. 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%)

  10. Knowledge of theNature of Matter Differentiate between pure substances, homogeneous mixtures and heterogeneous mixtures

  11. 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=P2V2P1 = P2V1 = V2 T1 T2 T1 T2 P1V1= P2V2 PV= nRT Values for R are given T1 T2 on your reference sheet

  12. Knowledge of theNature of Matter Apply units of mass, volume and moles to determine concentrations and dilutions of solutions. Molarity (M) = moles/Liter Molality (m) = moles/kilogram How many liters of solution are needed to make a 0.200M solution with 36.7g of Calcium chloride?

  13. 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= 0.331 moles CaCl2 110.984 g/mol 0.331 moles CaCl2= 1.65 L of solution 0.200 M solution

  14. Knowledge of theNature of Matter Analyze the effects of physical conditions on solubility and the dissolving process How do changes in the following affect solubility? pressure heat agitation

  15. Knowledge of theNature of Matter Evaluate problems relating colligative properties, molar mass and solution process Pactual = POXsolvent If 18g of Sucrose (C12H22O11) are used in a 250mL cup of coffee. (80oC), What is the vapor pressure of the sugared coffee?

  16. How many moles of Sucrose? (C12H22O11) • Molar mass = 342 g/mol • Moles = 0.105 mol • 1 mL = 1g of water, so 250g of water • 13.89 mol H2O 13.89 mol H2O = X 13.89 mol H2O+ 0.105 mol C12H22O11 X = 0.992 • Vapor pressure of water at 80oC = 355.1 (reference sheet) • P = (355.1)(0.992) • P = 352 mmHg

  17. Knowledge of theNature of Matter • 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 = -kfmoles solute kg solvent

  18. Practice problem What is the Freezing Point Depression if 2.84 moles of a solute are added to 0.687 kg of benzene? Normal F.P = 5.48oC Kf= 5.12 DTt = -kfmoles solute kg solvent DTt = -5.12(2.84/.687) DTt = -21.16 5.48oC -21.16oC=-15.68oC

  19. Knowledge of theNature of Matter • Solve problems involving an intensive property of matter • Density • Specific Heat D = m/V Cp= . Q . m*DT

  20. 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=Q 67.6J=Q

  21. Knowledge of theNature of Matter • Differentiate physical methods for separating the components of mixtures • Chromatography • Combined liquids • Extraction • Combined liquids • Filtration • Solids within liquids

  22. Break Time Take a 10 minute break!

  23. Knowledge of Energy and its Interaction with Matter • Distinguish between different forms of energy • Thermal • Electrical • Nuclear • Mechanical • Potential • Kinetic

  24. Knowledge of Energy and its Interaction with Matter The Kinetic Molecular Theory of Matter • Gases consist of large numbers of tiny particles that are far apart relative to their size • Collisions between gas particles and between particles and container walls are elastic collisions • Gas particles are in continuous, rapid random motion. They therefore possess kinetic energy, which is energy of motion • There are no forces of attraction between gas particles • The temperature of a gas depends on the average kinetic energy of the particles of the gas EK= ½ mv2

  25. Phase Diagram Pressure Temperature

  26. Points on Diagram A = Triple Point B = Normal Melting Point C = Normal Vaporization Point D = Critical Pressure Boiling Point E = Critical Point

  27. Knowledge of Energy and its Interaction with Matter Wood, A. (2006, May). CO2 info. Retrieved from http://www.teamonslaught.fsnet.co.uk/co2_info.htm

  28. As substance is heated, temperatures do NOT rise when it reaches a melting/boiling point. Temperatures remain constant until all matter reaches next state!

  29. Knowledge of Energy and its Interaction with Matter Calculate the enthalpy change for: C (s) + 2H2(g) CH4(g) Given the following equations: Equation DH C + O2  CO2 -393.5 H2 + 1/2 O2  H2O -285.8 CH4 + 2 O2  CO2 + 2 H2O -890.3

  30. 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

  31. Knowledge of Energy and its Interaction with Matter • 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

  32. Knowledge of Energy and its Interaction with Matter

  33. Knowledge of Energy and its Interaction with Matter DGo=DHo-TDSo Temperature must be in KELVINS!!! DHo- • + = endothermic • - = exothermic

  34. 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 v E = Energy of Quantum h = 6.626 x 10-34 J•s (Planck’s Constant) v = frequency of the wave C = lx v C = Speed of Light 3 x 108m/s l = wavelength v= frequency

  35. Homework • Diagnostic Exam in your AP chem Prep book- Page 17-26 • Only answer the questions for these Chapters & Questions • Ch 8 #21, 22 • Ch 9 #25, 28, 29, 30 • Ch 12 #55 • Ch 13 #60

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