1. FTCE Chemistry SAE Preparation 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 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%)

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 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%)

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

11. Knowledge of the Nature 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.) P 1 V 1 =P 2 V 2 P 1 = P 2 V 1 = V 2 T 1 T 2 P 1 V 1 = P 2 V 2 PV= nRTValues for R are given T 1 T 2 on your reference sheet

12. 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? Knowledge of the Nature of Matter

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 CaCl 2 = 0.331 moles CaCl 2 110.984 g/mol 0.331 moles CaCl 2 = 1.65 L of solution 0.200 M solution

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

16. Evaluate problems relating colligative properties, molar mass and solution process P actual = P O X solvent If 18g of Sucrose (C 12 H 22 O 11 ) are used in a 250mL cup of coffee. (80 o C), What is the vapor pressure of the sugared coffee? Knowledge of the Nature of Matter

17. How many moles of Sucrose? (C 12 H 22 O 11 ) – Molar mass = 342 g/mol – Moles = 0.105 mol 1 mL = 1g of water, so 250g of water – 13.89 mol H 2 O 13.89 mol H 2 O = X 13.89 mol H 2 O+ 0.105 mol C 12 H 22 O 11 X = 0.992 Vapor pressure of water at 80 o C = 355.1 (reference sheet) P = (355.1)(0.992) P = 352 mmHg

18. 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  T =k b m  T t = -k f moles solute kg solvent Knowledge of the Nature of Matter

19. 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 = 5.48 o C K f = 5.12  T t = -k f moles solute kg solvent  T t = -5.12(2.84/.687)  T t = -21.16 5.48 o C -21.16 o C=-15.68 o C

20. Solve problems involving an intensive property of matter – Density – Specific Heat D = m/VC p =. Q. m*  T Knowledge of the Nature of Matter

21. Practice problem What is the energy absorbed by an 8.32g sample of Gold that goes from 37 o C to 100 o C? (Specific Heat of Gold = 0.129) C p =. Q. m*  T 0.129 = Q/(8.3263) 0.1298.3263=Q 67.6J=Q

22. Differentiate physical methods for separating the components of mixtures – Chromatography Combined liquids – Extraction Combined liquids – Filtration Solids within liquids Knowledge of the Nature of Matter

23. Break Time Take a 10 minute break!

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

25. The Kinetic Molecular Theory of Matter 1)Gases consist of large numbers of tiny particles that are far apart relative to their size 2)Collisions between gas particles and between particles and container walls are elastic collisions 3)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 E K = ½ mv 2 Knowledge of Energy and its Interaction with Matter

26. Phase Diagram

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

28. Wood, A. (2006, May). CO 2 info. Retrieved from http://www.teamonslaught.fsnet.co.uk/co2_info.htm Knowledge of Energy and its Interaction with Matter

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

30. Calculate the enthalpy change for: C (s) + 2H 2 (g)  CH 4 (g) Given the following equations: Equation  H C + O 2  CO 2 -393.5 H 2 + 1 / 2 O 2  H 2 O-285.8 CH 4 + 2 O 2  CO 2 + 2 H 2 O-890.3 Knowledge of Energy and its Interaction with Matter

31. We want C (s) + 2H 2 (g)  CH 4 (g), so: C + O 2  CO 2 -393.5 CO 2 + 2 H 2 O  CH 4 + 2 O 2 +890.3 2(H 2 + ½ O 2  H 2 O)2(-285.8) -74.8

32. 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

33. HH SS GG Spontaneous? -+-Yes -- - @ low temps Yes @ low temps ++ - @ high temps Yes @ high temps +-+No

34. Knowledge of Energy and its Interaction with Matter  G o =  H o -T  S o Temperature must be in KELVINS!!!  H o - + = endothermic - = exothermic

35. 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 Js (Planck’s Constant) v = frequency of the wave C = x v C = Speed of Light 3 x 10 8 m / s = wavelength v= frequency

36. 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