1. Final Review Final Review Honors Chemistry

2. Classes of matter Matter Pure substance elementcompound mixture homogeneousheterogeneous

3. Physical Changes - are those in which the identifying properties of a substance are unchanged - Ex. : change of phase breaking, cutting breaking, cutting dissolving dissolving

4. Chemical Changes - are those in which different substances are formed - Ex.: burning rusting rusting decaying or spoiling decaying or spoiling acid reacting with metal acid reacting with metal

5. subatomic particles Atomic no.=# protons Atomic no.=# protons #protons=#electrons #protons=#electrons Mass no.=#protons + Mass no.=#protons + # neutrons # neutrons

6. Electron configuration notation Helium has 2 electrons, so its electron configuration would be 1s 2 Li 1s 2 2s 1 N 1s 2 2s 2 2p 3 Ne 1s 2 2s 2 2p 6 Na 1s 2 2s 2 2p 6 3s 1 or [Ne]3s 1 Principal quantum number sublevel No. of electrons

7. Every orbital can hold two electrons.

8. d orbitals

9. Predicting electron configurations from the periodic table.

10. Ways to represent titanium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 electron configuration __ __ __ __ __ __ 1s 2s 2p 3s 3p 4s 3d Ti: 22 2 8 10 2

11. How would calcium and chlorine combine? Ca Cl CaCl 2

12. Write the formulas- always put the cation first K + and N 3- K3NK3NK3NK3N Ca 2+ and N 3- Ca 3 N 2 Ba 2+ and NO 3 - Ba(NO 3 ) 2 Criss-cross rule of thumb

13. Naming molecular compounds Use prefixes 1mono- Use prefixes 1mono- 2di- 3tri- 4tetra- 5penta- 6hexa- 7hepta- 8octa- 9nona- 10deca-

14. My favorite no name NH 4 CO 3 Write the formula for Calcium chloride Nitrogen tetrahydrogen carbon trioxide CaCl

15. Ionic or covalent? NaBr CH 4 Fe 2 O 3 CO 2 CaO NH 4 Cl SiCl 4 Name ionic covalent ionic covalent ionic covalent Sodium bromide Carbon tetrahydride Iron oxide Carbon dioxide Calcium oxide Ammonium chloride Silicon tetrachloride

16. Types of reactions Decomposition: AB A + B Synthesis (or combination): A + B AB Single Replacement (or displacement): A + BC B + AC A + BC B + AC Double Replacement (or displacement): AB + CD AD + CB AB + CD AD + CB Combustion: CH 4 + 2O 2 CO 2 + 2H 2 O

17. Balance and classify Al + Cl 2 AlCl 3 Al + Cl 2 AlCl 3 2Al + 3Cl 2 2AlCl 3 Zn + HCl ZnCl 2 + H 2 Zn + HCl ZnCl 2 + H 2 Zn + 2HCl ZnCl 2 + H 2 Zn + 2HCl ZnCl 2 + H 2 Mg(ClO 3 ) 2 MgCl 2 + O 2 Mg(ClO 3 ) 2 MgCl 2 + O 2 Mg(ClO 3 ) 2 MgCl 2 + 3O 2 C 4 H 8 + O 2 CO 2 + H 2 O C 4 H 8 + O 2 CO 2 + H 2 O C 4 H 8 + 6O 2 4CO 2 + 4H 2 O

18. Writing Balanced equations (NH 4 ) 2 CO 3 + Ba(NO 3 ) 2 (NH 4 ) 2 CO 3 + Ba(NO 3 ) 2 NH 4 + CO 3 2- Ba 2+ NO 3 - NH 4 + NO 3 - + 2NH 4 NO 3 BaCO 3 Double displacement Net ionic: CO 3 2- + Ba 2+ BaCO 3

19. Chemical reactions involve energy Endothermic reactions absorb energy Endothermic reactions absorb energy ex.: photosynthesis CoCO 3 + 81.6 kJ CoO + CO 2 Exothermic reactions give off energy Exothermic reactions give off energy ex.: burning of wood C(s) + O 2 (g) CO 2(g) + 393.5 kJ

20. Reaction Tendencies Reactions occur spontaneously when a lower energy state is achieved. Reactions occur spontaneously when a lower energy state is achieved. In exothermic reactions, the products have less energy than the reactants.

21. Reactions occur spontaneously when a state of less order is achieved. Reactions occur spontaneously when a state of less order is achieved. Entropy is the disorder in a system.

22. Molecular mass (molar mass) NaCl 23.0 + 35.5 = 58.5 Ba(NO 3 ) 2 137.3 + 2(14.0) + 6(16.0) = 261.3 CuSO 4 5H 2 O 63.5 + 32.0 + 4(16.0) + 10(1.0) + 5(16.0) = 249.5

23. Solution concentrations Percentage by weight. g solute x 100% g solution g solution g solute x 100% g solute x 100% g solute + g solvent g solute + g solvent Molality moles solute = m kg solvent kg solvent Molarity M = moles solute L solution L solution

24. MV = moles Dilutions Titrations M 1 V 1 = M 2 V 2 M a V a = M b V b If I wanted 100ml of 2.0M HCl, how much 6.0M HCl would I need? 6.0M x ?ml = 2.0M x 100ml 33.3ml 33.3ml

25. Conversion Factors Molar mass Molar mass atomic mass in g = 1 mole atomic mass in g = 1 mole Volume of gas Volume of gas At STP, 1 mole gas = 22.4L At STP, 1 mole gas = 22.4L Mole-mole ratio coefficients from balanced equation Mole-mole ratio coefficients from balanced equation Avogadros number Avogadros number 6.02 x 10 23 molecules = 1 mole 6.02 x 10 23 molecules = 1 mole

26. How many molecules are in 2.0 moles of H 2 O? 2.0 moles x 6.02x10 23 molecules 1 mole 1 mole

27. What volume would 2.5 moles of H 2 gas have at STP? 2.5 moles x 22.4L = 56L 1 mole 1 mole

28. Stoichiometry g g mole mole mole mole L (gas) L (gas)molecules Molar mass 22.4L=1 mol 6.02 x 10 23 molecules = 1 mole

29. Multiple conversion factors may be needed How many grams of CO 2 are formed from 18.5 grams of O 2 in the following reaction? How many grams of CO 2 are formed from 18.5 grams of O 2 in the following reaction? 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 2C 2 H 6 + 7O 2 4CO 2 + 6H 2 O 18.5 g O 2 x 1mole O 2 x 4 mole CO 2 x 44 g CO 2 32 g O 2 7 mole O 2 1mole CO 2 32 g O 2 7 mole O 2 1mole CO 2 14.5g

30. How many liters of CO 2 are formed from 18.5 grams of O 2 in the following reaction? 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(g) 18.5 g O 2 x 1mole O 2 x 4 mole CO 2 x 22.4 liters 32 g O 2 7 mole O 2 1mole CO 2 32 g O 2 7 mole O 2 1mole CO 2 7.40L

31. Limiting Reactants In a chemical reaction, the reactant which gets used up first limits how much product is formed. In a chemical reaction, the reactant which gets used up first limits how much product is formed. i.e. If I had 5 moles of Al and 4 moles of Cl 2, which reactant is limiting? 2Al + 3Cl 2 2AlCl 3 5 mole Al x 3 mole Cl 2 = 7.5 mole Cl 2 2 mole Al 2 mole Al Cl 2 is limiting and Al is excess. Have Need

32. % yield = actual x 100% % yield = actual x 100% theoretical theoretical If 5.50g of hydrogen reacts with nitrogen to form 20.4g of ammonia, what is the percent yield? N 2 (g) + 3H 2 (g) 2NH 3 (g) Actual = 20.4g Theoretical = 5.50g H 2 x 1 mol H 2 x 2mol NH 3 x 17.0g NH 3 2.0g H 2 3mol H 2 1mol NH 3 2.0g H 2 3mol H 2 1mol NH 3

33. The amount of heat gained or lost depends on the amount of reactants used. 2Na 2 O 2 + 2H 2 O 4NaOH + O 2 + 215.76 kJ How much heat is released by the reaction of 5.0 moles of Na 2 O 2 ? 5.0 moles Na 2 O 2 x 215.76 kJ = 539.4 kJ 2 mole Na 2 O 2 2 mole Na 2 O 2

34. Q = mcT Heat = mass specific heat change gained or lost in temp Ex: How much heat is lost when a solid aluminum ingot with a mass of 411g cools from 660.0˚C to 25˚C? c Al = 0.903J/g˚C T = 660.0 – 25 = 635˚C T = 660.0 – 25 = 635˚C Q = (411g)(0.903J/g˚C)(635˚C)=236,000 J

35. Charles Law V 1 = V 2 P constant T 1 T 2 T 1 T 2 Boyles Law P 1 V 1 = P 2 V 2 T constant Combined P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 Gay-Lussacs Law P 1 = P 2 V constant T 1 T 2 T 1 T 2

36. If temperature increases, volume ______. If temperature increases, volume ______. increases. increases. Always change Celsius to Kelvin degrees. Ex.: If a gas expands from 125ml to 850ml, its original temperature of 15 C must have changed to what? 15 C=288K 288K x 850ml = 1958.4K 125ml 125ml

37. 52.5 ml of a gas is collected over water at 20˚C and 100.0 kPa. What is the pressure of the dry gas? 100.0 kPa = P gas + P water P water at 20˚C = 2.3 kPa 100.0 kPa – 2.3 kPa = 97.7 kPa What would be the volume of this gas at STP? P 1 = 97.7 kPaP 2 =101.325kPa V 1 = 52.5 mlV 2 = ? T 1 = 20˚C=293 K T 2 =273 K

38. P1V1 = P2V2P1V1 = P2V2P1V1 = P2V2P1V1 = P2V2 T 1 T 2 T 1 T 2 52.5ml97.7 kPa = 101.325kPa V 2 52.5ml97.7 kPa = 101.325kPa V 2 293K273K 293K273K V 2 = 47.2ml

39. PV=nRT P=Pressure P=Pressure V=volume V=volume n=moles n=moles R= gas constant = 0.0821 Latm R= gas constant = 0.0821 Latm molK molK 8.31L·kPa 8.31L·kPa molK or 62.4 Ltorr molK or 62.4 Ltorr molK molK T= temperature (K) T= temperature (K)

40. When given three of the four variables, we can find the fourth. A gas has a volume of 2.20L at 25 C. If there are.085 moles of the gas, under what pressure must the gas be held? A gas has a volume of 2.20L at 25 C. If there are.085 moles of the gas, under what pressure must the gas be held? P = nRT P = nRT V P = (.085mole)(0.0821 Latm)(298K) P = (.085mole)(0.0821 Latm)(298K) 2.20L molK 2.20L molK

41. Compute the relative rate of diffusion of helium and argon.

42. My favorite no What volume of 2.0M HCl do you need to get.5 moles of HCl? 2.0M x 22.4L = 44.8L mole

43. [H + ] = 10 -pH [OH - ] = 10 -pOH pH + pOH = 14.0 Sample problem: A 0.01 M sample of NaOH completely ionizes. What is its pH, pOH, [H + ], and [OH - ] ? [OH - ] = 0.01 mole/L = 10 -2 pOH = 2 pH = 14 – pOH = 12 [H + ] = 10 -12

44. pHpOH [H 3 O + ] [OH - ] 68 10 -6 10 -8 1 1310 -1 10 -13

45. Find the pH of a solution with [H 3 O + ] of 6.59 x 10 -10 M. pH = -log(6.59 x 10 -10 ) pH = -(log 6.59 + log 10 -10 ) = 9.18 = 9.18 Find the [H 3 O + ] of a solution with a pH of 9.18. Antilog(-9.18)

46. Equilibrium Shifts Concentration Removing products shifts equilibrium right Removing products shifts equilibrium right Adding a product shifts equilibrium left Adding a product shifts equilibrium leftPressure Increasing pressure on gases shifts equilibrium toward side with fewer number of moles Increasing pressure on gases shifts equilibrium toward side with fewer number of moles

47. Redox reactions Identify the reducing and oxidizing agents. Identify the reducing and oxidizing agents. Mg (s) + Cu(NO 3 ) 2 (aq) Mg(NO 3 ) 2 (aq) + Cu (s) 0 +2 +5 -2 +2 +5 -2 0 0 +2 +5 -2 +2 +5 -2 0 Mg is oxidized and is the reducing agent Cu is reduced and is the oxidizing agent

48. Types of radiation Alpha (α) low penetrating power Beta ( β) moderate penetrating power Gamma (γ) high penetrating power

49. Transmutations Complete the following nuclear equation. Complete the following nuclear equation.

50. Half Life If you start with 2.97 x 10 22 atoms of, How many atoms will remain after 62 minutes? The half life of is 15.49 minutes. 62 = 4 half lives 15.5 2.97 x 10 22 atoms (½) 4 = 1.86 x 10 21 atoms

51. Fission- breaking apart of a nucleus Can cause a chain reaction

52. Fusion- happens on the sun

53. TITRATION, NEUTRALIZATION…