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Final Exam Review Semester 2 Chapters: 8,9,10,11,13,14

Final Exam Review Semester 2 Chapters: 8,9,10,11,13,14. Chapter 8 Test Review. Vocabulary. Coefficient Product Reactant Combustion reaction Decomposition reaction Single replacement reaction Double replacement reaction Synthesis reaction Precipitate Aqueous Complete ionic equation

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Final Exam Review Semester 2 Chapters: 8,9,10,11,13,14

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  1. Final Exam ReviewSemester 2Chapters: 8,9,10,11,13,14

  2. Chapter 8 Test Review

  3. Vocabulary • Coefficient • Product • Reactant • Combustion reaction • Decomposition reaction • Single replacement reaction • Double replacement reaction • Synthesis reaction • Precipitate • Aqueous • Complete ionic equation • Net ionic equation • Solute • Solvent • Spectator Ion • Covalent bond • Endothermic reaction • Lewis Exothermic reaction • structure • Molecule • Pi bond • Sigma bond • Resonance • Structural formula • VESPR model • Polar covalent bond • Non-polar bond • Chemical equation • Chemical reaction

  4. Covalent Bonds • How many covalent bonds can elements in the following groups form: • Group 1 (alkali metals) • Group 2 (alkali earth metals) • Group 3 • Group 4 • Group 5 • Group 6 • Group 7 (halogens) • Group 8 ( noble gases)

  5. Properties of Covalent bonds • Bond length decreases as number of covalent bonds increases. • Bond strength increases as number of covalent bonds increases • Ex. • Bond length increases as number of covalent bonds decreases • Bond strength decreases as number of covalent bonds decreases. • Ex.

  6. Sigma and Pi bonds • Sigma- • Single covalent bond • Single bond- 1 sigma • Pi • Multiple covalent bonds • Double bond- 1 sigma, 1 pi bond • Triple bond- 1 sigma, 2 pi bonds

  7. Diatomic Molecules • List the 7 diatomic molecules:

  8. Diatomic Molecules • Molecules made up of two atoms. • There are 7 diatomic molecules. • H2, N2, O2, F2, Cl2, Br2, I2

  9. How many atoms in each formula? CH3OH CH4 PF3 OF2 NO2- BH3 SO42- CN- N2H2

  10. Common Prefixes

  11. Naming Molecules • SiS4 • PCl5 • CCl4 • NO

  12. Writing Formulas • Sulfur difluoride • Silicon tetrachloride • Chlorine trifluoride • Tetrasulfur heptanitride

  13. Steps to doing lewis structures with covalent bonds 1.Count the valence electrons for all atoms 2.Put the least electronegative atom in the center. Hydrogen is always on outside 3.Assign 2 electrons to each atom 4.Complete octets on outside atoms 5.Put remaining electrons in pairs on central atom 6.If central atom doesn’t have an octet, move electrons from outer atoms to form double or triple bonds

  14. Lewis Structures and Octet - + + - Practice by drawing H2 O2 N2 H2O CO2

  15. Lewis structures • CH3OH • BH3 • N2H2

  16. Lewis Structures with polyatomic ions • SO42- • CN-

  17. Tetrahedral Based Shapes Tetrahedral Trigonal Pyramidal Bent

  18. Non-Tetrahedral Based Shapes Linear Trigonal Planar Fewer electrons than octet!

  19. Summary of Common Molecular Shapes

  20. Molecular Shapes • CH4 • PF3 • OF2 • NO2-

  21. Chapter 9 Review

  22. List the following GENERAL equations • Combustion • Synthesis • Decomposition • Double replacement • Single replacement

  23. Identify the type of Chemical Reaction

  24. Name the type of chemical reaction • 2SO2 + O2 2AL(OH)3 + 3CaSO4 • 2Be + O2 2BeO • 2PbO2  2PbO + O2 • C2H6 + O2  CO2 + H2O • Li + NaOH  LiOH + Na

  25. What chemical reaction does this picture show?

  26. What chemical reaction does this picture show?

  27. Balance the following equations and write the ratio of coefficients:

  28. K Na Li No, Ni is below Na Ca Yes, Li is above Zn Mg Al Zn Yes, Al is above Cu Fe Ni Sn Pb Yes, Fe is above Cu H Cu Hg Ag Au Activity Series The activity series ranks the relative reactivity of metals. It allows us to predict if certain chemicals will undergo single displacement reactions. Metals near the top are most reactive and will displace metals near the bottom. Q: Which of these will react? Fe + CuSO4 Ni + NaCl Li + ZnCO3 Al + CuCl2 Cu + Fe2(SO4)3 NR (no reaction) Zn + Li2CO3 Cu + AlCl3

  29. Chapter 10 The Mole

  30. Define the following: • Hydrate • Molecular formula • Empirical formula • Percent composition • Mole

  31. Find the atomic mass for the following atoms: 9. C 10. H 11. Cl 12. O 13. Fe

  32. Converting Moles to Particles and Particles to Moles • You can use the mole as a conversion factor in dimensional analysis problems. • Since one mole is equal to 6.02 X 1023particles or things the conversion factors are: 1 mole__ 6.02 X 1023 6.02 X 1023 1 mole

  33. Using Molar Mass • Molar mass can be used in dimensional analysis to convert the # of moles of a substance into grams or vice versa. Molar mass conversion factors: 1 mol of substance= Molar Mass of Substance (g) Molar Mass of Substance (g) 1 mol of substance 1 mol of substance___ Molar mass of substance (g)

  34. MASS (g) MOLES (mol) Particles (atoms or molecules or F.U.’s)

  35. Find the molar mass for the following compounds: 14. CH2O 15. CaSO4 16. Na3PO4

  36. Solve the following molar conversions: 23. How many atoms are in 0.750 moles of zinc? 24. How many moles of magnesium is 3.01 x 1022 atoms of magnesium? 25. Find the mass of 1.00 x 1023 molecules of N2

  37. Percent Composition Determine the percentage composition of sodium carbonate (Na2CO3)? Molar Mass Percent Composition 46.0 g x 100% = 43.4 % Na = 2(23.00) = 46.0 C = 1(12.01) = 12.0 O = 3(16.00) = 48.0 MM= 106 g % Na = 106 g 12.0 g x 100% = 11.3 % % C = 106 g 48.0 g x 100% = 45.3 % % O = 106 g

  38. Solve the following percent composition problems: 26. Mg(NO3)2 27. (NH4)2S

  39. Formulas Percent composition allow you to calculate the simplest ratio among the atoms found in compound. Empirical Formula – formula of a compound that expresses lowest whole number ratio of atoms. Molecular Formula – actual formula of a compound showing the number of atoms present Examples: C6H12O6 - molecular C4H10 - molecular - empirical - empirical C2H5 CH2O

  40. Calculating Empirical Formula Example: A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a binary compound. Determine the empirical formula for this compound. 1 mol Co 4.550 g Co = 0.07721 mol Co 58.93 g Co 1 mol Cl 5.475 g Cl = 0.1544 mol Cl 35.45 g Cl 0.07721 mol Co 0.1544 mol Cl = 1 = 2 0.07721 0.07721 CoCl2

  41. Solve the following empirical formula problem: What’s the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen.

  42. Calculating Molecular Formula Example 1: A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of 283.88g. What is the compound’s molecular formula? Step 3: Multiply Step 1: Molar Mass P = 2 x 30.97 g = 61.94g O = 5 x 16.00g = 80.00 g 141.94 g (P2O5)2 = P4O10 Step 2: Divide MM by Empirical Formula Mass 238.88 g = 2 141.94g

  43. Solve the following molecular formula problem: A compound with an empirical formula of C2H8N and a molar mass of 46 grams per mole. Find the molecular formula.

  44. Which substances have the same empirical formula? Which samples have the same empirical formula?

  45. Hydrates Hydrated salt – salt that has water molecules trapped within the crystal lattice Examples: CuSO4•5H2O CuCl2•2H2O Anhydrous salt – salt without water molecules Examples: CuCl2

  46. Chapter 11 Review

  47. Define the following: • Stoichiometry • Mole ratio • Excess reactant • Limiting reactant • Theoretical yield • Actual yield • Percent yield

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