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Honors Chemistry Final Assessment Review

Honors Chemistry Final Assessment Review. Your Turn. Let’s see if you can do the last 4 orbital filling diagrams and the 4 quantum numbers for the last electron of #112, Copernicium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2

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Honors Chemistry Final Assessment Review

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  1. Honors ChemistryFinal Assessment Review

  2. Your Turn • Let’s see if you can do the last 4 orbital filling diagrams and the 4 quantum numbers for the last electron of #112, Copernicium • 1s2 2s2 2p6 3s23p64s2 3d10 4p6 5s2 4d10 5p6 6s2 • 5d1 4f14 5d10 6p1-6 7s2 6d1 5f146d10

  3. The Electrons of the 4 Configurations for Copernicium • 7s = ___ 6p = ___ ___ ___ • 0 -1 0 +1 • 6d = ___ ___ ___ ___ ___ • -2 -1 0 1 2 • 5f = ___ ___ ___ ___ ___ ___ ___ • -3 -2 -1 0 1 2 3

  4. The 4 Quantum Numbers for the Circled Electron (112th) of Copernicium

  5. Ionization Energy Trends

  6. Ionization Energy Trends

  7. Electronegativity Trends

  8. Electronegativity Trends

  9. PROBLEM: Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b)COCl2. SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair SAMPLE PROBLEM: Predicting Molecular Shapes with Two, Three, or Four Electron Groups The shape is based upon the tetrahedral arrangement. The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. <109.50 The type of shape is AX3E

  10. 124.50 1110 SAMPLE PROBLEM: Predicting Molecular Shapes with Two, Three, or Four Electron Groups (b) For COCl2, C has the lowest EN and will be the center atom. There are 24 valence e-, 3 atoms attached to the center atom. C does not have an octet; a pair of nonbonding electrons will move in from the O to make a double bond. Type AX3 The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar. The Cl-C-Cl bond angle will be less than 1200 due to the electron density of the C=O.

  11. PROBLEM: Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5. SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal. SAMPLE PROBLEM: Predicting Molecular Shapes with Five or Six Electron Groups (b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.

  12. Exit Quiz Al3+ combines with sulfate (SO4)2– to make aluminum sulfate. Write the chemical formula for aluminum sulfate.

  13. Exit Quiz Answer 3+ 2- Al (SO4) 2 3

  14. Naming Organics: The IUPAC Rules • Find longest carbon chain • Number the chain so the substituent groups have the lowest total number • Give alkyl groups attached to the longest chain a name and a number • Multiple alkyl groups named alphabetically. ex. 3-ethyl-2-methylpentane • Multiple groups that are the same: di(2), tri(3), tetra(4), penta(5), hexa(6), ex. 2,4-dimethylpentane • Halogens are named “halo” groups – fluoro, chloro, bromo, iodo

  15. The IUPAC Rules Double bonds = ene, triple bonds - yne Multiple double/triple bonds have a prefix just in front of the ending. ex. 2,3,4,5-octatetraene Cyclics = cyclo- ex. cyclobutane Benzene = 3 resonating double bonds

  16. The IUPAC Rules Alcohols – OH, -ol, many = -diol, must show the H of OH Acids - C=OOH, -oic acid, on the end, must show the H of OH Aldehyde - C=O on the end, -al Ketone - C=O in the middle, -one,

  17. The IUPAC Rules Ethers - O in the middle, R1-R2 ether, R1 = first in alphabet, ex. methylpropyl ether. Esters - C=O-O in the middle. Acid part is named last with -oate ending, other part is named as radical. ex. methylpropanoate First priority - functional groups above Second priority - double/triple bonds Third priority - side chains (radicals)

  18. Convert 15.2 m/s to km/hr • 54.7 km/hr • 0.912 km/hr • 4.22 km/hr • 5.47 x 107 • not listed

  19. Converting Word Equations into Chemical Equations #10 Strontium iodide + Lead (II) phosphate  Strontium phosphate + lead (II) iodide SrI2 + Pb3(PO4)2 ----> Sr3(PO4)2 + PbI2 3 3

  20. Practice Quiz Net Ionic Equations • Write the molecular, complete ionic, and net ionic equations for this reaction: Silver nitrate reacts with calcium chloride Molecular: 2AgNO3 + CaCl2  2AgCl + Ca(NO3)2 Complete Ionic: 2Ag+ + 2NO3-+ Ca2+ + 2Cl-  2AgCl + Ca2+ + 2NO3- Net Ionic: 2Ag++ 2Cl-  2AgCl

  21. Mixed Practice • State the type, predict the products, and balance the following reactions: • BaCl2 + H2SO4 • C6H12 + O2  • Zn + CuSO4  • Cs + Br2  • NaCl  Double Displacement Combustion Single Displacement Synthesis Decomposition

  22. Answers • BaCl2 + H2SO4 BaSO4(s) + 2 HCl • C6H12 + 9 O2  6 CO2 + 6 H2O • Zn + CuSO4  Cu + ZnSO4 • 2 Cs + Br2  2 CsBr • 2 NaCl  2 Na + Cl2

  23. Try This… • Predict the products, balance the following reactions and show the change in oxidation numbers : • Zinc reacts with aqueous copper (II) sulfate • Zinc is higher relative activity so… Zn + CuSO4  Cu + ZnSO4 Each Zn loses 2e- oxidation, reducing agent Each Cu(II)gains 2e- reduction, oxidizing agent

  24. Learning Check What radioactive isotope is produced in the following bombardment of boron? 10B + 4He ? + 1n 5 2 0 13N 7

  25. Write Nuclear Equations! Write the nuclear equation for the beta decay of Co-60. 60Co 0e + 60Ni 27 -1 28

  26. Write Nuclear Equations! In the following reaction, what is being emitted and what is the daughter nuclide? 59Fe 0e + 59Co 26 -1 27 Beta Particle

  27. x 6.02x1023 molecules 1 mol H2O x 1 mol H2O 18.02 g H2O Review Mass-Mole-Molecules: Determine the number of molecules in 73 g of water # H2O molecules = 73 g H2O = 2.4 x 1024 molecules H2O

  28. x 253.80 g I2 1 mol I2 x 3 mol I2 2 mol Al Try this one: Calculate the mass in grams of iodine required to react completely with 0.50 moles of aluminum. Al + I2 AlI3 2Al + 3 I2 2 AlI3 0.50 mol Al = 190 g I2

  29. x 253.80 g I2 1 mol I2 x 3 mol I2 x 1 mol Al 2 mol Al 26.98 g Al Try this one: Calculate the mass in grams of iodine required to react completely with 0.50 g of aluminum. Al + I2 AlI3 2Al + 3 I2 2 AlI3 0.50 g Al = 7.1 g I2

  30. What mass of ZnO is formed when 20.0 g of MoO3 is reacted with 10.0 g of Zn? ( you must first determine which one is the limiting reactant!) x 3 mol Zn x 1 mol MoO3 x 65.39 g Zn 2 mol MoO3 143.94 g MoO3 1 mol Zn 3 Zn + 2 MoO3 Mo2O3 + 3 ZnO 20.0 g MoO3 = 13.6 g Zn needed, have 10.0g. Zn is the limiting reactant

  31. Use 10.0g of Zn to calculate the amount of ZnO produced. =12.4g ZnO produced

  32. x 502.46 g SbI3 1 mol SbI3 x 1 mol I2 x 2 mol SbI3 253.80 g I2 3 mol I2 How many grams of antimony(III) iodide would be produced using 98.60g of iodine? Sb + I2 SbI3 2Sb + 3 I2 2 SbI3 98.60 g I2 = 130.1 g SbI3

  33. x 118.00 g SbI3 x 100 130.1 g SbI3 Using the previous problem, determine the percent yield if 118.00 g of antimony (III) iodide is produced. (130.1 g of Sbl3 should have been produced). What is the percent yield? 2Sb + 3 I2 2 SbI3 = 90.70 % Yield

  34. Empirical Formulas: 72% iron and 28% oxygen Determine the formula for this substance. moles of Fe 72 g Fe 55.85 g/mole = 1.29 moles Fe moles of O = 28 g H 16.00 g/mole = 1.75 moles O Formula:

  35. Multiply subscripts by 3 to get

  36. How much of 3.0 M HCl do I need to use (and diluet to 1 L) to make 1 L of 1.0 M HCl? 3.0 M HCl x X = 1.0 M HCl x 1 L X= .33L or 330mL

  37. Name HCl. • Hydrogen chloride • Hydrochloric acid • Chloric acid • Perchloric acid • Chlorous acid • Hypochlorus acid • I have no idea!! • Who cares??

  38. Name HClO4. • Hydrogen chloride • Hydrochloric acid • Chloric acid • Perchloric acid • Chlorous acid • Hypochlorus acid • I have no idea!! • Who cares??

  39. Name Fe(OH)3 • Iron (III) Hydroxide • Iron Hydroxide • Ironic Acid • Iron (I) Hydroxide • Iron Oxyhydride • Not listed

  40. Which of the following definitions of an acid includes conjugate acids? • Arrhenius • Bronsted-Lowry • Lewis • Kenzig • Woods • Toburen • Sabol • Sanson

  41. Identify the conjugate base in the following equation. • NH3 • H2O • NH4+ • OH-

  42. H2O + CO32- OH- + HCO3According to Bronsted-Lowry theory, in the above reaction, H2O is a(n) • Acid • Base • Conjugate acid • Conjugate base

  43. According to Lewis theory,PCl3 is a(n) • Acid • Base • Salt • Conjugate Acid • Conjugate Base

  44. The pOH of a 0.0030M solution of H2SO4 is: • 2.52 • 11.48 • 2.22 • 11.78 • 0.99 • 13.01

  45. 100.0 mL of 3.000 M nitric acid neutralizes 3.000 M of aluminum hydroxide. How many mL of the base did you use? • 100.0 mL • 50.00 mL • 33.33 mL • 16.67 mL • 8.333 mL • Not listed

  46. Learning Check A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

  47. Calculation • P1 = 0.850 atm V1 = 675 mL T1 = 308 K • P2 = 1.06 atm V2 = 315 mL T2 = ?? P1 V1 P2 V2 = P1 V1T2= P2 V2 T1 T1 T2 T2= 1.06 atm x 315 mL x 308 K 0.850 atm x 675 mL T2 = 179 K - 273 = -94 °C = 179 K

  48. Zinc will react with hydrochloric acid. What are the 2 products for this reaction? • ZnCl + H • ZnCl + H2 • Zn2Cl + H2 • ZnCl2 + H2 • Not listed

  49. Zinc will react with hydrochloric acid. What kind of reaction is this? • Double replacement • Single replacement • Synthesis • Decomposition • Not listed

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