1 / 60

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

raja
Download Presentation

Honors Chemistry Final Assessment Review

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related