Chem 1C Exam 2 Ver 1

Chem 1C Exam 2 Ver 1 - First letter of your last name Chem...

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Unformatted text preview: First letter of your last name Chem 1C Midterm 2, Summer 2006 Name (Print): Perm: PRINT YOUR NAME AND PERM NUNIBER ABOVE AND PRINT YOUR NAME ON THE LAST PAGE OF THIS EXAM. This exam should contain 5 pages (this cover sheet and 4 pages of exam). Please count to make sure your exam is complete with no blank pages. You have 60 minutes to complete the exam. . You are allowed one 8.5” x 11” sheet of paper. No books. Show all work for questions which reguire calculations, otherwise points will be deducted. Box your final answer to receive maximum possible credit. The exam is out of a total of 100 points. Ver. 1 Chem 1C: Summer 2006: Exam 2 (Version 1) Perm 1(a). Give the oxidation state of the central metal atom and the coordination number: (4 points) Coordination number [COCNH3)4(N02)C3]C1 + 3 6 (b) Give the formula for the following compound: (4 points) Oxidation State Bis (ethylenediamine)dinitroiron(III) sulfate 1: Fe- w“): C” an.) 2 :i 909 (c ) Give the systematic name for the following compound: (4 points) [Co(NH3)5Cl]C12 PMG. Owns-Jain e We ca'baaklfll Wide (6 points) 1 2(a). Draw the optical isomers for [cis-Co(en);C12]+ ILL CMA co “3 l/ l \N (b) Draw geometrical isomers of Wt(NH3)4II]2+ and label them appropriately (4 points) I I 53%] I if?” W3 PE it” NH: by: N H; i cusl “ “3 I 3(a) . Circle true or false for each of the following statements concerning the eompiex ion [CoCNH3)Br (can? en : ethylenediamine = NH2CH2CH2NH2 (10 points) a) The complex ion contains Co(II) True b) The complex ion has cis and trans isomers but no optical isomers True c) The complex ion has cis and trans isomers and optical isomers True d) Since ethylenediamine is a strong field ligand (large A), the complex is paramagnetic True e) The geometric isomers of the complex ion have identical chemical properties True smseleummerO6ver1 1 False ase Chem 1C: Summer 2006: Exam 2 (Version 1) Perm b) Which of the following ligands are capable of linkage isomerism ‘2' (6 points) (strike out wrong answer) N02'(yes fame-)— I' 6-yee-/ no) SCN’( yes {—96—}- NHQCH2CH2NH: l 110 ) N3' (yes-l no) OCN‘ ( yes I no.) 4(a) Draw the d-orbital diagram, including the elections, for low spin [C0(NH3)5 ] 3+ (3 points) Is the ion diamagnetic or paramagnetic? Label orbitals. 3+ ..__._. 6—1 Co gamma“: 42* elf-3"" Alwyn T I N»- al A 3 2/3 xa- H'- {DFQ-mag neHC! (b) Draw the d-orbital diagram, including the electrons, for tetrahedral, high spin [CoCh]2". (3 points) Is the ion diamagnetic or paramagnetic? Label orbitals. 4‘ ’r ’t‘ 1+ -——- a . CO 7 Clem “’1, 6'” T‘L 9"" r1 Far-Am HE)“ -——J'-—- H L's“ “La-1’ “lull-5"" (c) Complexes C0(NH3)53+, Co(CN)63‘, and C0F63' absorb radiation at in no 3 ecific order 290 nm, 440 P 11m, and 770 nm. Match the complex ion with the absorption wavelengths: (6 points) Wavelength Complex 290 nm CO CON) {:3 H 440 11111 CO CW3) L3 f 770 nm Co F6 3 ’ 5. What volume of ethylene glycol C1H502, a non-electrolyte, must be added to 12.0 L of water to produce an antifreeze solution with a fi’eezing point of —40°C ’? The density of ethylene glycol is 1.11g/em3 and the density of water is 1.00 g/cm3. Kr: 186°C kg/mol solvent. Molar mass of ethylene glycol = 62.07 g A T 4 0C (10 points) 1C 0 m = . _ :3 a :1; 5 a C' 0 : 2|.5‘x62.o7=— 1335‘? CzH-édL/kfl gait/w = tasrj slag/L 01L max-w F" 12L éi-Uakw)malul =132rxi2. : 14.0?— H QWOL I . l I kj/ L smse] Csummer06ver1 2 ?‘ dé‘kSl'b; C2 Chem 1C: Summer 2006: Exam 2 (Version 1) Penn 6. For these liquids: (i) pure water, (ii) a solution of a non-electrolyte of mole fraction 0.01 in water, (iii) a solution of NaN03 of mole fraction 0.012 in water, (iv) a solution of CaClg of mole fraction 0.01 in water, choose the one with the following: (10 points) (a) Highest osmotic pressure: ac: L 5 etc}; a»... (1‘)) Highest boiling point: Cg, (LI 1 5 010.3% (c) Highest freezing point: “(blew ((1) Lowest boiling point: Cd Mar (8) Lowest freezing point: Cac’! 2. s ‘5“. m 7. A 1.6 g sample containing a mixture of 25% naphthalene Clng and 75% anthracene CHI-1m by mass was dissolved in 20.0 g benzene Calls. What would be the freezing point of the solution? The freezing point of benzene is 551°C. Kf= 512°C kg/mol solvent. (10 points) “MM-$4.4), Ms ;_- 0'” 3 cm”? + 1‘7»? cm 9:0 129 I79 = 0'00‘1‘97wd .nn 2.09 beam Thu—rs) 0.00fi97s5—v _-_- 0,43 ma} m {kg tutti-me. :. m ATP = kl; m : Sillbc log/mat sdvcwl-XO-Lfifl’t/Kj 5W x: 2-5100 0 Ti: :: 5.3—1.6. —Z-STI°C = 3.0 c L. 8. Calculate the osmotic pressure at 25°C for an aqueous solution of 1.0 of a protein (molar mass 9.0 x 104 g/mol) if the density of the solution is 1.0 g/cm3. (10) A : 2.7M?I 917;. smsel Csummer06ver1 3 Chem 1C: Summer 2006: Exam 2 (Version 1) Penn 9. At temperature T, the vapor pressure of pure benzene C5H6 is 0.930 atm. A solution was prepared by dissolving 10.0 g of a non-dissociating nonvolatile solute (mol. Mass M) in 78.1] g of C5H6 at that temperature T. The vapor pressure of the solution was found to be 0.850 atm. Assuming ideal behavior of the solution, determine M. (10 points) ° - Pf 095041; P 3' P J X‘ — fl 2 _: ‘2“: (ch (6": Qh" Eh 093045.. 0 1'" 78‘.“ : 6.9] :' '18.“ KIWI]: q '15:” 1° ___.. + .nfld 79-]: M m: NAME (PRINT) Questiorfitm W Points smseleummerOéverl 4 ...
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Chem 1C Exam 2 Ver 1 - First letter of your last name Chem...

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