key_t3_07 - Chemistry 1035 General Chemistry Test 3 Form A...

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Unformatted text preview: Chemistry 1035 General Chemistry Test 3, Form A You have 75 minutes to complete this test. A periodic chart and reference tables are attached to this test The Honor Code is in force on this test: you may neither give nor receive aid on this test Turn in only your opscan sheet when you are finished. Scores will he posted tomorrow morning on Blackboard ‘ 1. Which of the following molecules incorrectly dis _a s the bond polarity? {rm-if (1) N ~ 13 (.2)N — o (4) (:1 - o 2.. What type of bonding occurs in the molecule SCIZ‘? Cl) Ecvalent ”— 3 ‘2 =7” ”-- 7‘95 ® Polar covalent éfi/V“: A £33,9-": 0, 5 .3 Metallic C/l/é] 3 ,O (4) Ionic 3. Which of the following is not a physical characteristic of a solid metal? (1) Conductor of heat 7 7O (2) Shiny (3 Malleable @ Flammable 4. Ozone can be depleted in the stratosphere by reacting with Cl atoms (released from a C—Cl bond of a halogenated hydrocarbon). If the bond energy ofa C~Cl bond is 339 Miami, what wavelength of light is necessary to break the . . 9 ' a h c; 9 ~ ‘ . 25 4 QB 6533-53 l0”7 }\ ‘— __..,.:_.- ‘F’CO‘CUS No i V ‘ X m b ,_ '35, 8' 22.3 VarlI/ilo" (2) 3.39x I0“5m A: M,¥ écngfi/O (3) 3.53 x 10“m 3325mm (4) 5.86 X What ‘ ._ 7 N; 3.53 We 5. According to the information presented in our lectures, how many states of matter exist? (1) One 6’ 455 (:2) Two (3) Three .Four 6. What is the formal charge on the nitrogen atom in the molecule CN" ‘3 __ minus one "- ,.... _ - / 6 zero ,_. i )7: _': 5 "h" [Z‘iééj [an (3) plus one [C E/i/fl 4/ (4) plus two '2” 0 page 1 of 4 7.. Please determine the Aan=1 classification and shape of‘ the molecule SFgClg. c: ‘ l AX4, tetrahedral I If; erg“) Q‘ (2 AX4E, trigonal bipyramidal a, _ / (3) AXgfig, square planar o—“S. \ if: m (4) AX4E, tetrahedral { )( g C I A ,4 l, 8.. How does BCig violate the octet rule? I 0 j 9' {LL B has a total of‘5 pairs of‘bonding and nonbonding electrons. I yarl' 431g,\@13 has only .3 valence electrons, and can accommodate only three pairs of bonding electrons. (3) One Cl forms a double bond with an adjacent Cl. (4) The molecule only has 4 atoms. The octet rule dictates it should have 8 atoms. 9. Arrange the following three molecules in order of increasing bend polarity: F2, SCig, C134. 1:2 (c CF. < set: _/ - .._ . (2?) t7 5'5 ‘ r r “— C“ 5.,"- l ' (2 133<8c313<cr21 (lb/1’ 2, J O (3) SC12<CF4<F2 45/1/55,“ :7. CM“ ”'4 (4) CE; < F3 < SC]; ' ,...»- ' A W a, F 2 it 5 ® I0. What is the hybridization of chiorine in C103_ ? “b { i 1) S 2 . l (2 3:3d 3 Ema/:1) fwrfj / l 7 € (3))593 l ;,,g:1)3:.27#'6’””j (4) szd ‘W l 1‘ What type of'atomic orbitals of the central atom mix to form hybrid orbitals in SFg ? F I: (1) one 5, three p, and one d .._ l )3 CIQSS one s and three p 4 7< gr; ’7,» g:— I": -' (3); ne 5, threep, and two d '3) l2, 1 three p and three :1 57"" I9 'V H\ x“ 12. many 0 bonds are in the molecule Irortiazt'd‘? //C*—-C\\ eventeen ‘N 0‘0"“ "N—H . \C= / u I l (2) tlnrteen / \ :0: H H ' ' (3 twen -two ” H i L Li L l W (4) ten 13.. What is the hybridization of nitrogen in beplrasate? ii ’i‘ l‘ i 0 “l “*i—t—i-N-i-Ci’ ML”? (2) SP2 14—9: H H {DI—H SP page 2 of 4 14“ What unit is used to express temperature in gas law equations? ‘ (l) Celsius 64m % (2) Fahrenheit ( (3 ‘Kelvin 4) Klingon 15.‘ A sample of SFG has a voiume of‘2.50 L. at 150°C.) If the pressure remains constant, what volume will the gas upy at 75AO°C? O V) 6:2” 638? 2«04 L . Y :3— :"33 0 2'5— ; ._“""""._.a:w (2)1251“. h 12/ 4,11% B'ffi (3) 3D? 1.- fl (4) 2450L V3; "’ 2,0‘-'}L_ 16. Another sample of SFO with a volume of'Z..50].. (and a temperature of l50°C) is compressed from 110 atm to 50 atm 1f the temperature of the gas remains constant, what will the volume of the gas be at 5 .0 atm of pressure? Earl” (1) 50L ' p. P? VIZ-fl La “/ (2) 1.261.. Ply “33¢“ 3 0.10 L {I )( 45:4 =3; V2, @7050 L 2 5- V2» 2 6)}.5 C.) 17“ P ease determine the density of Kr gas at ST?“ 695' {50 t3 .. 6) 3.74m J mad : 224 L. 4 ; C % — ~ 3 (1 “'3- 2, cl 9 c3. (2) 830 g/L- I ”NC l5?” s; ,5 \3 V5“ (3) L75 g/L 5: '7’" "jib-L ( (4) 536 g/L 18.. How many moles of gaseous phosphine (PH3) will occupy 0 OSOOL at STP? l ) (1) 44.8 moles fl Q 05,1, Jr ”€11,“ (2% 0032 moles Y1 140/5 (p 0 7-2 H L 5L; 0 . S O'COZZSWJS ,1 an (9:00.222; Vafigllm (4) l‘i‘é-meies (9,00 “2me 19‘ Aluminum reacts with excess HCl to form aiuminum chloride (AiClg) and H2 gas“ The hydrogen gas was coiiected over waters Exactly 35.8 mL of gas Were collected at 25i0°C1 The pressure was 750 mm Hg (tort). How many moles of Al reacted with the excess acid? (Be sure to first balance the equation) 92923 £9 42 end me We r%/+2. .32 x 10 moles (3) 2.097 x 10'3 moles H70": '2 3- 7"? (4) 699x 10" moles { SO _25 7 Q 0 05 :78;L>-l/l h/CZC’EJ—l>2,¢;8 1V)“ .. VI Al=c9r00 3%; 7% "3%sz page3of4 5/) 2— ((3:32. 7(0th .20. What is the ratio of effusion rates for He and UFfi? (Note: FM of UFG W 35.2) (1 8.86 75—; CD 567 @793 dilaflc’ .7, vflw i ’ (3) L32} Mite- “‘08?!” (4) 87-9 r24: 62/ 21,. The vapor pressure of ethanol is 115 torr at 34. 9°C. The heat of vaporization in 40.5 kJ/mol. Using the Ciausins— Ciapeyron equation, calculate the boiling point of ethanol (vapor pr—essure m 760 torr at boiling). , m ‘— 5W’Wl“ we: - 229—3 ~_.-_ .— warm ;1_ ML ?reb (2) 230°C in Hag" I” T/ T '" 33625 11i| (3) 57°C 5'32! Md? L - (4) 100°C m . ___ . o 22. What is the strongest interparticle force in CCLg? C‘ I )2‘3 7 (I) ionic Forces 7K '2 7i“ Dipole—Dipoielnteraction 7“ /C F. C l 01" | (3) ndon Dispersion Forces (“( ” v Hydrogen Bonding 2K Determine which of the moieeuies form intermoiecuiar H-bonds NF3 "2* (iii mml (3) HC] (4) so2 X 24. Which statement best describes boiling points of L-iCl and H0]? (1) HCI has the higher boiling point because of Hydrogen Bonding K )2 W HCI has the higher boiiing point because of Dipole—Dipole Interaction K (3) fliCi has the higher boiling point because of Ionic Forces / (4) L iCi and HCl have the same boiiing point because of London Dispersion Forces K 25. Choose the correct ordering of the following molecules by surface tension. (1) HO~CH2vCH2~OH < HO~CH2~CH~OH~CH2—OH < HaC—CngCme-i ‘Zél 1-)- ; OH { , 3% (2) Hac—CHg—CH~OH <HO—CH2»CH-OH~CH2-OH <HO~CH2-CH2—OH (OH. 2 (3) HO—CHz—CH~OH«CH2~OH < HO-CHz—CHZ—OH < Hsc—CszCH~OH '@HaC-CH2—CH—OH < HO—CHz—CHg—OH < HO—CHTCHOH—CHz-OH page 4 of 4 ...
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