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Exam 1 - Vera-5011040 w Exam 1 NSulclifl'e-—(53120 1 This...

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Unformatted text preview: Vera-5011040 w Exam 1 NSulclifl'e -— (53120) 1 This prinlrouL should have 30 quits-lions. Multiple—choice questions may continue on the next coluum or page — find all damper. before amverlng. THIS EXAM SHOULD ONLYBE TAKEN BY THOSE EN THE 23PM SECTION” 001 10.0 points Which of Lhc {allowing utalxmonts is true m gardlug electrornametlc radiation? 1. Elontmmngfietic radiation with a wave- length of 400 sun has a. frequency that is lawm- than that with a wavelength of 600 nm. 2. Electrwnagne‘l-lc radiation With a wavo length of 609 am has a {requcncy that ls lower than that with a. wavelmgth of 400 run. our- reel: 3. Electromagnetic radiation with a waw length of 600 rim travels faster than that with a wavelength of 400 um. d. Electromagnetic radiation with a wave« length of 400 um travels faster than that with a wavelength of 600 nm. 5. The frequency of electromagnetlo radio» Um: dctmmlnm how fast“. travel“... Explanation: v w; fA, 50A 2: )1; . Higherwavclenmscor respond to lower frequencies, and vice Verso. 002 10.0 points Which element exists as a diatomic molecule in ils most stablcctatc? i . boron 2. hydrogen correct 3. water 4. phmrpliorus 5. sulfur Explanation: H, N, O, 1“, Cl, Br. and I all exist. in nature as diatomic molecules. Note.- Walzer isn’t even an element: it. is a compound. 003 10.0 point-Q Which of the following electron oonfiguratlons would repth a halogen? 1. [3425“ 23.3 2. {Ne} 35? w correct 3. {Ne} 332 4. [Ar] 451 5. mucus? Explanation: Halogms are membm of Group 7, so this means that. they have 7 valoncc écctrous. Flam the noble gas configurations above, (maul «.2 and {M33523} boo. have mm» electrons. But if you chuck your periodic tan ble, you out: Lhal. mangmmo has the config- uration [M13452 and dam not belong to group men, so the only halogen present is [Ne] 352 3175. 004 10.0 points Carbide: the radial distribution function (EDP) plot shown below. which listen-mat would be: much? 0. 15 30.10 traced}; 0 5 10 15 20 radius (7.1.59) 1. The elm-Lieu can Somullfllfii be found at Version 040 —- Exam 1 "Subcllfl'o ~- (53120) 3 greater electron repxflslens in the outer or- bitals, thus expanding the electron cloud. correct 3. 19" has one loss clearer: $0 them is. more effective ohiclding of the nuclear drargc. 4. F' has one more electron. which causes loam electron repulsion]: in the outer orbitals, thus mcpandjug the electron cloud. 5. F“ has one low electron so there is lac: Effective shielding of the nuclear charge. Explanation: Since electrons all have the some doc-go {negafivc), their natural tendency is to move asfur apart from such cum:- astheycau. Since F” has one more electron than F, the proton; are less able to draw the electrons towards the nucleus, so the radius mmda. 009 10.0 points Wlild: of the follmvmg atoms has 13:01:11,395: radius? LMP; 2. Si 3. Br 4. K correct Ii. Li Explmflcn: Halli toad to increase to the lcfi of and doWn the poriodlc table. Aunts a potted, Luwartis the right. lhe number of protons increases, but the num— locr of 5119115 does not. This increased positive charge pulls electrons to the nucleus. There» fore across a period atomic radii dmuaoes. Going down. a. group, electrons in the inner shells shield the positive charge of the nucleus from the outermost ulcctmns. Thurcloro the more allelic you have the largo: the atomic radii will be. in thus problem, K is the.» farthest down and the fmdzwlefl. of all the chOiCBS. 910 10.0 points '1133 (‘1th of gold is 19.3 g/mL. What is the mom: ofu gold nugget which has a volumu of 3.23 ml}? 1. 0.170 g 2. 5.33 g 3. 30.4 g 4. 68.3 3' correct. Explanation: 01] 10.0 points Find theminlmum uncertainty in the postilon of a cannon ball with 3 mm 0150,000 g with an uncertainty in velociw of 100 m-s" 1. l. 2.1.0 x 10‘” m 2. 4.52 x 19*” m 3. 1.05 x 10'“ m comet 4. There is not enoughinformation. 11. 1.05 x 10”“ m Explanah'un: mmmmllgmfiokg Amman.“ ll ESAmApmAzmAv Ft. > A“: —' EMA-u _ 1.05457 x 10'“ J - s " 2(50 kg)(100m-s") -W 1.415457»: 10““ m 012 10.0 points 0f the following elements, the one which is: NOT a. racial Sr. 1. atomic. correct Version 0-10 —~ Exam 1 w SuLcilfi'o M (53120) 2 dist-ulna lrfaq from the nucleus correct 2. ’I‘hiisisnol. a valid RDF plot as it must be zero at both ends of the r-axis '3. ’i‘he mom likely distance to find the cloc- mm from the. nuclws is 155er 4. The ductmn will never be found closer to the mlclcusthan firing 5. This orbital has two sphericul node: Explanation: 905 10.0 polka; An electron in a hydrogen atom could un— demo any of 93m Hashim listed below, by omitting ligfiit. Which transitkm would 33w. light ofthe shortest wavflmgth? Hint: Think Planck, romanber Bohr. You may also want. to think about the Rydberg equation, but it is 1101: memory to do the calcalafioms. Lumflbcnml anamwnma 3.11mdbonz3 mnmdtonmleorrecl’. flan-310nm} Explanation: Photons will: shortur wavelengths have more energy than those with longer wave- !mgtlas. n. x: 1 is the lowest energy level, the chem worm as :1 increases. A tranldllon from n m It to n m 1 will release more energy than any other option ijm. 006 10.0 points An electron occuphs an atomic orbital that has the shape 00. All you can say about its four quantum mmbcns is 1.na20r;1iglaer;£ml;mem ~1, (3, +2; 111. Fur-$5. correct 2. 1’: c2.- 2 or higher; 3 m 2; m, "—'= “2, A1, 0, +1, 4-2”... :9: ll;- 3.n:1;£T1;m(:.~1,0,+1; ”um-“dig. l 4.nm0;€=0;mm0;m,m§;-§, 5.n~—vlcr20rhighnr;€ 1; m; r: —-1, 0, +3971, 2 i9” Explanation: Since 00 is the shape of u p orbital, .8 must be i. l canonly 1:0 1 when n a 2 or more, and the potsiblc values of m; are + 1, G, and w 1 when E is I. 00'1” 10.0 points An electron in a 3d orbital could have which oldie following quantmn moors? int-:3; em 1; rub-=22 2.n=3;£:.«3;m‘=:1 -3.nm3;ém0;m¢=0 4.11:3; 3:2;1mmw3 5.n=3; £12; mgm-Ocorrcct (Smooth £m2; mgm 7.n.=3;ex1;m.mw1 Explanation: 3 wafers to tlm principal quantum number n. d corrcnponds to tho subS'rdlnry quantum munbor 3 a 2. Since 3 ms 2, rm could be ~2, —-1, 0, lorfl. 008 10.0 points l‘“ is bigger than F because 1. P“ has one more electron so there is less talkative shielding of thenuclm charge. 2. l?“ lmo one more electron which caused Version 040 w Exam 1 - Sutclil'fc— (53120) 4 5. 3 KJ' ion. 2. mm . p Explanation: 3. cobalt. 015 EDI) points 4' iron. The first ionization potmtiol of the elements B, C, and N (atomic: mmbcrs 5. 6, and 7) 1‘5. silver. steadily mcmfism, but that of 0 $51953 than that of N. The lam. interpretation of the lower Explanation: value for O is that Iron, cobalt, silver. and potacslum are all metals since they are Do the Iofi. of the line dividing metals and non-metals. Arsenic is the only clemom; to the right of that. line. 013 10.0 points Whid: Went concerning the electrons WWW the nucleus of an atom is NOT correct? 1. When 2 electrons palr together. they have opposite Spins. 2. The second main mum level «so: have any number of viectrons up to ton. correct. 3. The number of cloctrons in an isolated ncuual atom is equal to the atomic uumbu. 4. No more than 2 electrons can occupy a single orbital. 5. The attraction of the electrons by the nucleus helm determine thesmc ofthe atom. Explanation: 014 10.0 poinfl Tho electron ooufimation 15a 223 21'»5 32 312% Guild rcprocenr. I. a neutral atom of Ar. 2. norm of the ions or atoms mentioned. 3. a Cl” ion. 4. any of the ions or atoms mentioned. our- ml: 1. the elem-ton removed from 0 commends to a different value of the quanttm umber 3 than that affine electron removed from B. G. or N. 2. the halfifillccl set of? orbitals in N makes it, more dilficult to remove an electron from N than from 0. correct 3. the ionization potential of N is :5 maxiu mum and the values decrease steadily for the elements 0, F, and No. 4. there 2:.- more shielding of the nuclear chargeino thaninB,C, or N. 5. the electron removed from 0 is farther from the nudcus and lheccfiure [ms tightly bound than that in N. Explanation: The ionization potential is a measure of the ease with which electrons are loot by an atom. The ionization potential mcmm from left to right across; the table. The nutm- doctronic commutation of N atom is 2322593 {half-filled orbital), which fives it extra «ability. This inu‘cams the summit of energy needed to ro move the first electron frum N as compared to O. mo 10.8130iuts Choose: the correct nuclear symbol and by [alien notation for the isotope which has a mass number of 28 and atomic. number of 14. 1. fiSi, silicon-28 con-wt 2. 2&3, sulfur-28 VasionlMU-Examl— 3. i351, 5111001114 4. {33, 5111mm 0. $36, 5015.“. 1-1 a. {39, phosphom 14 7. $331. salmon—14 s. 1&1), phosphum :4 9. 2113?, hosylwrus- 28 Explanation: mass rumba m 15.8 atomicnumber m 14 melon! symb 01 and hyphen notation for given unciide .., '1‘ atomic malice! x umber of protons element with 14 protons: slllcon (Si) nuclear symbol m... LESS hyphen notation ——-1 silicon-28 017 10.0 poinls For whichof those elements would 2310910:an all'mity process shown he more favorable? z. N + a" -o N" 2. He -i~ c" —-a He" 3. Kr + e" «m Kt" 4. C + e‘ —-t C— cox-red. 5. Bo «l- e" —+ Be" Explanation: The one for C as it resultgin a half-fill set of 29 orbitals. All oftbcollzer gunmen either require the electron to be placed into a new orbital 0! higher energy, or to be placed into an already half-full (and so already more stable) set of orbits-la; either of which in nnl'avorablc. 018 10.0 points How many protons are present in one Ca:+ ion? anlon 040 -- Exam 1 M 2. 2.0818 x 10'37 m 3. 1.47244 :4 10“27 [:1 Come 4. 2.0313 x104" m Explanation: 0.001 kg 13 m m 0.000% g x =2 4.5 x 10"7 kg ,1 .5 E z .33.. 1) ”1'1: 6.020 x 10-34 “£21 a (4.5): 10-?1ch1 m/s) 2 1.117(2er x 1047:» 025 10.0 points Convert 586 ounces to ldlogram. 1. 0.9602113, 2.15.5kgmmce 3.4250 kg 4. 4.25 x 10°10 5.80.7113 0. 207001cg 1.1.05 x 1071“: 0.002073“; 9. 3.07 x 10"“ kg Explanatioml lb 1: 1 I: 5360mm x 1600' x 129511; -.- 16.61 kg 026 10.0 paints According to the “mulls of Einntm'n‘n photo- electric effect experiment. 1. than is a threshold frequency; at higher SMtdm‘e» (53120) 5 1. 22 2. 21 3. 19 4. 2D carter}: 5.18 Explanation: The atomic number of an clemmt is equal to the number of protons in that element’s Nucleus. The mmbcr of protons define-s mel- erneot. and cannot be changed without mang- ing the element. Gallium has an albumin {rumba 0920. 019 10,0 points If a particle is confined I10 3 modimensioaal box of length 300 pm, for 123'; the particle has zero probability of luring fmmc! m. 1. “find 200nm, respectively. curred 2. 75, 125, 175, and 225 pm, rapoctiwly. 3. 150 pm only. 4. 50 and 250 pm, respmztivuly. 5. 50, 150, and 25012111. reqwclively. Ex planati um :10” 290:300 020 10.0 points What is the con-mt urdm' of decreasing ire- quency'? 1. Intraviolet radhtioiu. visible light, in» frared radiation, radio wavm can-(act 2. infraroti raéinfion, radio waves. visible light, ultraviolet. radiation Sutcliffem (53120) 7 frequency, phntma'icclxom are produced while at lmver frequent-y they are not. correct 2. there is a threshold temperature; at 12531111:- tampmamres, photoolmlrm are pro. duced while at lowes- umperamucn they are not. 3. there is 0 thzahold wavdength; at longer wavelengths, photoelocuom are pro tiueed while at shorter wavelengths they are not. 4. there is a threshold Interm'ty; at higher intensity, phoweiextrom are produced while at lower inlwmily they are not. Explanatium When almimmagndic tadiaLicn (light) of sull‘xdent minimum (mm-3g Strikw a metal cathode, electrons art.- knudmd off its sur- face, travel to an anode, and form a current throw]: a circuit. TWO important ohm-nations were: 1)Electr0us can he (1de only if the light is sulficimtly energetic. Electron election 1.». independent of Lime or intensity. The mini- mnme'nergy varies by element. 2) The current increases with incrmsing inbumity and is independent. of color. Therefore, light lama-1106p 05110010115, and: having a. partic‘flaramomt ofenetgy thatcan be tranfl'erred to an eioct‘mn during a 001% Elm. If the energy is equal to or granul- than tlw. amount needed to liberate the electron it can escape to join the photoelectric unmet. Eulezfily is the {number of photons lfitting asurfanc per unit time. 027’ 10.0 palms A plmmn 1m a. wavelength of 532 mm. What is the frequency of this photon? 1. 5.61: x 10” Hz cox-red. 2.1.5 x 10“ Hz 3. 3.53 x 10‘“ Hz 4. 5.54 x 1015110 Version (MOMExamE .. 3. radio waves, visiblellght, ultravioletmdb alien, infiaroci radiation 4. radio waves, infrared radiatim, visible light, ultraviolet radiation 5. vim‘hle light, ultraviolet radiation, in- l‘rareci radiation. radio waves Explanation: ultraviolet radiation > visible light > 311me radiation > radio waves 021 101?th Which one of the following pairs of elements would he migrated to form an ionic com. puumi? 1. A13, B: 2.3, F 3.C,l 4-Rh.Cl eon-cot 5. 11,0 Explanation: Metals and nonmetals react to form ionic oompoumilt Nonmetalfi and nmxmt'tals react to form corral-mt 00121110112103. Rb is a metal and Cl is; :1 nomnezaf, so they will form an ionic compound. S, F, H. C. As. Br, and l are all nonmetalz. 032 10.0 points Suppose that ahypotlaeticai element «mists of a mixture of two isotopes. One isotopic, hafinga mass of 44.0, ispresenl: in 70 percent atomic abundance, while the other isotope, hm a mass of 45.0, account: for the other mpaemt. W130i; shouldwc expect. the wer— imonzally detmmined atomic wing“. far this lxypotlwt‘wal element to be? l.. 46.0 2. 44.6 curred; Version 040 “Exam 1 .. 5. 3m x 10“” 112. 0. 0.000504 Hz 7. 3.74 x 10-93 Hz 3. 3.74 x 104* Hz 9. 56-1000 Hz Explanation: A L- 532 mm The equation that connects frequency v to waveth )1 is n f; u —-- A whwacinllxe Speed ufljmin amllm (3.00 x 103 m/s}. anem'be? that Hz is equal to inverse seconds {9“} and that 1 nm iseqml to I x 10"” m. 3.06) x loam/s . x 10‘” m ‘53? “WT...“ an 5.6391 x 10” Hz v: 028 10.0mm13 What is the «itch-0100 mfigmatlon of yt- trium (We! Y)? 1. 1:” 2s! 2p“ 352 3125 MW 452 «p3 552 spl 2. is? 2;? 2p” 352 39“ 3.1!“ as? up“ 5:3 3.1.2 25’ 21.5 352 3p“ ad” 459 4,15 4.13 d. 252 23: 27,9 332 3‘95 49’ 4:35 3d“) 33 5. 1s? 2s? 21.3 3.? 3:05 a.“ 301° 4;.“ as? 4.11 correct Explanation: 029 10.0 points The ground state electron configuration of the sliver atom is [Kr] 55‘ dim insteaé of fKr} 552 4d“. The observation can be explained (theoretically) by the fact that 301.0%» (53120) 0 3. 44.0 4. 45.4 5. 45.0 Explanatiom Atomic weight in the waiglileé average of the isotope mas-wees: AW E (mafimflfractiong) + (massflamcclong) m (IMHO/f) + (46)(0.3) : 4-1.6 023 10.0 with: The following three statements refer to the Bohr theory of the atom. Z?) An electron can remain Sn a particular orbit as long as is: confinually absorbs radiation 0T1: definite frequency. Z2) Tllelowest energy orbits are those closed. to the nucleus. 23} An doctroncanjumpfiwn Minuet orbit to an cum orbit by «001010; radiation of 3 dm‘ frequency. Which raponsc «0an all the stammts that are confistent with the Bohr theory 0f the atom and no others? 1. 32 only correct 2. 23 only 3. Z2 and 33 only 4. 21 and 22 only 6. 21,22 and Z3 Explanatiom 024 iG-Gpoims What is the tie Broglie wavelwgth of a flea midwew through its jump? Amme the mass offlm ficais 4.50 x 10'“ g and it is moving at. 1.00 m/s. 1. 1.4725 x 10‘JG m Suttlifia - (53120} 8 1. the 4d subway laurel always has 10 elec.» trons. 2. Llae dd mbmwgy lave! has higher energy than the 5.9- Ku'behflg'y level. 3. an mllancad stability is mod-ital. with filled sets of equivalmt orbitals. correct 4. only one electron can: occupy a 55 or- him]. 5. the magnetism measureme‘m; shows one unpaired electron. Explanation: 030 10.0pOints How many 12 electrons does Se (atomic num ber 34) FW? 1. 10 ‘2. 2% 3. 12 4. 5 5. 0 6. I1 7. 264301"th Explanation: Se has 6 electrons in the 2p and 3p orbitals plus 4 more in the dip orhlml. This gives it a. mini of 15 p ulmtmm. ...
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