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exam1-key - Exam #1 Name Read the instructions carefully...

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Unformatted text preview: Exam #1 Name Read the instructions carefully and follow them. Show your calculations to receive maximum credit. Significant figures should be correct. You have until 8:50 am to complete this exam. Formulas and constants: T‘OlOl m l i E=hv=h'£ P a 905:3 V _ A —2kze4n:2m l l O E = nzhz 2 20 ' _ 4kezar2m 3 20 A=L > m' v Ll L10 A(m-v)-Ax2—fi— 4m 5 S h=6.63x10'34Js c = 3.00 x108ms'l C? 5 l<=8.99x109JmC'2 me=9.11x10'3lkg =71.60x10"9C Avogadro’s number = 6.02 x 1023 atoms mol'l if 22‘“ l) Geologists find a large stone, and want to determine if it consists of a compound or if it is a \90 mixture. They take several samples of the stone and perform mass spectrometry, obtaining the spectra shown below. 20 8 8 E ‘._.,.i- u “um”.......i.w\w.u..,»..n .. mu"...Mann...“ .i. = “Eu-4 w “...o. fig 3 z: i i "< in ........... .. . .,.,.i...i.-.i i. .._,_.,. .._V..~....,,. .......i.,.v,....v .., SJ _..., w ,. .. _ t... - .v.....m-.,,...v._., ,_.. ) g f g '3' E 1%." ....,,,,..‘ :4 c: 0 20 40 ()0 u 20 4o (,0 3 il ’1 Mass mm is Mass umu Samplel ( ) Sample_2_ H _ ( ) i a) Is the specimen (the entire stone) one pure compound or a mixture? Why? mixture mt Pixéxi erto 6? elements (Fe) part?“ credii’ “<3 éxflamajflo b) How would the spectra differ for the type of matter not chosen in part a? Sketch examples of what the two spectra might look like. 100 O O 80 ,. .. 80 I :1 =3 '1 I! ‘I :1 -i 5| :l w i Relative Abundance (%) Relative Abundance (%) () 20 40 (50 (l 20 - 40 60 Mass (a mu) Mass (am u) Sample 2 2 .mofiS spec, fixati' look ‘Hfieasawrvc - 2 5 5 2) Below is a representation of the ionic compound, KCl, in which spheres of a different shade represent ions of a different element. a) Label one K+ ion and one Cl' ion, and provide a rationale for the labels. ion : \4+ ; Ql’ Small lowx%€_ tom cxplaoal‘lbm " ions \nowf/ same, it Cl: elem/linens » “(Shows 50 await? + K loo 3 more Pull on elec‘l‘mei b) Complete the table: Element of of of rotons neutrons 1:32st 2?“ 3523}? W Number Number Ground state electron configuration electrons 3501' 3) A compound, carbonyl bromide, is composed of 1 C atom, '1 O atom, and 2 Br atoms. Isoto e - 78.92 80.92 49.3 —‘_ 16.99 b) Calculate the atomic weight of Br from the isotopic data. 78.012 avm 0.507 + 80.012amq'O.L/Q3 3 7oltcl 8“ch c) How much would all of the Br contained in 0.25 mol of carbonyl bromide weigh? 2 Br Q+Omf3/ l COK‘AoOxm/l bromide, ——> 050 m0\ Er /O,ZE) \mol oarEOnyl lov‘omida d) The mass spectrum of carbonyl bromide is shown below (any isotope with less than 2% abundance has been neglected). Note that the compound is not broken down into the separate elements. Why are there three peaks? i" ? ‘JI O .1.” m.“ ..,,._....‘..M,,.. ,,... .N..t.....wt--v..., ,. ,.,., .... MMNWUW, .t. .. Ix) 0 Relative Abundance (%) H) ‘ o r r . . . mo 18?. 134 1,86 I 190 192 194 ; Mass (amu) 6Xp\0tpc<+tons d’ba’b descr{\oed mass spec+rom ‘3‘5060‘ there wem 3 dticgerew+ weCcaSMJTS OR: COW‘\OOY\>/\ \OV‘Omt‘Ae— WQV‘F» mfiduwtflH u 4) Consider a neutral atom in an excited state with the electron configuration 1322522p53p1. a) Which atom is this? Ne, b) Label the energy level diagram below with the name and ionization energy for each ® orbital (in the appropriate box), given the following ionization energies: Ionization eneries for the 5 lowest ener; orbitals of this atom 6.6x10'7J 6.4x10"J 1.2x10‘J 7.3x10'l J 1.4x10" J Energy of the orbital Q) E = 0 (reference) the orbital Fifi»!le— ¢7'3X\O’|QS ’llx 10"93 “ca “0‘33 mono"? Depict the electron configuration of the atom on the energy level diagram above using arrows to indicate the electrons. The highest energy electron depicted above can “relax”, that is become lower in energy, 'by two different transitions. Draw arrows on the energy level diagram above depicting these two transitions. Label the larger energy transition “A” and the lower energy transition “B”. l0 2 Ll e) Give the wavelength of light emitted when the electron undergoes the larger energy transition, transition “A”. E;- —.(uxio"q3- ~ 1.2xio"“‘3 = I. I Mom‘s y : : C9.(o$"loagj‘s -3.00>< {Ogm/s HMO‘m3 : l.Ct KIOWW» “Pull Credll‘ f (:th‘Ono lgu’l‘ corredqtordxhvgmxm 06$ me m f) Draw an arrow 0 the energy level diagram a ove representing ionization from the ls orbital. Label it “C”. he. )2 g) Circle the atomic orbital that the highest energy electron occupies in the excited state electron configuration and write the name of the orbital next to it: - 11) Give the 11 and l quantum number values for the orbital you selected in part g. C Y‘ed 1+ %\ “en ‘1: WPOV) or lo I la l C 058)") qloOV‘e/ bul (Lorre CV in 7 a $5 {gt/led u 5) We have previously discussed the photoelectric effect as evidence for the particle nature of light. This effect is employed in a technique called photoelectron spectroscopy to measure the energies necessary to remove electrons from a specific orbital. Below is a plot of the ls orbital ionization energies versus atomic number. (p. . ,.. g l z 15 Orbital Ionization Energy (J) l l l l l l l i l l l l l l l l l O 20 A v 40 60 Atomic Number Describe the general trend relating atomic number and ionization energy of the ls orbital. " because. 01C the elecjtmns clue apnoions (313+), . M (i iii (A repaired 1'1 GdNi 573 Sc 1 Provide an explanation for this behavior (1-2 sentences). ~0k£3 Q+Omic fill: f) lOfllZOt‘HOA eneroéy T 6) Rank the following elements in order of the number of unpaired electrons, or number of singly occupied orbitals (clearly label your rankings): ut7SA' WV C73 : [Xe] @352 / ‘ ,tuu unison red ho’le - remit 53‘ flicir \r a CC 005 leAClezszés e353 .5 you olsdn‘t ‘ ' ; f‘ Cl N. [flaws/r WWW 5:) timej 5 c) OPloi’lfilS 30 gel elecl‘l‘ows “‘7 3 unpafr‘ee) 5C :CAJUsZ BA’ 8 l umpou‘an {- dam—.38 m_ Emmi, 3E0; an 32: E ES 50:72.58 uESfi 353.52 35:22:26 a aim: an aD was .5.» .59 $qu£0 5:.» 8:: 333°: .EuEBu 2: ho «aeofl uu>=¢mum=2 230 .anE $2: 2: 3522.; £385 5 38.05 92—; 2: .333: 035m 0: 25—. Eco—=29 3m fink—mu ESE—av yé .EB 56;" m_ 32: 3:55 935; 888 $858 .5 52 .E :5an .33 as“. E MD_ZE.U< hmét. mh wodt‘ ch made mw wNKwF ww Mméww hm om.Nw_. ww made. mo mNHmF we wmémw mw mmdmw Ne 6V5 fie vméwr cw 5d: mm Nfio: wm wmdmr hm 3:550 Em Bum @ EmEEoo MDHZ<IPZ<A a5 2: Eu 2: 8ng mi 58 2: m3 9 o wz_._..<kw< s=.=ZOA_Om EDEWE 23313:... >m30mw§ EEZm—Im thwwze EDJ<FZ<P :m E on 5 mm. ANNNV ww AOFNV mm 803 Wm mmdow MW NEON Nw mm.¢oN um mmeN cw hmdmv ah mcdmr wh NNNm.‘ hb mmdmw @h FN.mw_. mh #wdwe Vb mmdmw mh 9‘.th Nb 0 Q szQ SEED-Em... >ZOEFZ< mN._.n_. wm om.oN_. mm omHNF Nm wh—‘NF Hm Ed: om Nwé: av SIN: wv hmfiow bv Nwdor 3.» ENE. mv N099 3a 39 ZOFn—>Mv_ MM wmhdw WM woman mm wméh vm NNméh MM vaN NM mwfimw Hm maimm cm 9.9mm ¢N mmmdm WN mmmdm EN www.mm wN wmmém MN mmm._.m VN Nwm. a w m=> h m; w m> ____> IL Mv ¢m.mm NV mcmNm uv vwmém 0v womdm mm Nofiw Mm www.mm FM on MN Nowflv NN mmm.vv HN whodv GN wmodm ma wvmdm w— mmvfim bu mmon mm vsmdm mm wawN VA Nmmdw mm m§<ZPZmqum zOmOm monvu Nfi ommdw = m 4823 233.5% 5.25: . 0 m 3 $52 0:20? mafia”. :m 9 m mmmzaz 0:20? Mm . 1H N 83w 3 83: a $6.3 w 30.3 5 Ear w EMS m S: \ MH/ NNFo.m v Sumd m 22m: <=> : $5 3 <> 2 <>_ 2 <5 2 3:: 5% S. mosmm—m ._.U<M._.mm< ._<U=2m_mU ZOEKDZmEEOUmd 03:: um mum—9232 LDOMO mammZDZ LDOMO N Zmoomnci E 28... M g m._.zm_>_m._m m_._._. “.0 m._m<._. 059mm; Mg“ 88+ N \=m\58fifiow~mm§3§mfl§ H GOIUEld ...
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This note was uploaded on 10/30/2010 for the course CHM 111 taught by Professor Stanitski during the Fall '08 term at F & M.

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exam1-key - Exam #1 Name Read the instructions carefully...

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