Exam 2 - Version 092 — Exam 2 This primrout should haw 32...

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Unformatted text preview: Version 092 — Exam 2 - This primrout should haw 32 questions. Multiplochoioe qutstjono may cont-{mm on the next column or page _ find all choices heron: answering. This exam is ONLY for stndwrs in the MW? 23pm soctionl! on: EM! points atom going from 1 Do 4 in order? 1.-1,o,o,1 3” ~11 . l 2. —1,2, 1. o ' s. 1, —1,—1. 1 4.6,0,0,0 Q, 0, 1, —-3 correct 6. M1, 0,1,9 Expiaxmh'on: The formal charge is calcuiated by FC in group #- —(# bonds + # unflw‘roti o“) 'l‘lmo Fostwwua‘owo F‘sz6w>{2+4}m0 FCJm5~{-l+0}ul FC4m6~{i+6}—~:w1 902 EDI} poi'nm Of the following combinations of comizounds : names. the one which is incorrect. is 5.. 503 : sulfur trioxide 2. K28 : dipotassium sulfide correct 3. NagO : sodium amide Vomion 092 w Exam 2 ~ The C in CF; will have 4 RHED’s: three single bonds to Cl ammo. and one lone pair. Neither f" or G violate the octet rule. 000 10.0 points Choose the fomula for the compound man» ganeaofll) alum-ale. 1- Mam: 2. MnGlOg 3- MMCIDzle 4. Mn(C1{))p 5. MnC104 6. M58303): curred: 7. Mn{()]04)g 8. MnClOg Ex Narration: 010 36.0 points Balance the equation ?PBr3+Tl{gO ma?ligi’03+?HBr. using the smallest. possible integem. What is the sum of the coefficients in the balanceci equation? 1.? 2.3 correct: 3. 10 4. 11 5.9 Explanation: A balanced equation has the. Alamo IIUJXI‘ bar of each kind of atom on both siéefi of the equation. We find the number of each kind of atom using equation confident-r. and com- position stoichiometry. For example, we finti \ ‘_ {/(f ‘- -Su2.cli.fl‘e- (53120) 1 4. MC] : lithium chloride 5. N205 : diniixogm pentoxide Explanation: i’rdlxos one}: as di, Lri-, eLc. are used what more than one compouno can be made from the elements involved. This commonly happen; when two or more nonmetal: come together to £01m a compound. Hence pm- ftxes are used for sulfur trimdo and oinin jtoogen penloxidc. When metals bond with nonmetals, typically only one compound can lie made bemoan thorn m pmfixrs are not needed. This is the case with the compound K25, which is properly named potassium sul- Side. 303 10.0 points If the fdlowiog crysmliiae in the me type of structure, wliixfia has the 10mm lalLlcc merry? Explanation: Lowwt; charge liaisiflcs due to larger moo 00¢ 20.0 points (formida- urexeaction MgQSiOr) + 4 1120(3) —-v 2 M5(0H}2(aq} + Sil‘lflg) . flow mud: grams of silanc gas (Six!) is formed if 25.0 g of Mgfiii rm with mom H90? 1.. 0.419 g 2. 0.095 g 3. 10.5 3 correct Sutcfiffo w {53120) 3 there are 5 H alarm; on the reactant side: 2 ll '7 m . m ‘ .l-laborno ‘il‘lngngO {ill The balanced equation is Why + 3&0 we» H3P03 + 31113]- , “dmil’JBrfiHandzommmemh side. 7 sumooel'ficlonm: 14.34. 1 +3fls or: room Which of the following metal ion; has 1.1m momdnstam electron configuralion {Ar-lads? 1. Po” curred: 2. Ca“ 3. Mn” 4.. Cu+ 5. Ni” Explanation: Wriae the dmrtron configurations for the ions llsboé to see thI one matches the outer clearer» configuration given. The Aufbau or» do? of doc-tron filllmz is: is, 25, Tip, 3:, 3p, 63, 3a.. 49, 5s, 4d, 53;, 55, 4;, 5d, 51), etc. 5 orbitalr can hold 2 0105110115,? orbitals 6 electrons, and d orbimh 10 electrons. Non: some Moeptlons do occur in the electron coo- fimzration of atoms became of the stability of oitlm- a full or half-full outermost diorbltal. so you my hood to 303mm. for this by ‘str‘ fling’ an Flatiron Elam the (1': — 1):; 0):le- Wlim electrons are rcmnvod from orbitals in a neutral atom to create a positive ion they are taken in this order: outermost. (highest value of n.) p, outermost. s, then. outermost d, 012 36.0 points In the molecule N0. the antlbondlng moleeu. lax orbitals have more 1-. Nitrogenvlike charazber. correct; Version (392 - Exam 2 — Subcllfi‘e — (53120) 2 4. 9.16.; Explanation: mugs] “ 25.0 g Fin-.2. we calculate the males Mags; present: } mol MggSi ’26.? g Mg-gSi r:- 0326 moi MggSi '! :Tml Mggsi 2 25.0 g M3231 x The balanced equation for the reaction Endl- caba that 1 mol Siiid is producer! for read: mole of MggSi reacted. Reacting 0.326 mole; of Mggsi would produce 0.326 molar: Sin. We amount from male; to grunts: 32.138ll‘14 7 gag-i4 :: 0.325 mol 31H. x r morsm. w £0915 g Sill-1 905 10.0 points Which of the {allowing is a radical? 1. on; 2. BF; 3.0}1; 4. 13:0 unrrect Explanation: Odd valence e“ coum. B 10.0 points CHFg is (1 mo polar than CHI; because Qmore; the. Ge!“ bonds are more polar than e C-[inmda correct; 2.lv&s;t.l1e C—H bond in CH9; is a non—polar bond. 3. km: tine Whahalral 1200mm dooream the polarity of 0-!“ bonds. 4. less; the filter: polar 0F bonds are gym. metrical anti cancel the dlpolc moments. 5. more; the GI! bond in CHF; is 3 non- polarbond. Explanation: Since F is more clecuonegaijve than I, o (l? bond will have a greater dill’ercnce in éoch-onogafivifies that: {I C-l bond. OD? EM} poinlo Whit}: of the following musics exhibits roar (mace? 1. N03“ correct: '2. H20 3. None of the cholera has more than one possible bonding scheme. 4. H95 Explanation: :5: Jo'- l H . In. ” . .N . M £9.” \ 9: IQ.” ‘9: :5: l N . / ., {9/ 39>. (308 10.0 points. Moor constructing the Lewis dot. formula for CF“; , give its clochnic arrangement and molecular olmpc. 1. trizooai bipyramldol, seesaw 2. trigonal planar, trigOnal planar 3. tetrahedral, tetrahedral 4. tetrahedral, angular 5. trigoml bipyramldal. 'llohaped 6. tetrahedral, trigoml pyramidal correct Exp] anafi on: Version 092 w Exam ‘2 ~ Sutcliflo w (5312(3) 4 2. Oxygen-like diameter. Explanation: The bonding molecular orbitals reflect. le more decimal-negative filament. 0 and the am tibonding molmular orbitals ntfloci. the {m écclxonegafivo element; N. 013 10.0 points Which of the following is TRUE about: anti. bowling orbitals ofthe molecular orbital the. cry? 1. Antibondiog orbitals are made from the overlap ofulrpna and pi orbitals. 2. The strongman book between atoms have no electrons in 311th and ing orbitals. 3. Alfimugh mjbomling orbitals may ac. oopt. dam-om, the electrons never remain for long. iboncl‘mg orbitals are higher in onorgy than their currlxpmdlng bonding orbitals. commit 5. Unslnrod eioclrons are 1215ch in auti- bonding orbitals. Explanation: Anfihondfinfi orbitals are hifim: in mvxgy .than their corresponding bonding orbitals. Molecules can most. with electrons in 3min bonding orbimls, as loozas the molecule does not place as many electrons in antibondlng orbitalo as there are in bonding oriyltals. on 10.0 points How many carbon atoms are in 1.00 X 1.072 [grams oftrlnltrotoluene, an explosive? 'll‘lniv trotolueue is 07H; N305. 1. 4.22 X :0" atoms 2. 1.86 >< :0“ atoms cur-met: 3. 1.75 x :0” atoms 4. 2.65 x 1523 atoms 5. 4.24 x 10” atoms Explanation: manage. =z: r x {693 Each C7H5N309 molecule contains seven carbon Moms. There are. Avogodro’s num- ber of QfisNaOs moleculw in one mole o£ CTH5N305- We need the molecule mass of (”a-HrNaOa so we can convert. grams of CTHsNSOa to moioo CTHstOg: Molecular mass of ammo. z 7(12.01 glmol) + 50 .01 33/1110!) + 304.0: gfmol) ‘5. 6(16 .00) .—.— 22T.25 g/mol We can use this molecular mass to convert g 0-;H5N305 to 11101 C7HsN305: '1‘ mo] CrlisN305 1 mo] (77H5N305 z 100 N W E G’H‘ 30“ x 227.15 a 073mm. =.: 0.4402 mol C’rHsNiiOo We can now we Avosgodro’s nmnlacr and the ratio of C atoms to C7H5N305 molecules to find the number of carbon atoms: ? atoms 0 w 0.4402 mol Cyl'lfi‘v‘305 x 5.022 x 10“ CyHrNsOs 2 mol mama). 7 atoms C X 1 melee C7HrN305 — 1.556 >4 10M atoms 015 10.0 1:051:18 Whid: of the following is the correct Lewis formula for hypofiuorous acid (HOF)? 1- = ilk-om : Version 092 ~Exam 2 — 5. 6. 7. 3' H-—0%F : 19- Hwéwl? Explanation: The Lewls formula for laypofluorous acid (HOF) is P] _0_p : 016 20.0 points Which one hasthe groom number of atoms? 1. 3.05 moles of argon 2. 3.05 mole-s of CH4 curred; 3. 3.05 moles of helium 4. 3.05 mole; of water 5. All have the same number of atoms Explanation: For 305 moles of water: (3.02 x i023 moloo '? atoms m 3953110ng0 x lmo} 3 atoms lmolecule 225.51 K 10% atoms Pb)- 3.05 melee of (31-14: . 2 33 . r atoms _ 3.05 mol on. x 9M lmol 5am i-molocule . 9.28». 10“ atoms Venr‘lon 092 —E)cam 2 — 3.4l mol C .= 1 me] C 3.4110 3.42 mo W ... 1 mol 0 4.14 mu T m 1.33molH To find the lowest whole number ratio, do termine the lowest number the above can be multiplied by so that all will be whole mum bets. in this caoe the number is 3: lmoleiimlimolC lmoiOX3m3J-n010 1.33am] H x3r-w-3.99 mall-I Therefore our lower-t wholenumber ratio giver: us His empirical formula angHqu. 021 10.0 palms If a molecule has square plamr molecular ge~ ometry, what must be its hybridlwtlon? 13. war? «Au-rod Explanation: A molecule with square planar mole-mic: geometry has oclaboclrai electronic geometry Whisk ooororpontio to mad? hybridization. 022 10.0 points Whlch Mateo-rant is FALSE about 03? 1. lt is parmnaguetic. cox-moi: 2. 1:: would be expected to ham: 3 slaw-Der bond length than C!" . 3. It 113m four electrons in wry, orbitals. 4. It has a bond order 01'2. Explanation: SutcliEe-u (53120) m For 3.05 molar of helimm 6.02 x :0” atoms ? atoms no 3.05 mol lie x 1 mo] $1.34 x 10?" atoms For 3.5 moles of argon: 5.02 x i023 awn-rs ? atoms —,—. 3.05 mol A:- x 1 mol 2» 1.84 x 10““ atoms 017 10.0 points Comidor the water molecule. The i‘l-OvH bond angle is " little less than the M‘trahodraé mg}. - lone palm are more repulsive than 2. 90° bonanza: the electron dot structure is drawn that way. 3. a little greater than the tetraheérfi :11:- glu became lone pairs are not as repulsive as bonding pairs. 4. 180° because the twa pairs of bonding electronsan to get. as far apart aspomlblu 5. aqua! to themMcdml Wedlovfi". Explanation: 018 10.0 points What is the percentage of carbon by weight in table mgar {ngflnogll'e‘ 1. Not enough information is given. 2. 3.5% 3. 48.5% 4. 411% cox-red: 6. 51.5% Explanah‘un: .. part. Percent Whol x 100% Sutdil't‘e .. (53120) 7 023 10.6 prim Consider the diatomic moieculm CD, which hag a bond order of 3, and ()2, whirl: has a. bond ardot of 2. One can state Elm 02 would be (easier, ham-tier) to break apart than C0. and would also have (longer, more. Shorter) bonds than CO. L easier; shorter 2. easier; longer correct 3. harder; more 4. harder; shorter 5. easier; more 6. border; longer Explanation: The larger the bond order, the more Ruble the molmule. This corresponds to mmngor and ehorter bonds. 021 10.0 points The electronic geometry of SnClg' is I. triganal planar. 2. linear. 3. tetrahedral. 4. octahedral. E. trigonal bipyramidal. correct Explanation: 025 {(3.0 points What is the shape of CS?“ 7 Lsoesaw 2. trigonal planar correct 3. T—shAped Version 032 - Exam 2 - In this case, the part would be the mass of C in A certain mass 01' Gui-1220“ {the whole). The simplest amount of 0121122011 to use is the mass of one mole ofcnl’an“. Eacbmnle CICuHmO” conL-alru 12 mol at C, 22 mol 0”], and 11 mol ofO. Welmowtlic atomic mm of each of these elements from the pm-lodio table. Ufingtheso mile moses we calculate the grams of each element in one mole of Cul‘luOu: 12.010? p; (I f (I :.-= 12 5; rom molC >< lrnol C == 144.1281; C , - 1.0mm g u gftoml‘l -—22moll-l x lmoll-l .. 22.1741 [5 H , and 15.999450 gli-omOmllmole imolO n ”5.9933 0. To find the mass of one mole oregano” we add the masses of the component parts: molar mm 012322011 m 114.128 3'. + 22.1147 {:4- 175,993 g 3 342.295 g Cial’lggOu mul 01232202: some percentage ofC‘ in (312113901; is 144423;;0 mum. - (53:20) a 020 10.0 point.“- A sample of vitamin C was analyzed and found to contain 40.9% carbon. 4.58% hydro» aw. and 55.5% oxygen. Whale the empirical formula for vitamin C? I. (333403 correct 2. C2840: 3. CHO 4. Cgl‘lgo Explanation: % C m 40.9% % 0 m 54.5% The empirical formfla ls thefimplest whole number ratio ofatoms present in themolcculc. The ratio of atumaln a compound is thesamc as the ratio of moles. of atoms in a molecule of the compound. The compound is 40.9% C. 54.5% 0 and 4.58% H by mass. These peroentagw are true no matter how large or small our sample of the compound. Aluuming we have a 100 5 mph of the compound, our sample would contain 40.9 g C, 54.5 g 0, and 4.58 g H. Using; the atomic weight. of each elcmmt, we find the moles of each elmcm % K m 4.53% ‘1 Cam—m..— prwmtin 100 goftho sample: 95 342.293 a cannon " 100% =1 42.106333. l mu] 0 ? m . ..._...__ rnolC I509g0>< 12-011 EC 019 10.0 points m 3.41 mol C Which oompolmd in likely to haired low melh in“ ml? 1 1 0 m“ ?molO=’vl.5 0 —-——~— Lcnlciuo: chloride ° 3 x 15.999 3 o m 3.42 mo] 0 2. None oftluzse molomicr. it: likely to have a low melting point. ?moll~1 ”4.53;; H..- giggilili. 3. magnesium oxide ' g z 4.54 mol 3 4' sull'ur moflde correct We now have a mole ratio, but wc want the . lowest. whole-minim: ratio. gividing each of 5' b‘uhm fluoride the above moles by the lowest number guar- Exptmmtinn: unites us at toast one whole number: Version 092 — Exam 2 - Sutciiffe .. (53120) 8 4. mahedral 5’ I?!“ f 5.Mgoml pyramidal H—(II—g EC ”H Explanation: H b There are three rogo'om of electron [lonely (Witlxno lune pairs) around the central atom: a. C if \.91- 028 10.0 palms On the maul Poultry; scale ofeloctronegativl~ ties, the electronegativity o§ Selenium (Se) is 2.4, while that of chlorine (Cl) is 3.0. Barbed on these values we should expect the bOndlng between Se and CI to be i. ionic. 2. wvalmf.‘ and polar with the Cl and of the bond ul'uzlnfly negative. eon-act 3. covalent and nonpolar. 4. covalent auri pole: with the So end of the bond slightly negative. 5. unstable in any eircmmtancm. Explanation: Since the (tlectrouegafivity difference is final! (3.0 — 2.4 a 0.6), we tamed: and this bond Will becovalmt and polar (because the «dectwnegatifity difference it: not eel-o). Cl has a larger declxunegalivily than So, so it will draw electrons toward itself and away from So. Ttu's will make the Cl and of this bond slightly negative. 027 10.0 point-s Carbon ins a valence of four in nearly all ot' iLs compounds and can form aiming and ring}; of C atoms. Consider the propync structure. What hybridizatious Would you expect for the carbon atoms idmfified by a. b, and c. respectively? 1.529. 52>. or" 5/3359 2. 51):. 5?. s2 l/"g 3» £112. 5105'. 5323 ,3 4- ans. so. 3P3 ‘ fi- 893. Av. s1) curredc 0- 3:73. 51>”. $293 7- spa. ml. or“ Explanation: 1i 3 5P 5? EJ 3. ENG —C EC —-il l 'E H 5? 028 10.0 points Choose the species that is incorrectly matched will: electronic geometry about the cent-ml atom. 1. H20 : tetrahedral 2. Fifi ‘. pyramidal correct 3. CE: : [pushedral vi. NB: : tetrahedral 5. BeBm : linear Explanati on: 029 10.0 points Vmion 092 -' Exam 2 wSutdiffcfi (53120) 9 Um electronvdot notaticn to dm'lonstram the formation of ionic compounds involving the filaments Ca and l, L.Ca'+Ca'+Ca'+-I'——+ -. 3." Ca+ + Ca”? + (13+ + - ——> C331 2. Nona ol' {hm 3.61-4- él‘4n-l:+-I=+-l:~—. {I : 4.: II : ——~» Cazia 4.Ca'+-12—-> of + = = m» Ca] 5- Ca3+ Claw Ca=+~1 -+-1 - m. .. .74 _. a». Ca“ +Cng++CaH+3 i : +2} = _.» C3312 6. Ca=+'1=w~a “‘12 C3:+.’]‘:+.I:Mm, Ca“ 4- =1 = w-wCaI Bxplmatiom Ca: give; up two elecmms to form the (3'32" catiomand - I 53cquiwsanelmront9fmm the = I = anion, Two iodine sniom combine with one 03L alum cation to form an electrically neural cfr++=1 : +=1 : "um-$03.12 030 10-0 poinw How many sigma and pi bonds are in the molecule NHQC KOCH2? 1.85'tgzmaml3pi 2.655‘gmaand2pi 3. Seamaandflpi 4.8Wmd2pi correct 5.352311“: and2p': 6.6smaandtlpi Explanation: 031 20.0 point; The ion BeF' has a total of '3 shared dam-om and Z nombomied daemons. 1.2;2 2.438 3. 2;4 4. 6:2 5.4;4 Version 092 M Erma; 2 w Sutcéifiu - (53120) B. 2; Seen-tact 7.4; 2 8.2; 3 9.6;3 3XpEanah-un: Beis one ofthe exocrptions to the octet xule. The to»! umber ofvermce arc-mm 'm Tm: Lewis Dot mucme 5:: [WW 5:] 032 10.0 points How many nan-bonding Electron pairs are amumi I in iF; ’? 1.4 2.0 3.2 4.1 5. 3 correct Explnmtiun: Iodine has seven valence. Plectrum. Flourine has seven valence electrons. Each flourlne atom wili form a covalent 50ml by sharing one of iodine’s electrons. This leaves iodine with 5 electrons 1)le the additional electron that. glam; the molecule a not, negative énargc. That makes-.53): hon-bondim electrons around iodine, or three W .. ,.._ .. wl—F: 10 ...
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