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Unformatted text preview: Vera5011040 w Exam 1 NSulclifl'e — (53120) 1 This prinlrouL should have 30 quitslions.
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‘llc 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 pointQ
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 listenmat
would be: much? 0. 15 30.10 traced}; 0 5 10 15 20
radius (7.1.59) 1. The elmLieu 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 docgo
{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 ms" 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“: —' EMAu _ 1.05457 x 10'“ J  s
" 2(50 kg)(100ms")
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 010 —~ Exam 1 w SuLcilﬁ'o M (53120) 2 distulna lrfaq from the nucleus correct 2. ’I‘hiisisnol. a valid RDF plot as it must be
zero at both ends of the raxis '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’. ﬂan310nm} 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, 42”... :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; mgmOcorrcct
(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 nonmetals. 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 elemton 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 {halffilled
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, silicon28 conwt 2. 2&3, sulfur28 VasionlMUExaml— 3. i351, 5111001114 4. {33, 5111mm 0. $36, 5015.“. 11 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 silicon28 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— coxred. 5. Bo «l e" —+ Be" Explanation: The one for C as it resultgin a halfﬁll 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 halffull (and so already more stable)
set of orbitsla; 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 1034 “£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 deﬁnes 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 conmt 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 frequenty 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 (mm3g Strikw a metal
cathode, electrons art. knudmd off its sur
face, travel to an anode, and form a current
throw]: a circuit. TWO important ohmnations 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 lama1106p 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 coxred.
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 4Rh.Cl eoncot 5. 11,0 Explanation: Metals and nonmetals react to form ionic
oompoumilt Nonmetalﬁ and nmxmt'tals react
to form corralmt 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 1093 Hz
3. 3.74 x 104* Hz 9. 561000 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 «itch0100 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 maswees: AW E (maﬁmﬂfractiong)
+ (massﬂamcclong)
m (IMHO/f) + (46)(0.3)
: 41.6 023 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 iGGpoims
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 modital. 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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