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Unformatted text preview: Vomicn 015 "Exam 3 m Suwliﬁ'e  (53120) i This printout should have 30 questions.
Mlﬂtiplodmoioe questions may continue on
the ncxt {2011111113 or page w ﬁnd all choij bnfure answering. This exam IS ONLYtobcLallm bysmdmls
in the 2~3pm MWF class. DO NOT START
THIS EXAM if you do NOT attend the 2»
313m MWF class. AVOGADRO‘S NUMBER 6.022 x 10“ 001 $0.0 points
Name Llw compound H4310; {H45i04 is
sllicio acid.) i. hydrogen ﬁlicale 2. hydrogen persilicate 3. polillicit: add 4. hyposillcous acid 5. hypocihcic acid 6. hydrogen silicon oxide 7. ailicoua acid correct 8. sll'zclc acid
Explanation: 002 10.0 points How much Cl’lgO is medal 1.0 prepare 445 mi.
of a 2.65 M solution of CHgO? 1. 35.4 3 correct ‘2. 79.6 g
a. 17.5 g
4. 179 g
5. N01, enough information is given. 6. 45.63; 7. 0.03%.; Ex‘plaualiiom MmaGﬁM Vrnd'lﬁmbaﬁllﬁéb 2.65 Anal Cﬁgo ’1’ :2: .
.gCl‘lzo (0445L)>< mm x 30g who 1 mol
m 353775 3 00's 10.0 points A metal bull: in ﬁlled with 10.2 g; of CH0]:
and 1.67 g of CH4 (two gases). What is the
total pressure in the bulb if the temperature
is 345°C and the volumeis 50.0 mL? 1. 107 am 2. mm) mm 3. 12.1 alsm 4. There is no pressure betawe the metal
bulb exerts more force than the gas dons. 5. 1450 atm a. 193 atm correct 7. 0.1mm.
Explanation:
ram $102 WEEImWOOESGMOl
Ch" '311953— '
11201
new mUS'Fg EG—gmalmmol W :2 0. i898 11161 V 50 ml. m 0.05 l.
T ﬂ 345°C+ 273.15 ‘7: 618.15 K Applying the ideal gas law equation, P V m 11R?"
P : nﬁT
p £9i§§§llﬁ§i¥ﬂ§llm
0.05
"'!'~ 192.554 atm 004 E01} points
if 3.16 liters of ammonia react with an excess
of oxygen at 500°C and 7 31m lame, 4 mug) 4. 5029;} _. 4 N0(1z)+ a H90(g) Vmoion 015 w Exam 3 wSutcliﬂ‘cu (53120) 3
15 X11010: Which of (in; following is expected a. bail at
0549213 “101 GEES K 2 mol Calls the higirmst tempoalum?
3: 4.11009 mol 02,
which is more than what is actually present. ]. CPL; Therefore Lhalinﬁljngrwmtant must boon.
The molecular weight of C02 is 44.9095
gfmol.
The liminng reactant (02) will yield
2.3376: mo] C04; whirl; is equal to 127.1 g of 002. :2 i G
1:20 02
6.549213m03 cans x m 44.0095 3 CO, _
x 13310100: “‘ m” ‘5 CO“ 007 10.0 points
A mixture of (3.75 mol H2 gas :md 0.75 mol
N2 gas is introduced into a 15 liters container
havinga pinhole leak at 30°C. Amer .1. period
of time. which of the following is true? 1. The partlal prmuc of N2 exceeds that of
['[g in the container. wineat 2. The partial prcmure of Hg exceeds that of
N2 in the containun. 3. The partial prmuw of bath gases in~
crease above their initial values. 4. The partial pressures of the two gm
clmzzge, but Ill2y remaln equal. Explanation:
X“, :3: 0.79 am. 1 am
Because of the pinhole leak, the con
talnm will become equillbmted with the al.
mos’pherc. Air 2's composed of 753% N2, 20%
On, and 1% other 59:595. Evmtually the gases
in the container will have the 5mm ratios w
:13! and the prwsmc will be 1 mm. X14, a 0.79
pm X“, am. 0.79 (1 atm)
m 0.79 am. The partial presume of N2 (murals all other
gases. 008 10.0 points 2. Cgl‘lm correct
3. C4Hw 4. Calla 5. Czl'la Explanaﬁon: The boiling point tends to inﬁrm with
increasing molomlar weight. 05H” has the
biggest MW, 50 it. should boil at the highest
temperature. 009 10.0 points
Carbon dioxide forms crystals best described
86; 1. Norm afﬁrm 2. metallic. 3. molecular. correct
4. ionic. 5. covalent. Explanation: CO: is a comical molecule. Any Wed
lattice will be held together by intermolemﬂar
forces: Hvbonds, London Fm, and dipole
djpole interactions. Cowlnnl. bonds am not
found bmvoem CO; molecules, only mm
the molecule; therefore, the crystal will be
molecular. 010 39.0 points
CLUE: All three C atoms in lit)le structures
are in a line: CCC
Glycerol (0311301) has a lﬁgher viscosity
than proparrot (CallaO). 1. Thu. correct Version 015  Exam 3 w what volume of steam will br: produced at tho
same temperature and lame? I. 41341. correct
2. 3.555 I.
3. 7.1 l. L
4. 2.37 L
5.. 5.925 L
Explanation:
’1‘ r: 500C : W3 K Vmg, m 3.16 L
L  atm
P ~T_ aim R .. 0.08206 mmmam'mﬁ
PV 2 n RT
n ~ H
W RT (7 atm)(3.16 L} m a In
"m (0.08206 m?) (m K)
3 0.348?18 mo] NH; 6 mol H10 who 2. 0.348718 mul NH3 x W t: 0.52307?“ moi K30 RT
Vino ==«" 335$?”
__ (0.523077 mol}(773 K) Talm
x 0.03206 L ' all!) ma]  K.
:2 4.71 L OR A 61.1120
5:50 LNH: " 4 LHBO This. can ONLY be done if both substances
are gases. our. 10.090in15
which 335 would you expect to have the Vemion 015 ~Exam 3 — 2. 19.1339
Explanation:
'llmreis more ﬁulxmdlng inglycorol. 011 30.0points
Which of the following summers is NOT a olrom; elmtrulylc?
1. Mg<N03h
2. NH‘C‘.
3. HNOs
4. HCIOQ 5. N33 correct Explanation: A 51mm; electrolyte completely ionizos in
water. All of the substances are electrolytes
but. Mi; is only partially ionized in water. 013 10.0 points
Write thebalamed molemalar equation for the
reaction of barium hydroxide with perobloric
acid. What is the Deﬁcient of H20?
1. three
2. ﬁve
3. two corroct 4. four 5. one Explanation:
Moll» 4v 2 P103104 wr
13.461009 + 2 HgO(€) 013 10.0 points
Real gases behave moot nearly Eikeidcal gases
al. 1. high pzw‘mms and low molar masses. 2. low temperatures and high pressma. swim A (53220) 2 largmt value for the van dot Waals constant
can? 1. Ho
2. No
3. (1114 d. NHs correct
Explannﬁom 096 10.0 points
For the reaction mm. 4. r0. m. 700. 4. r 14.0 $2.9 yarns of Calls are nﬁxal with 115.5
warns ong and allowed to react. How mud;
CO; could be pruducml by this reaction? 1 . :01] g 2. 245.1 g 3. 90.79 g
4,153.9 g
a. 3971;
0. 169 g
7. 265.9 g, 8. 127.1 gcorrect' 9,111.3 g 10. 249.2 g
Explanation:
maﬁa : on 3
“flat: balanced oquatlon is
2 can. + 15 on m 12 C09 + 6 mo. The molecular weight of Calls is "(8.1113
g/mol, mm; 0.549213 moi Cells. The molecular weight. of ()2 is 31.9938
g/mol, giving 3.60951 mo! 02. 1.1101 115.5 g Sutcliﬂ’ew {53129) 4 3. high temperatures and low prmes.
correct 4. high temperatures and high prawnes, 5. law temperatms and low pressures. Explanation: At. big: temperaturm the gas moltzmxlcs are
moving more rapidly and the effects of the
attractive forces are less sigmﬁcant. At low WW tho molawlw No on nvwage mo?)
lumbar apart and the effects of thus attractive forcel: am lost; signiﬁcant Whose than: are
fewer ‘clooe mcmmlm’. 014 10.0 points
For the reaction 2 H01 + NnnCOs —+ 2NaC] + H20 4 002 what volume of 0.45 M HCl solution would be
needed to produce 2.5 L of C709 at STE”? 1. 2.2 x 102 1111..
2. 4.5 x 10‘ mL
3. 7.5 x :09 mL
4. 2.5 x :0” ml.
5. 5.0 K 10“ ml. correct 6.1.2 x :92 ml. Explanation:
mum 0.45 mol 7
CO x 1311101002 2molllCl
“ ' 2 22.4mm " lmnl (:02
1 L 1000 ml. Xmas. mo: HG] ” 1 L
= 495.032 mL. 015 10.0 points
All of the following statememo. mccepL om,
are important postulate; of the kinetic»
molecular Ihuory of ideal gm. Which one is
not a part of tha kinetic molecular theory? V 2.5 Lof C02 1. The time during which a collision 2m Version 015 — Exam 3 ~Sutch'll‘eu (5312(3) 5 Moon two melonqu scam: is nedirp‘bly short
compared to the Limo berm2m collislom. 2. The average kinetic energy of the
moleculen i5 ﬂavorrely proportional to 9.1m ab.
solute tremﬁerature. curred: 3. Cases consist oflarge xuzmbers ofpartielos
in rapid random motion. 4. There are no attractive nor repulsive
forces between the individual moleculw. 5. The volume of the molecules of a gas is
very mall compared to the total volume in
which the gas is contained. Explanation: The average kinetic; mergy of gas molecules
is DlREC'i'LY (not indirectly) proportional
to the absolute temperature. Aa temperature
increases, so does kinetic energy. (313 10.0 points
Sag X has a larger value than Gris Y for the
van der Waals constant "1)". This indicates
that 1. the molecules; of gas K have a higher
velocity than do the molecules of gas Y. 2. the molecules of X have Stronger intcrv
molecular attractions for each other than the
molecules of V have for each other. 3. the molecule; of X are larger than the
molecules or 'r'. correct 4. the molecules of gas X repel other X
molecules. Explanation: DE? 10.0 points
We observe that 0.14 mole of Clg gas occupies
a volume of 14 L at some temperature and
pmo. What volume would be occupied by
0.14 mole Calla gas at the same temperature
and prme‘! I. 28 L
2. 42 L 3. Thereisnot enough information given to
answer the question. 4. M L correct Explanation: Avogadro‘s Law state; that at the some
temperaturem promire, equal volumes of all
gossamerAmie the some number ofmolecuhs. 013 10.0 points
Write the total ionic equation for th'm reac
tion occun'uag in water: Barium perchlorate
and sodium mrlfatcnroinixcd to form «radium
perchlorate and barium sulfate. 1. 2 113+ + $0; + Ba“ + 2010" w.
21%”? + 2010* + BaSOd 2. 2 Na" i 303' 4 332+ + 2610: w»
2 Net‘qu + Ba" + sci" 3. 2M + so? + a.“ + 2610; _.
2N3+ + 2610; 4 133304 4. 2 red 4 sci“ + mum); m.
2 :w + 2010; + 3350. 5. 2 NJ .1. 503" 43;“ + 2 CEO; _.
2 Na+ ~i 20K); + Ba“ «o 303" rs. 2 913+ + so}? + 1339+ + 2 Clo; ms
2 Na“ + 2 CEO; ~~ Bast); coned: 7. 2 Nﬁ + 2010,; —~> Naomi a. 50?,“ + an” a 3350.
Explanation: 019 £03 potato
Adhmlve forces are forces 1. between the molwuiee of a liquid. 2. between the molecules of the liquid and Version 015 ~ Exam 3 W Suteljll‘e — (53120) 7 yield, we divide the actual yield by the theo—
retical yield: 35 E ‘3on
103.475: Cr)ng
4.5% '? percent yield K 100% 023 10.0 points When 2.83 p; of phosphorus was burned in
chlorme, the product was a phoophorus chlo
ride. [is vapor took 1.7? times as long to
dim as the same amount of (102 under the
same conditions of temperature and pressure.
What is the molar mass of the phosphorus
rillorlde? 1. 138 g/molcorrect
2. 155 g/mol
3. 102 g/mol 4. 37.7 glmol Explanation: Because the phwphorus chloride tool: 1.7?
times as long to cil'usc, the rate 05 9033510)) of
002 must be 2.77 timm that of the plwsphe
rte; compound. Eﬂ’co, 177 Wren: MWco, a 12.01g/rnol +206 g/mol} 44.01 g/mol Eff o: «‘f FEW: 60
Eileen 2 MWPCL.
Wren. cho: ’3 44.0111/le ( = 137.879 g/rnel . 024 10.0 points At eonstant temperature if the volume of a
gar. sample is tripled the pressure will i. be increased by a factor ofﬁ. 2. be one third the original. correct 8. be tripled.
4. remain the me. 5. he cut in half.
Explanatiom
3 V1 :2 V2 Boyle's Law relates the vellum and gum
of is sample of gas: Fri/1 :sza
P1 W. t P1(3V1)
P1 :39. 1
Pz—EPI 025 10.0 points
‘What are the strongwt ﬁrmwnolocular forces
present in a liquid sample of PCI3?
i. (scialarm
2. hydrogen bonding
3. dipoledipole correct
4. dispersion 5. interionie Explanation: 026 10.0 points
At 800°C and 32.0 Torr, the density of mm
phor vapor ls 0.0829g L“. What is: the
molar m of carpilot? 1. 34903 nml"" 2. magmar" Version 615 vExamilw another allface. eoneel. 3. Neither of thwe is correct. Explanation: om 10.0 points
A drop of liquid Limits to have a spherical
shape due to the property of l. vapor pressure.
‘42. close pairing. 3. surfaw melon. correct 4. capillary action. 5. vimo's‘ty. Explanation: Molecule2 0n the surface of a liquid are
inﬂzicnned by intermolecular attractions 61>
warda the interior; these attractions pull the
mfmlayw termd allocater. The moutetav
ble situation is one in which the surface area
is minimal. Fm a given volume, asphere has
the least possible mn'faoe area. 021 10.0 points
Which of the following can form innermolecu
lar hydrogen bonds?
CESCIhCE‘lgCHﬁHn CSECHQOCHQCHg ' chugcwm
\CHgCH2C(0)CH3 aCHaNHa correct
' 1) tion: Only molecules with H attached to the
Electromagnle atoms N, 0, and P can hy
drogen bond. of the molecules ﬁver. only
CHKGHINHQ has the H directly bonded to N,
soil. can undergo hydrogen bonding. 022 10.0 Vendor: 015 —Exam 3 w 3. 243 g1 mar" ‘
4. 152 g  Incl”! correct 5. 20.33 I moi" Explanation: T m 80°C 4 213.:35'K = 353.15 K a m
P m (12 Torr) m 0.0157395 atm
p a 0.0329 31' L The ideal gas law is PvmnRT
n .P "I? t "if?" with unit of measure moi/L on each tide. Multiplying trade by molar mm (MM) giver;
P n
gMMmrﬁ.mmp, with units of g] L. __ pRT
MM w—P W (0.0829 g/L) (0.08206 £51m)
" 0.0151395 atrn
x (353.15 K) «s 152.152 g/mol 027 10.0 points In which of these mpmmds would you ﬁnd ONLY dimer) forces existing between the
molmllt's?
I. cal.“
11. liBr; Ill. C‘Hzcli: IV. C02. 1.11 and lllonly
am only 3. I only 4. [I only 5.1 and In only Sutdiﬂ‘e — (53120} 5
Consul): the reaction
acguﬁo —> cairn + smo. If 15.0 g of Colin is obtained from 150 g of
0314150, what is the percent yield? 1. 4.5% 2. 10% 8‘ 14.595 curred:
4. 32.1% 5. 24% Explanation:
mean: = 1505; memo = 1503
Our ﬁrst step is Lu determine the theoretical yield of the reaollun. The reaction started
with 150 g of Cgﬁaﬂ. We convert from grams
to males using; the molar mass: ’3 mol (331150 m 150:; 031150
1 mol 03H50 x W 2.5826 mol Cgl‘lao Elmira the lidmood equation we see that. 3
moles 03350 are requimé to produce 1 mole
CplIn. We use this mole ratio to calculate
the maximum numbers of moles of C9ng that
could he proéuced: ? mo] Coll 12 a 2.5826 moi Cal‘lso
x 1 mol (193:2
3mol CgHso
a 0.3609 mol Cg,ng We use the molar rnaw of Cgliu to covert
from moles be grams: '1' g Cpll1: : 0.8609 moi Colin
X 32019 8 Cal‘lu
1. mo} Colin
3 103.4? g Cnlllﬂ This is the theoretical yield; the maximum number of grams C9er of that could be po
duced from 150 g 63350. '13; ﬁnd the parent Sumlille  (53520) 8 B. I and IV only correct
1’. II and IV only 8. ill and [V only 9. l and [I only 10. 1V only Explanation: ‘ A nonpolar covalent molecule would have
only dispersion forces with another nonpular
covalent molecule. 028 10.0 points
A 22.4 L Vmsei comm 0.02 mol H; gas,
0.02 mol N gas, and 0.1 mol N113 39:5. The
total prosewe is 700 Mr. \Vhaz is the partial
pressure of the ll; gas? 1. 7 ton 2. 30!} ton correct
3. 28 ten 4. 14 torr 5. None of these Explanation:
mom rm 0.14 moi
m, m 0.09. moi Floral 1'00 tort x“, ﬁle. 2 “'92 1“") a 0.142857 new“ 014 mol " Pm * xii: Pam:
v: (0.142857) (700 Lou)
: 100 tort 029 10.0 pulnw
You have a. and: oolulien mumbling (3.01.2 M
gimme. How many ml. of this solution must
be diluted to 50 mL to prepare a 0.005 M
gluoosesolution? Version 015 A Em 3 ~ 3.1mm». w (53120) 9 I. 12.0mL 2.26.3111}; 3. 20.8 mL coned.
4. It cannot be done. 5. 2.6 mL Explanation:
M 00012 M
V2 m: of) ml. A porléon of the 0.012 M glucose wluﬁon
will be diluted with water to form 50 ml. of
the 0.005 M glucorsc solution. All of the mole:
of glucose in the new dimte saluan must.
come from the more wnccula'amd solution.
(There is no glucose in the water!) We use
the damed volume and molarity of Hm (iiluto
EOluLion to determine the number of mules of
glucose needed to make this wlution‘. 7 Incl glucose a Ell) mLsoln
x 1 Lsoln
1000 ml; 50111
0.005 mol glucose
1 L soln
==r 0.00025 mol glucose We need 13110113}; of 2.3113 0.012 M solution to
pmvidu 0113025 mol glucose. We mm the
molarity to convert from moles b0 volume of
the solution: ? mL soln w." 0.00025 mol glucose
x 1 Lsoln 0.012 moi glucom
1000 mL min 1 L soin 20.8 ML soln 20.8 ml. «ﬁlm 0012 M solution clilumcl 1K)
50 mL with Waller would give us the limited
:milxtion. Mg m 0.005 M x 030 10.0 points
A sample of nitrogen occupies a volume of
546 mL at. STP. Whal. volume would :ﬁmgm
campy at 177‘0 under aprcsxure 01‘380 tort? 1.900mL 2. 150 mL 3. 662 mL 4.179951 mL correct Explanation: P; v: 760 rm 7‘: 0’0 + 273.15 273.15 K
T, = 177°C + 273.15 x 450.15 K P3 m 380 be" Vi m 546 1111..
Using the: Combined Gas Law,
I": V: A P2 V2
T1 m T2
P1 V11":
V2 _ T1 P2 and recalling that. S’I‘P implies standard mm»
paratuxe (0°C or 2?3.15 K) and prmxe (1.
aim or 760 ton), we have V? m (760 ton) {546 mL) {450.15 K) (27335 K) {380 to”)
w: 1799.52 mL ...
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 Fall '07
 Fakhreddine/Lyon
 Mole

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