Sound brains Tutorial.

SoundbrainsTutorial.

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ABSOLUTESOLUTIONSTOTHE2017/2018CHM101MID-SEMESTEREXAMINATION

ByAdebayoMayowaA.(p.k.aICONMAYOR)

CoordinatorofSoundBrainsTutorials(SBT).

PotentTutorofPart1CHMandMTH.

Hellodearfriends,

Iimploreyoutoensureyou’reableto solveallthetestquestionsbefore your examination as somequestionswillsurelyberepeatedword-for-wordwhilesomewillberepeatedwith changedvalues– thereason Ibeseechyounottocramtheanswersbutunderstand how to go about thecalculationsandconceptswell,asthesolutionsherehavebeenwell-delineated

Youcandowelltoreachme@08068486171 via call or viaWhatsAppIfyouencounteranyprobleminanyofthesolutions,orforanyotheracademicreasons.

SUCCESS!…thisisjustthetrailer;thefullserieswillbeabomb.

Theabsoluteoptionsaregivenbelow;checkthenextpageforthe

thoroughsolutionstoeachquestion;

TYPE1

1.A

2.D

3.B

4.A

5.D

6.D

7.D

8.B

9.A

10.C

11.B

12.A

13.D

14.D

15.D

16.B

17.D

18.C

19.C

20.C

21.D

22.D

23.D

24.D

25.D

26.C

27.C

28.C/D

29.B

30.C

31.B

32.D

33.B

34.C

35.D

36.D

37.C

38.C

39.D

40.A/C



1.Whichofthefollowingstatementsgivesthecorrectmolecularinterpretationofthereaction:

Fe2(SO4)3(aq)+BaCl2(aq)→FeCl3(aq)+BaSO4(s)afterithasbeenbalanced?

A.1moleculeofFe2(SO4)3reactswith3moleculesofBaCl2toproduce2moleculesofFeCl3and3moleculesofBaSO4.B.1moleofFe2(SO4)3reactswith3moleculesofBaCl2toproduce2molesofFeCl3and3moleculesofBaSO4.C.2moleculesofFe2(SO4)3reactswith3moleculesofBaCl2toproduce2molesofFeCl3and3moleculesofBaSO4.D.2molesofFe2(SO4)3reactswith3moleculesofBaCl2toproduce2moleculesofFeCl3and3moleculesofBaSO4.

GiventheequationshowingthereactionbetweenIron(III)tetraoxosulphate(VI)andBariumchloride.

Thisequationhastofirstbebalancedinordertomakeitcomplytothelawofconservationofmass:

Onbalancingwehave;Fe2(SO4)3(aq)+3BaCl2(aq)→2FeCl3(aq)+

3BaSO4(s)

consideringtheirmoles,wecouldseethattheFe2(SO4)3combineswithBaCl2intheratio1:3,andtheproductFeCl3andBaSO4

areintheratio2:3.So,themolecularinterpretation

willcorrectlybe;1moleculeofFe2(SO4)3

reactswith3moleculesofBaCl2toproduce2moleculesofFeCl3and3moleculesofBaSO4.

OptionAiscorrect.

2.Youperformedanexperimentinthelabandfoundoutthatthereare36.3inchesinameter.Usingthisexperimentalvalue,howmanymillimetresaretherein1.34feet?

A.43.05x102mmB.4.43x105mmC.

44.3mmD.4.43x102mm

Tosolvethisquestion,itisimportantyouknowtheconversionofthevariousunitsoflength,ifyoudon’ttoendsolvingbroadlyintheexamhall.Hereistheconversion1mile=5280ft=63360inches=1609m=1.609km.wearebeinggiventhatthereare36.3inchesinameter,atthesametimebeingaskedthemillimetresthatin1.34feet.Thisisn’thard,justdoyourconversion.Firstconvertthe1.34feettoinches.i.e5280ft=63360inches1.34feet=

(63360inchesx1.34ft)/5280ft=16.08inches.Soitisnowpellucidthat1.34feet=16.08inches.Thenwecannowsay:Since36.3inches=1mthen16.08inches=(16.08inchesx1meter)/36.3inches=0.443meter.

Aftergettingtheequivalentof1.34feettobe0.443m.We'renowtoexpressouranswerinmillimetres:Don’tforgetmilliis



justaprefixhavingamultipleof10-3,butintroducingitwillmakeanappearanceof

1/10-3.

i.e.0.443m=0.443x1/10-3=0.443x103

=4.43x102mm.OptionDiscorrect.

3.Whentheisoelectronicspecies,K+,Ca2+,

andCl-,arearrangedinorderofincreasingradius,whatisthecorrectorder?

A.K+,Ca2+,Cl-B.Ca2+,K+,Cl-C.K+,Cl-,

Ca2+D.Cl-,Ca2+,K+

Ionicradiiaretobeconsideredherenot

atomicradii,sincethespecies(i.e.Ca2+,k+,

Cl-)providedareionsandnotneutralatoms.Rememberthat;Ionicradiiofanionsarealwayshigherthantheirrespectiveatomicradii,becausetheyacceptelectronstoformtheirions.WhiletheIonicradiiofcationsarealwayslowerthantheirrespectiveatomicradiisinceelectronsarelostforthemtoformions.Analysingtheprovidedspecies;

K+→Neutralpotassiumatompossessfourshells,whereonlyoneelectronisoccupiedinthelastshell.Butinthispotassiumion,oneelectronislost,thereby

theoutermostshell(i.ethe4thshellwhichisalsoregardedasN-shell)isevacuatedsimultaneously;having3shellswith18electronsleft.

Ca2+→Neutralcalciumatompossessfourshells,wheretwoelectronsareoccupiedintheoutermostshell.Butinitsion,thetwoelectronspresentintheoutermostshell(i.e.N-shell)arelost,whichareevacuatedalongsidetheshellasnoelectronisleftintheoutermost(beingaslast)possessing3shellswith18electronsleft.

Cl-→Neutralchlorineatompossessthree

shells,inwhichthe3rd(i.e.outermostshell)possesssevenelectrons(whichremainonetoattainoctetconfiguration.Butinits

ion(i.e.Cl-)oneelectronhasbeenattracted,agglomeratingthenumberofelectronsintheoutermostshell,wecannowsaychlorineion(i.e.Cl)possess3shellswith18electrons…Now,youcanseethattheyareallisoelectronic(i.e.theyallhavethesameelectronicconfiguration.)Thiswillmakeuslayemphasisontheireffectivenuclearchargesincetheycannotbejudgeddirectlyfromtheirelectronicconfiguration,asit’snowvividthatthey’reisoelectronic.Thethreeionshavetheiroutermostelectronslocatedonthesameenergylevel,buttheydonothavethesameionicradius.That,happenedduetohowthenumberofprotonseachspeciespossessinitsnucleusvaries.Now,potassiumhasanatomicnumberof19,has19protonsinitsnucleus.Calciumhasanatomicnumberof20,has20protonsinthenucleus,chlorinehasonly17.

Inotherwords,Ca2+hasbiggereffective

nuclearchargethanK+whileK+hasbigger

effectivenuclearchargethanCl-,whichisproportionaltoabiggernetpositivechargefeltbytheoutermostelectrons.Thiswillcompresstheenergylevelsabitandmaketheionicradiussmallerforcalciumcationthanthatofpotassiumcation,atthesamephenomenonmakingthepotassiumcationsmallerthanthatofthechlorideion.Therefore,Wecannowsaytheincreasing

orderoftheseisoelectronicspeciesisCa2+,

K+,Cl-.OptionBiscorrect.

4. Asamplecontaining33.42gofmetalpalletsispouredintoagraduatedcylinderinitiallycontaining12.7mlofwater,causingthewaterlevelinthecylindertoriseto21.6ml.Calculatethedensityofthemetal.A.3.76g/mlB.3.75g/mlC.8.9g/mlD.



37.5g/ml

Usingtheformuladensity=mass/volumeThemassofthemetalhasbeengivento

be33.42g.Thevolumeofthemetalcanbegottenbyextricatingtheinitialvolumeofthewaterfromthevolumeobtainedwhenthemetalpelletsarepouredintowater.[thewayvolumeofirregularsolidsaregottenunderdensityandupthrustinphysics.]therefore,Volumeofthemetal=21.6ml–12.7ml=8.9ml.Nowdensityofthemetal=33.49/8.9ml=3,75506.But,mindyou,youranswermustberecordedinthelowestsignificantfigureoccurringinthegivendata.Lookingbackatthevaluesbeinggiven(i.e.33.42g,12.7ml,21.6ml),theleastsignificantvalueoccurringis3significantfigure.Thisimpliesthatourfinalanswerhastoberecordedin3s.f.=3.75506=3.76g/ml

OptionAiscorrect.

5.Acarbon-oxygendoublebondinacertainorganicmoleculeabsorbsradiationthat

hasafrequencyof6.0x1012s-1.Whatistheenergyofthisradiationpermoleof

photons(NA=6.02x1023mol-1,c=3.00x

108ms-1,h=6.63x10-34Js)

A.2.00x10-14JB.3.978x10-21JC.7.18x

1011JD.2.4x103J

Energyofaphoton=hf

E=(6.626x10-34)x(6x1012)

E=3.9x10-21JButwe’rebeingaskedtocalculatefortheenergypermole,Ibelieveweknowthatofallconstantbeingprovided,onlyNAcould

beseeninmol-1now,toattain‘Energypermole’we’llhavetomultiplythevalueforenergygottenbytheAvogadro’sconstant,asin;

Energypermole=3.98x10-21Jx6.02x

1023/mole

=2.4x103JOptionDiscorrect.

6.aK4Fe(CN)6+bH2SO4+cH2O→dK2SO4+eFeSO4+f(NH4)2SO4+gCO.Ifthechemicalequationaboveisbalanced,whataretherespectivevaluesofa,b,c,d,e,f,g?A.6,2,3,1,1,6,3B.2,1,6,6,1,6,3C.2,3,4,8,1,6,6D.1,6,6,2,1,3,6Youcandowelltoopentopage123infundamentalchemistrybyAbassOlajire,asthesamequestionisbeingset,andthesolutionbeingwell-delineatedinthismaterial.Although,wehavethreemethodforbalancingchemicalequationswhichare(I)Trialanderrormethod(ii)Mathematicalmethod(iii)half-cell/halfequationmethod.Themethodsuitableforsolvingthisquestionwithoutwastingtimeandatthesametimeavoidinganyformoferroristhemathematicalmethod,whichisthemethodalsobeingusedinthematerialbeingreferredto.Pleasedowelltostudyitwell,asit’spossibletobeseenintheexamexactlyoratypeof.It’llbeseenthatthecorrectoptionisD.

7.DeterminationofthechargeaswellasthemassoftheelectronwasfirstmadebywhichofthefollowingscientistA.J.J.ThomsonB.ErnestRutherfordC.Louisde-BroglieD.RobertMillikan

Thisquestionseemsconfusing,butit’snot.Pleasenotethat“aswellas”isnotthesameas“ratio”,Ibelieveweknowtheaccompanimentruleofconcord.Theinterpretationofthisquestionisthatoutofallthesescientistsintheoptions,whodeterminedthechargeofelectronandalsothemassofanelectron.Analytically,themainvaluegottenbyJ.J.Thomsonis1.76



x1011C/Kgwhichisthechargetomassratioofanelectron(notthespecificchargeormassofanelectron)inhiscathoderaytubeexperiment.WhileRobertMillikancamethroughlaterintothephenomenonanddiscoveredthespecificchargeofan

electrontobe1.6x10-19Cwhichisthemainthinghedeterminedinhisoildropexperiment.Although,noscientistisbeingrecordedtodiscoverthemassofanelectron,butR.AMillikanisrelativelybelievedandregardedtobetheone,sinceheprovidedthelastthingneededforthemassofanelectrontobeknownasfollows;chargetomassratio=charge/mass.Imputingtheirrespectivevalues,we'llthereforehave

1.76x1011C/Kg=1.60x10-19C/Massofelectron.

Massofelectron=1.60x10-19/1.76x1011

Massofelectron=9.1x10-31Kg.

OptionDiscorrect.

8.Anexperimentrequires43.7gofpropan-2-ol,insteadofusingabalance,astudentdispensedtheliquidintoameasuringcylinder.Ifthedensityofpropan-2-olis-

0.785g/cm3,whatvolumeofpropan-2-olshouldbeused?

A.5.567cm3B.55.7cm3C.34.3cm3D.

1.80x10-2cm2

Massofpropan-2-olisgiventobe43.7gthedensityofpropan-2-olisgiventobe

0.785g/cm3

densityofpropan-2-ol=massofpropan-2-ol/volumeofpropan-2-olThereforevolume=mass/density

=43.7g/0.785g/cm3

=55.6688cm3

Don’tforgetthattheanswerhastoberecordedintheleastsignificantfigureoccurringinthedata.(whichis3s.f.)

So,55.6688=55.7cm3.OptionBiscorrect.

9.Anexcitedhydrogenatomemitslightwithawavelengthof397.2nmtoreachtheenergylevelforwhichn=2.Inwhichprincipalquantumnumberdidtheelectron

begin?(RH=1.097x107m-1)A.7B.9C.5D.3

Tosolvethisquestion,don’tforgetthat

hc/λ=R∞[(1/n2f)–(1/n2

i)]issimilitude

to

1/λ=RH[(1/n2f)–(1/n2

i)].

SotheRH=1.097x107provideditisasaresultofR∞/hc,becausecomparingthesetwoformulae,wecandeducethatRH=R∞/hcwhereR∞ isRydbergconstant

(2.179x10-18J),histhePlanck’sconstant

(6.626x10-34Js)

cisthespeedoflight(3x108m/s).imputethesevaluesintotheformulaRH=R∞/hc,

you’llseethatRHistruly1.097x107m-1.WhatI’mtryingtofigureoutisthatanyofthetwoabovecanbeused,becauseIknowtheformerisquitecommonthanthelatter.Butsincewe’veseenthattheyarethesame,wecandowelltousethelatter,sinceit’sfaster.

Now,using1/λ=RH[(1/n2f)–(1/n2

i)]

λ(wavelength)=397.2x10-9m

RH=1.097x107m-1.nf(finalenergylevelreached)=2.

thereforewehave;1/397.2x10-9=1.097

x107[(1/22)–(1/n2i)],

Then1/397.2x10-9=1.097x107[(1/4)–

(1/n2i)],

0.2295=[0.25–1/n2i],

Therefore,1/n2i=0.25–0.2295=

0.0205,

n2i=1/0.0205=49,

√n2i=√49Therefore,ni=7.

OptionAiscorrect.



10.Anatomofaparticularelementistravellingat1%ofthespeedoflight.ThedeBrogliewavelengthisfoundtobe5.784

x10-6nm.Theelementislikelytobewhich

ofthefollowing?(NA=6.02x1023mol-1,c

=3x108ms-1,h=6.63x10-34Js)

A.40CaB.39KC.23NaD.9Be

DeBrogliewavelength?!!!!It’scommonthatdeBrogliewavelength,λ

=h/mcwherehisthePlanck’sconstant

(6.026x10-34Js),cisthespeedoflight(3

x108m/s)andmisthemassoftheatom.Travellingat1%speedoflightmeans,the

coftheatom=1/100x3x108

=3x106ms-1

Let’scomputeforthemass(i.em)firstm=h/λc[fromλ=h/mc]

m=6.626x10-34Js/(5.784x10-6x10-9m

x3x106),

m=3.818x10-26kg.Now,don’tforgetIatomicmassunit=

1.66x10-27kg.

if1a.m.u=1.66x10-27kg

x=3.818x10-26kg.

x=(3.818x10-26kgx1A.M.U)/(1.66x10-

27)x=23amuDon’tforgetthattheatomicmassunitofofanelementisalsoitsmassnumber.Theelementpossessingthemassnumberof23issodium.[usethesamemethodforthesimilarcalculationyou’rebeinggivenasassignment,you’llget40……whichiscalcium)

Theanswerhereissodium,23NaOptionCiscorrect.

11.TheideaofapositivelychargednucleusinthecentreofanatomwasdiscoveredthroughA.cathoderaydischargeexperimentB.alphascatteringexperimentC.atomicemissionlinesD.wave-particledualpropertiesofelectron.

TheideaofapositivelychargedNucleusinthecentreofanatomwasdiscoveredthrough“alphascatteringexperiment”whichisalsoregardedas“gold-foilexperiment”asErnestRutherfordaswellashisstudentsGeigerandMarsdenmadethisdiscoverybybombardinggold-foilwithalphaparticleemittedfromaradioactivesource(likeUranium),wherebyrelativelyallthealphaparticleswerefoundtopenetrateintothegoldfoil,whichmadeErnestRutherfordbeingscientificallyconclusivefromthefulldetailsoftheexperimentthatabout99.95%ofthemassoftheatomisconcentratedinthecentrallyplacednucleus.

OptionBiscorrect.

12.Whichofthefollowingstatementsarecorrectaboutdeterminateerrors?A.TheyareknownassystematicerrorsB.DeterminateerrorsarealsoknownasrandomerrorsC.determinateerrorsareirregularandvariableinmagnitudeandsignD.Theyareduetofluctuationoftemperatures.

Outofalltheoptionsprovidedhere,onlyoptionAiscorrectaboutdeterminateerrors,astheyareknownas‘systematicerrors’.OthercharacteristicsintheoptionbelongtoIndeterminateerrors.

OptionAiscorrect.

13.Listthepropernumberofsignificantfiguresinthefollowingnumbers:0.216;90.7;800.0;0.0670;0.0060A.3;3;4;4;4B.3;3;1;2;2C.3;3;3;3;2D.3;3;4;3;2

Let’sconsiderthemoneaftertheother.I)0.216:3significantfigure[becausealldigitsaresignificantexceptzeroesbeginningnumbers]ii)90.7:3significantfigures[don’tdoubtthezeroinbetween,aszero(s)beingnon-



zerodigitsaresignificant.iii)800.0:4significantfigures[thisoneismostlythoughttobe3significantfiguresduetohowthezeroattherighthandsideofthedecimalpointisadmittedtobeinsignificant,butNO!itissignificant,asthereisarulewhichsays‘terminalzerosattheRHSofdecimalpointsaresignificant.]iv)0.0670:3significantfigures[becausealldigitsaresignificantexceptzerosbeginningnumbers].v)0.0060:2significantfigures[thezerosbeginningnumbersareinsignificant;whiletheterminalzeroattherighthandsideofthedecimalpointissignificant].So,theyhavethenumberofsignificantfiguresas3;3;4;3;2respectively.

OptionDiscorrect.

14.Whichofthefollowingrepresentthesmallestmass?

A.2.3x103µgB.2.3x10-2C.23cgD.0.23mg

Toknowtheonewiththesmallestmass,youhavetoexpressalloftheminjustgramwithoutprefix;

2.3x103µg=2.3x103x10-6g=2.3x10-

3g.

2.3x10-2kg=2.3x10-2x103g=2.3x

101g.

23cg=23x10-2g=2.3x10-¹g.0.23mg=

0.23x10-3=2.3x10-4g.

so,2.3x103µg=0.0023g.2.3x10-2kg=23g.23cg=0.23g.0.23mg=0.00023g.Ibelieveitisnowvividthat0.23mgisthesmallestofall.

OptionDiscorrect.

15.Inagroundstatepatominthegasphase,howmanyelectronshavequantumnumbersn=3,l=1,ml=-1?[Atomicnumberofp=15]A.2B.0C.3D.1

Consideringphosphorusatominitsgroundstateinthegasphase.Itsatomicnumberis15,Let’scheckitselectronicconfiguration

15P=1s22s22p63s23p3.Now,let’sdealwiththedesignationspossessingthequantumnumbers(n)of3

(i.e.3s2and3p3).Let’snowdealwiththedesignationspossessingtheazimuthal

quantumnumber(l)of1.(i.e.3p3,becausethedesignationispwhentheazimuthalnumberis1….Don’tforget(s→0,p→1,d→2,f→3).Now,we’vebeenabletofigureoutthatthedesignationwe’reto

focusonis3p3(sinceitpossessesn=3,l=1).Don’tforgetthatthevalueofml(i.e.magneticquantumnumberis-1to+1).So,theMlherewillbe-1,0,+1,reasonwhythereare3-orbitalscontainedinP-block.

We’redealingwith3p3wherebythreeelectronsarepresentinthepblockasfollows;ml=

↑ ↑ ↑

-1 0 +1(x) (y) (z)

Wecannowseethatonlyoneelectronispresentinml=-1,undern=3,l=1.

OptionDiscorrect.

16.CalculatetheactualnumberofSO42-

ionsin14.3gCr2(SO4)3.(O=16;S=

32;Cr=52;NA=6.02x1023).

A.4.39x1022SO42-ionsB.6.59x

1022SO42-ionsC.6.59x1023SO4

2-

ionsD.4.39x1023SO42-ions

Noofmole=mass/molarmassforasubstanceofgivenmass.So,wecanfirstfindthenumberofmoleofCr2(SO4)3.MolarmassofCr2(SO4)3.=2(52)+



3(32(4x16))=392g/mol.NoofmoleofCr2(SO4)3=14.3g/392g/mol.=0.03648mol.

Cr2(SO4)3.↔2Cr3++3SO42-

1 2 :30.03648 0.07296 0.1094

So,themoleofSO42-is0.1094mol.

Don’tforgetthatNoofmole=Noofparticles/Avogadro’sconstant.Forelementaryparticles(i.e.Atoms,ions,molecules).Now,noofmoles

ofSO42-=NoofSO4

2-ions/Avogadroconstant.

Therefore(0.1094)/1=(NoofSO42-

ions)/(6.02x1023)

NoofSO42-ions=6.59x1022

OptionBiscorrect.

17.Mercuryhasadensityof13.6g/ml.Whatvolumeofmercurymustbetakentoobtain225gofthemetal?A.1.65mlB.165mlC.0.060mlD.16.5ml

Density=mass/volume;volume=mass/densityVolume=225g/13.6g/mlItsvolumeis16.5ml(3s.f.)

OptionDiscorrect.

18.Howmanygramsofzinciodideis0.654mol.Zinciodide?[Zn=65.41;I=126.90]A.208.67gB.20.88gC.208.76gD.22.98g

Beingaskedthemasswhenthenoofmoleisgiven?!Easy,noofmole=mass/molarmassMolarmassofZinciodideZnI2=65.41+2(126.90)=319.21g/mol.Therefore,mass=Noofmolexmolarmass.massofZnI2=0.654molx319.21g/mol.=208.76g

OptionCiscorrect.

19.Whichofthefollowingis/aretrueabouthomogenousmixtures:I)Theyarealsoknownassolutionsii)Theyhaveuniformpropertiesthroughoutagivensampleiii)containphysicallydistinctpartwithdifferentproperties.A.IIonlyB.IIIonlyC.I&IIonlyD.I,II&III

Alwaysrememberthatthedifferencebetweenheterogeneousandhomogenousmixturesisthedegreeatwhichthematerialsaremixedtogetherandtheuniformityofthecomposition.Homogeneousmixturesisthemixturewherethecomponentsthatmakeupthemixtureareuniformlydistributedthroughoutthemixture,meaningthatthecompositionofthemixtureisthesamethroughout.Onlyonephaseofmatterisobservedinhomogenousmixture.Theyarealsoknownassolutions.WherebyHeterogeneousmixtureisthemixturewherethecomponentisnotuniformorhavelocalizedregionswithdifferentproperties,therearealwaystwoormorephasesinaheterogeneousmixture,whereyoucanidentifyregionwithpropertiesthataredistinctfromthoseofanotherregion,eveniftheyareinthesamestateofmatter(e.g.liquid,solid).So,consideringthoseexplanation:Wecandeducethathomogeneousmixturesarealsoknownassolutions.Theyhaveuniformpropertiesthroughoutagivensampleanddonotcontainphysicallydifferentpartwithdifferentproperties(whichisacharacteristicofheterogeneous



mixtures).I,IIaretrueabouthomogenousmixturewhileIIIisnot.

OptionCiscorrect.

20.Toa0.254gsampleofanimpurelead(II)ethanoatewasaddedexcessK2CrO4inwhichcase0.130goflead(II)chromatewasprecipitated.Whatisthepercentagecompositionofleadintheorganicsalt?[O=16;Cr=52;Pb=207].A.42.80%B.28.80%C.32.80%D.38.80%

Massoflead(Pb)inlead(ii)chromate,PbCrO4=207/323x0.130gmassoflead=0.0833g.%compositionofleadintheorganicsalt(CH3COO)2Pb=0.0833/0.254x100%=32.8%

OptionCiscorrect.

21.Fourgrams(4g)ofamixtureofCaCO3andsandistreatedwithanexcessofhydrochloricacid,and0.880gofCO2isproduced.WhatisthepercentofCaCO3intheoriginalmixture?A.30.0%B.10.0%C.70.0%D.50.0%

CaCO3+2HCl→CaCl2+CO2+H2OnoofmoleofCO2=0.880g/44g/mol.=0.02mol.RatioofCaCO3toCO2is1:1ThemoleofCaCO3inthereaction

isalso0.02mol.ThemassofCaCO3

inthereaction=0.02mol.x100g/mol.[the100g/mol.IsthemolarmassofCaCO3]ThereforethemassofCaCO3=2g?Now,thepercentofCaCO3intheoriginalmixtureis2g/4gx100%=50%

OptionDiscorrect.

22.Whatisthemolarconcentrationofasolutionpreparedbyadding750.0mLof0.25MNaOHto250.0mLof0.10MNaOH?Leaveyouranswerto2significantfigures.A.0.25MB.0.15MC.0.20MD.0.21M

Themolarconcentrationherecanbegottenbyusingtheformula;n1+n2(inmol)./V1+V2(inL)n1isthefirstnumberofmoleofNaOH.n2isthesecondnumberofmoleofNaOH.V1isthecorrespondingfirstvolumeofNaOH.V2isthecorrespondingsecondvolumeofNaOH.n1=0.1x750/100=0.1875mol.n2=0.1x250/100=0.025mol.

Molarconc.Ofthesolution=(0.1875+0.025)mol./(0.75+0.25)L=0.2125mol./1L=0.2125M

OptionDiscorrect

23.Whichhalogenhasthegreatestfirstionizationenergy?A.ClB.IC.BrD.F

Withoutwastingtimeonthis;weallknowthationizationenergyincreasesacrosstheperiodfromLEFTtoRIGHT,anddecreasesdownthegroup.ReasonwhyHalogen(Group7elements)possesshighionizationenergy,buttheycan’tpossessthesameI.E.asitdecreasesdownthegroup.ThedecreasingorderofthehalogengasthusisF>Cl,Br,I,As.So,itcanbeconcludedthatfluorine,Fhasthegreatestionizationenergy.

OptionDiscorrect.



24.Whatisthemolecularmassofanorganiccompoundhavingamolarmassof60.00amuifitconsistsof39.9%C,6.72%Hgiventhatthethirdelementintheacidisoxygen?[H=1,C=12,O=16].A.C8H6O4B.C6H10O4C.C3H6O2D.C2H4O2

C=39.9%H=6.72%O=[100-(39.9+6.72)]=53.38%C H O39.9/12 6.72/1 53.38/163.325/3.3256.72/3.3253.34/3.3251 2 1CH2O[CH2O]n=60(12+(2x1)+16)n=6030n/30=60/30n=2Themolecularformulaofthecompoundis[CH2O]2=C2H4O2.

OptionDiscorrect.

25.Howmanyunpairedelectrondoesa

groundstategasphaseCr2+ionhave?[AtomicnumberCr=24]A.0B.6C.2D.4

TheatomicnumberofCris24.Now,itselectronicconfigurationis

1s22s22p63s23p64s13d5atthisgroundstate.We’rebeingaskedthatoftheground-stategasphase

ofCr2+ion.

ThereforeCr2+isactuallytheexcitedstateofCratom,butataground-stategasphaseofitself.

Cr2+chromiumatomhaslost2electronstheelectronic

configurationofCr2+willbe

1s22s22p63s23p64s°3d4.Mindyou,thetwoelectronswillbelostinsuchawaythatoneelectronislostfromthe4s-orbitalhavingnoelectronleft

(whichwillresulttothissub-levelbeingevacuatedandtheotheronefromthe3d-orbital.Now,toknowthenumberofunpairedelectronsinthisground-

statephaseofCr2+ion,wehavetoconsiderthe3d-orbitalbeingtheoutermostorbital.Wehave5spatialorientationsind-orbital(consisting4electronsinthiscase),asin

↑ ↑ ↑ ↑

Wecanvividlyseethatallthe4electronsintheoutermost3d-orbitalareUNPAIRED.

OptionDiscorrect.

26.Quicklime(CaO)isproducedbythermaldecompositionofcalciumcarbonate(CaCO3).CalculatethevolumeofCO2producedatSTPfromthedecompositionof60.8gCaCO3.[C=12;O=16;Ca=40;Molarvolume=22.4L].A.31.4LB.26.3LC.13.6LD.34.1L.

CaCO3↔CaO+CO2

NoofmoleofCaCO3=massofCaCO3/molarmassMolarmassofCaCO3=100g/moleTherefore,noofmoleofCaCO3=60.8g/100g/mol.=0.608mol.ComparingCaCO3toCO2:theyarein1:1so,thenoofmoleofCO2islikewise0.608mol.Findthevol.ofCO2atSTP;NoofmoleofCO2=molarVolume/22.4L/moleVolumeofCO2=0.608molx22.4L/mole=13.6L

OptionCiscorrect.

27.TheonlyproductwhenCuSO4reactscompletelywithNH3isCu(NH3)4SO4.WhatmoleofNH3isrequiredto



produce568.8gofCu(NH3)4SO4?[H=1;N=14;O=16;S=32;Cu=63.5]A.6mol.NH3B.4mol.NH3C.10mol.NH3D.8mol.NH3

CuSO4+4NH3→Cu(NH3)4SO4

moleofCu(NH3)4SO4canbegottenasthemanisprovided.NoofmoleofCu(NH3)4SO4=568.8g/227.5g/moleMolarmassofCu(NH3)4SO4=63.5+4(14+(1x3))+32+4(16)=227.5NoofmoleofCu(NH3)4SO4=2.5mol.TheratioofCu(NH3)4SO4tothatofNH3is1:4.So,themoleofNH3isrequiredtoproduce568.8gofCu(NH3)4SO4is2.5x4=10mol.

OptionCiscorrect.

28.AnorganicacidcontainsC,HandOonlyinonemoleculeoit.Whensubjectedtocombustionanalysis,onemoleofitwasdiscoveredtocontain68.8%Cand5.0%H.Whatisitsempiricalformula?[H=1;C=12;O=16].A.CH2OB.C3H5O2C.C7H6O2D.C7H6O2

Sincethepercentofallconstituentofacompoundmakeatotalof100%.Thepercentofcarbonis68.8%,thatofHydrogenis5.0%,thenthatofthelastelementbeingoxygenis100–(68.8+5.0)=26.2%C H O68.8 5.0 26.212 1 16

5.733 5.0 1.63751.63751.63751.63753.05 3.05 1Multiplyby2

Theratiowillherebybecome7:6:2.TheEmpiricalformulaisthereforeC7H6O2

OptionCandDarecorrect.

29.When20gofLireactswith30gofoxygentoformLi2O,whichofthereagentsisinexcessandbyhowmuch?[Li=6.94;O=16.0]A.Lithium,by13.00gB.Oxygen,by7.00gC.Lithium,by7.00gD.Oxygen,by13.00g

LithiumreactswithOxygentoformLi2O.Wehave4Li+O2→2Li2O.NoofmolesofLi=20/6.94=2.88mol.NoofmoleofO2=30/32=0.94mol.ConsideringthosetworeactantsLiandO2comparingtheirmoleratioLi O2

4 12.88 0.94–avail.moles2.88 0.72–req.moles0 0.22ThismeansthatwheneverLithiumisusedup(beingtheLIMITINGREAGENT).0.22mol.ofO2willbeleft(beingtheREAGENTINEXCESS).TonowknowthemassofthisO2inexcess;wecanusetheformulanoofmole=mass/molarmass0.22mol.=massofO2inexcess/32g/mol.ThereforemassofO2inexcess=0.22x32=7.04g.So,wecansaythat7.04gofO2isinexcess.

OptionBiscorrect.

30.A25.0cm3portionof0.30Msodiumoxalate,Na2C2O4,solutioniswarmedandtitratedagainstanacidifiedsolutionofKMnO4.If

45.0cm3ofKMnO4solutionis



requiredforthetitration,determinetheconcentrationoftheKMnO4in

g/dm3.[O=16,k=39,Mn=54.9]A.13.59B.15.79C.10.53D.12.58

Consideringthereactionbetweensodiumoxalate,Na2C2O4andacidifiedKMnO4inredoxdimension.Acidicmediumtobepreciseasoneofthereagentisgiventobeacidified.

2Na++C2O42-+K++MnO4

-→2Na+

+2CO2+K++Mn2++2O2.

Thespectator(Na+&K+)ionsthereby

cancelout,leavingbehindthemainionsasfollows;

C2O42-+MnO4

-→2CO2+Mn2+.Balancingthisequationinacidicmediumwillhave

C2O42-→2CO2(Oxidation)

MnO4-→Mn2+(Reduction)

BalancingusingH+andH2O,asH+isaddedtothesidehavingmoreoxygenatomswhileH2Oisaddedtothesidehavinglessoxygenatoms.

C2O42-→2CO2

MnO4-+8H+→Mn2++4H2O.

addappropriateelectronstoeachoftheabovehalfequations,theappropriateelectronsaremeanttobeaddedtotheRHSoftheoxidation-sidesinceelectronsarelostWHILEtheyareaddedtotheLHSofthereductionsideaselectronsaregained.

C2O42→2CO2+2e-

MnO4-+8H++5e-→Mn2++4H2O.

Theelectronsaremeanttobebalancedsoastocanceloutlater,thereforethefirstequationistobemultipliedby5andsecondequationby2.Wetherebyhave:

5C2O42-→10CO2+10e-

2MnO4-+16H++10e-→2Mn2++

8H2O.

Addingthetwoequations,wethereforehave;

5C2O42-+2MnO4

-+16H++10e-→

10CO2+10e-+2Mn2++8H2O.Balancingtheequation,we’llnowhave

5C2O42-+2MnO4

-+16H+→10CO2+

2Mn2++8H2O

Don’tforgetthatC2O42-isthe

componentofNa2C2O7andMnO4-

beingthecomponentrepresenting

KMnO4.

Now,sincewe’vebeengiventheconc.Andvolumeofsodiumoxalate,wecannowcomputeforthemole:Noofmoleofoxalate=conc.XVol.=0.30x25/1000=0.0075mol.GettingthemoleofcorrespondingKMnO4solution,wecomparingtheirratiofromthebalancedequation;

5C2O42-+2MnO4

-

5 : 20.0075:2/5x0.0075

=0.003mol.Now,wehave0.003mol.ofKMnO4

solution.Themolarconc.OftheKMnO4

solution=moleoftheKMnO4

/Volumeofthesolution(indm3)

=0.003mol./0.045dm3

=0.0667M.Massconc.OfKMnO4=Molarconc.OfKMnO4xitsmolarmass.MolarmassofKMnO4=39+54.9+16(4)=157.9g/mol.So,themassconc.OfKMnO4=

0.0667x157.9=10.53g/dm3.OptionCiscorrect.

31.Inanexperiment,20.0gofmethanoland15.0gofcarbonmonoxidewereplacedinareactiontoproduceethanoicacidonly.Determinethepercentageyieldoftheethanoicacidproducediftheactualyieldofthe



ethanoicacidis30.1g.[CO=28;CH3OH=32;CH3COOH=60]A.89.1%B.93.6%C.81.9%D.96.4%

ReactionbetweenmethanolandcarbonmonoxidetoproduceethanoicacidCH3OH+CO→CH3COOH.NoofmoleofCH3OH=20g/32g/mol.=0.625mol.NoofmoleofCO=15g/28g/mol.=0.536mol.Checkingforthelimitingreagent;thelimitingreagentisvividlyCOasthey’rein1:1WecanknowthecorrespondingnoofmoleofCH3COOHbycomparingitsratiotothatofthelimitingreagent,CO.Theyareinratio1:1,sothenoofmoleofCH3COOHisalso0.536mol.WecangetthemassofCH3COOHbyapplyingtheformula,noofmole=mass/molarmass.massofCH3COOH=NoofmoleofCH3COOHxmolarmassofCH3COOHMolarmassofCH3COOH=60g/molemassofCH3COOH=0.536x60=32.16g.ThemassofCH3COOHisactuallytheoreticalasit’sgottenfromcalculation.While30.1ghasbeengiventobetheactualyield;Percentageyield=actualyield/theoreticalyieldx100%=30.1/32.1x100%=93.59=93.6%

OptionBiscorrect.

32.Whenbalancedinabasicmedium,theionicequation:

CN-+MnO4-→CNO-+MnO2,

exactlybecomes

A.2MnO4-+4H2O+6e-→2MnO2

+8OH-

B.3CN-+6OH-→3CNO-+3H2O+

6e-

C.3CN-+2MnO4-+4H2O+6OH-

→3CNO-+2MnO2+8OH-+3H2O

D.3CN-+2MnO4-+H2O→3CNO-+

2MnO2+2OH-

We’retobalancethefollowingionicequationinbasicmedium.

CN-+MnO4-→CNO-+MnO2.

Wefirstsegregatethisequationintooxidationandreductioni.e.halfequation.

CN-→CNO-(Oxidation)

MnO4-→MnO2(Reduction)

WebalancebasicequationusingOH-andH2O,asOH-isaddedtothesidehavinglessoxygenatomswhileH2Oisaddedtothesidehavingmoreoxygenatoms.

CN-+2OH-→CNO-+H2O

MnO4-+2H2O→MnO2+4OH-

Appropriateelectronsareaddedtoeachofthehalfequations,theappropriateelectronsaremeanttobeaddedtotheRHSoftheoxidationsidesinceelectronsarelostWHILEtheyareaddedtotheLHSofthereductionsideaselectronsaregained.

CN-+2OH-→CNO-+H2O+2e-

MnO4-+2H2O+3e-→MnO2+4OH-

Theelectronsaremeanttobebalancedsoastocanceloutlater,thereforethefirstequationistobemultipliedby3andthesecondequationby2.Wetherebyhave;

3CN-+6OH-→3CNO-+3H2O+6e-

2MnO4-+4H2O+6e-→2MnO2+

8OH-

Addingthetwohalfequations,wethereforehave;

3CN-+6OH-+2MnO4-+4H2O+6e-

→3CNO-+3H2O+2MnO2+8OH-+



6e-

The6electronsinbothsideswilldefinitelycancelout,andotherspeciescommontobothsideswillbebalanced.

3CN-+2MnO4-+H2O→3CNO-+

2MnO2+2OH-

OptionDiscorrect.

33.Fifteengrams(15.00g)ofhydratedcalciumsulphate,CaSO4.nH2O,weighs11.85gafterbeingheatedtoaconstantmass.WhatisthevalueofninCaSO4.nH2O.[H=1;O=16;S=32;Ca=40]A.1B.2C.7D.

ThemassofCaSO4.nH2Ois15g.Thewaterofcrystallizationisdriedupafterheating,leavingbehind11.85gofCaSO4;Themassofthewaterofcrystallization(i.e.nH2O)is(15g–11.85g)=3.15g.Thenumber(i.e.n)ofthewaterofcrystallizationcanbeknownbyapplyingtheformula;(massofthehydratedcompound/itsmolecularmass)=(massofthewaterofcrystallization/itsmolecularmass):(15/136+18n)=(3.15/18n)crossmultiply270n=3.15(136+18n),270n=428.4+56.7ncollectliketerm270n–56.7n=428.4(213.3n/213.3)=(428.4/213.3)n=2.

OptionBiscorrect.

34.Themaximumenergyofelectromagneticradiationthatcancompletelyremove(ionize)anelectronfromthehydrogen

atomis1.36x10-19J.Determinetheinitialexcitedstatefortheelectron(n=?).(RH=

1.097x107m-1,c=3.00x108m/s,h

=6.63x10-34Js)

A.1B.7C.4D.2

Wedon’twanttowastetimehere.WeknowthatRH=R∞/hcandE=hc/λ

So,theformulaE=hc/λ=R∞[(1/n12)-

(1/n22)].

Canbewrittenas1/λ=RH[(1/n12)-

(1/n22)].

Therefore,E=hc/λ,λ=hc/E

λ=(6.626x10-34x3x108)/(1.36x10-19),

λ=1.46x10-6m.

Now;(1/1.46x10-6)=1.097x107

[(1/n12)-(1/n2

2)],

Dividebothsidesby1.097x107

[1/(1.46x10-6x1.097x107)]=(1/n12)-

(1/n22)

(1/16)=(1/n12)-(1/n2

2)n2willtendto∞

(1/16)=(1/n12)-(1/∞)

1/16=1/n12

n12 =16

n1=4.OptionCiscorrect.

35.When2.86gofamixtureof1-butene(C4H8)andbutane(C4H10)wasburnedinexcessoxygen,8.80gofCO2and4.14gofH2Owereobtained.CalculatethepercentagebymassofC4H8andC4H10

respectivelyintheoriginalmixture.[H=1,C=12]A.49.0and50.7%B.37.2and62.8%C.60.8and39.2%D.39.2and60.8%

Letmrepresentthemassof1-butene&nrepresentthemassofbutane.m+n=2.86.Thereforen=2.86–m--------------(I)Wecangetthemassofhydrogenfromthe4.14gofwaterby2/18x4.14g=0.46g.Massofhydrogenin1-butene+massofhydrogeninbutane=massofhydrogen=(8/56)xm+(10/58)xn=0.468m(29)+10n(28)/1624=0.46



232m+280n=747.04-------------(ii)Substitutingequ(I)intoequ(ii),wehave;232m+280(2.86-m)=747.04232m +800.8–280m=747.04(-48m/-48)=(-53.76/-48)m=1.12gn=2.86–1.12g=1.74g.So,themassof1-butenehasbeenfoundtobe1.12g,andthatofbutanetobe1.74g.Thepercentagebymassof1-buteneintheoriginalmixture=(1.12/2.86)x100%=39.2%Thepercentagebymassofbutaneinthemixture=(1.74/2.86)x100%=60.8%Thepercentagebymassof1-butene(C4H8)andbutane(C4H10)respectivelyintheoriginalmixtureis39.2%and60.8%.

OptionDiscorrect.

36.Whatisthenetionicequationforthebalancedequation:16HCl+2KMnO4→2MnCl2+8H2O+5Cl2+2KCl?

A.8H++MnO4-→2Mn2++8H2O

B.8Cl-+8H++MnO4-→Mn2++4H2O

+4Cl2

C.8H++MnO4-→Mn2++4H2O

D.10Cl-+16H++2MnO4-→2Mn2++

8H2O+5Cl2

Breakingtheequationdownintoionsandcancellingoutthespectatorions,we’llhavethetwohalfequationstobe;

Cl-→Cl2(Oxidation)

MnO4-→Mn2+(reduction)

followingtheprocedureshighlightedinQ30…

2Cl-→Cl2

MnO4-+8H+→Mn2++4H2O

2Cl-→Cl2+2e-

MnO4-+8H++5e-→Mn2++4H2O

Multiplyingtheoxidizedhalfequationby5,andreducedhalfequationby2,we’llhave

10Cl-→5Cl2+10e-

2MnO4-+16H++10e-→2Mn2++8H2O

The10e-willcanceloutwhenaddingthetwohalfequationstogether;

10Cl-+2MnO4-+16H+→5Cl2+2Mn2++

8H2OOptionDiscorrect.

37.Calculatetheamountofquicklime(CaO)thatcanbepreparedbyheating200kgoflimestonethatis95.0%pureCaCO3

(CaCO3=100,CaO=56.1)A.127kgB.117kg.C.107kgD.100kg

95%pureCaCO3=(95/100)x200kgofCaCO3

=190kgofCaCO3

noofmoleofCaCO3=190kg/100g/mol.=1.9kmol.CaCO3↔CaO+CO2

ThemoleratioofCaCO3toCaOis1:1,sothenoofmoleofCaOisactually1.9kmol.NoofmoleofCaO=(massofCaO/itsmolarmass)1.9kmol./1=massofCaO/56.1g/moleMassofCaO=1.9kmolx56.1g/mol=107kg

OptionCiscorrect

38.Whattypeofreactiondoesthisequationrepresent?

(CH3)3C-OH+Br-→(CH3)3C-Br+OH-

A.CombinationB.DecompositionC.ReplacementD.Combustion

WecanseethatBr-issubstitutingfor

OH-inthecompound,(CH3)3C-OH,therebyreplacingitbygettingitextricated.TheprocessisundoubtedlyReplacementreaction,whichisaprocesswherebyanelementorioncombines/reactswithacompoundbysubstitutingforanelementorioninthecompound(NOTE:Theelementorionreactingwiththecompoundmustbe



higherintheelectrochemicalseriesthanthecomponentofthecompound

it'ssubstitutingfor….Br-could

substituteforOH-duetothethefact

thatBr-ishigherthanOH-intheelectrochemicalseriesofanions(aseriesshowingthearrangementofanionsintheorderofabilitytoacceptelectrons,correspondingtotheirdecreasingorderofreactivity…lookat

theirarrangement;F-,SO42-,NO3,Cl-,Br-,I

-,OH-).OptionCiscorrect.

39.CalculatethemassofBaCO3producedwhenexcessCO2isbubbledthroughasolutioncontaining0.205molofBa(OH)2,[Ba=137.3,C=12.1,O=16.0,H=1.0]A.35.5gB.25.5gC.45.5gD.40.5g

Ba(OH)2+CO2→BaCO3+H2O.SincethemoleofBa(OH)2is0.205mol.ThenthemoleofBaCO3isalso0.205mol.astheyareinthemoleratio1:1NoofmoleofBaCO3=massofBaCO3/itsmolarmassmolarmassofBaCO3=137.3+12.1+16(3)=197.4g/mol.MassofBaCO3=0.205mol.X197.4g/mol.=40.5g

OptionDiscorrect.

40.Accuracyinmeasurementis:A.EstimatedintermsofabsoluteerrorB.Agreementbetweentworeplicatemeasurements.C.ClosenessofmeasurementtothetruevalueD.Alloftheabove.

Accuracyisthedegreeofclosenessofasinglemeasurementorthemeanofsetofvaluesofreplicatemeasurementstothetruevalueoracceptedreferencevalue.

Whileprecisionisthedegreeofclosenessofasetofvaluesobtainedfromidentical/replicatemeasurementsusingthesameapparatus.Fromthese;Wecandeducethataccuracycanbeexpressedintermofabsoluteerror[whichisequaltox(forsinglemeasurement)orẋ(i.e.meanofreplicatemeasurements)–T(i.e.truevalue)]Precisionistheagreementbetweentworeplicatemeasurements.Accuracycanberegardedtobetheclosenessofmeasurementstothetruevalue.So,

OptionAandCarecorrect

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