AP Chemistry Homework Packet Chapter 1AK · AP Chemistry Homework Packet Chapter 1 1. State whether...

APChemistryHomeworkPacketChapter11.Statewhethereachofthefollowingobservationsisaqualitative(qual)orquantitative(quant)observation. Thepieceofmetalislongerthanthepieceofwood. Solution1ismuchdarkerthansolution2. TheliquidinbeakerAisblue. Thetemperatureoftheliquidis60°C.Aquantitativeobservationcontainsanumberandaunitwhileaqualitativeobservationusesdescriptors.Answer:qual/qual/qual/quant 2.Aquantitativeobservationa) containsanumberandaunitb) doesnotcontainanumberc) alwaysmakesacomparisond) mustbeobtainedthroughexperimentatione) isnoneoftheseAquantitativeobservationcontainsanumberandaunitwhileaqualitativeobservationusesdescriptors.A)TrueB)False—aquantitativeobservationdoescontainanumberC)False—anobservationdoesnothavetomakeacomparison—itoftendescribesjustasingleisolatedoccurrenceofaprocessD)False—weobservenumerousprocessesintheworldarounduseveryminutewithoutperforminganexperimentE)False—Aisacorrectstatement.AnswerA3. Generally,observedbehaviorthatcanbeformulatedintoastatement,sometimesmathematicalinnature,iscalleda(n)a) observationb) measurementc) theoryd) naturallawe) experimentIfobservationsofacertainprocessareconsistentlythesameeverytimetheprocessoccurs—forexample,ifanobjectacceleratesatthesamerateeverytimeitisdropped,howwedescribetheprocessdoesnotchangesowecanstatethatitisanaturallaw.Inparticular,wemaybeabletocreateamathematicalstatementthatdescribesthebehavior.Usingtheaboveexample,wewouldsaythatundertheinfluenceofgravity,massesaccelerateat9.8m/s2nearthesurfaceoftheearth.AnswerD4. Thestatement“Thetotalmassofmaterialsisnotaffectedbyachemicalchangeinthosematerials”iscalleda(n)a) observationb) measurementc) theoryd) naturallawe) experimentThestatementgivenisobservedbehaviorinallcircumstancesofallchemicalreactionsandsoisanaturallaw.AnswerD5. TFAchemicaltheorythathasbeenknownforalongtimebecomesalaw.Alawissimplyastatementordescriptionthatacertainprocessorphenomenonisknownto

Chapter Error! Unknown document property name.: Error! Unknown document property name. 2

occur.Atheoryisastatementofhow/whyaprocessorphenomenonisthoughttooccur,anddrawsonmanylinesofevidencetojustifyit.Asatheoryexplainsalaw,itcannotbecomealaw.AnswerFalse6. Whichofthefollowingmetricrelationshipsisincorrect?a) 1microliter=10–6litersb) 1gram=103kilogramsc) 103milliliters=1literd) 1gram=102centigramse) 10decimeters=1meterA)True.Theprefixmicrorefersto10-6—therefore,amicroliterwouldbe10-6L.B)False.Askilorefersto1000,103kilogramswouldbe1000x1000,or1,000,000grams.Ontheotherhand,1gwould1/1000thofakilogram,so1gwouldbe10-3g.C)True.millimeans1/1000thso103milliliterswouldbe1000x1/1000thofaliter,or1Liter.D)True.Centirefersto1/100thso102(or100)wouldbe100x1/100thofagram,or1gram.E)True.Decirefersto1/10thso10x1/10thofameterwouldequal1meter.AnswerB7. ForwhichpairistheSIprefixnotmatchedcorrectlywithitsmeaning?a) mega=106b) kilo=1000c) deci=10d) nano=10–9e) centi=0.01A)True.Megareferstomillionand106isonemillion.B)True.kilorefersto1000.C)False.Decirefersto1/10th,not10.D)True.nanorefersto1/1,000,000,000th(onebillionth)—specifically,nanomeans“9”—and10-9isonebillionth.E)True.centi-refersto1/100thand.01is1/100th.AnswerC8. Ametricunitforlengthisa) gramb) milliliterc) yardd) kilometere) poundA)False.gramisametricunitofmass.B)False.milliliterisaunitofvolume.C)False.yardisanEnglishunitforlength.D)True.kilometerisametricunitmeaning1000meters.E)False.ApoundisanBritishunitofforce.AnswerD9. WhichofthefollowingisnotaunitintheSIsystem?a) ampereb) candelac) Kelvind) metere) calorieA)anampereisanSIunitofelectricalcurrent.B)acandelaisanSIunitofluminousintensity(thebrightnessoflight).C)theKelvinisanSIunitoftemperature.D)themeterisanSIunitoflength.E)thecalorieisaBritishunitofenergy.AnswerE

10. Orderthefourmetricprefixesfromsmallesttolargest.a) nano-<milli-<centi-<kilo-b) milli-<nano-<centi-<kilo-c) kilo-<centi-<nano-<milli-d) kilo-<centi-<milli-<nano-e) centi-<nano-<kilo-<milli-Youshouldquicklylearntothinkofnumberprefixesintermsoftheexponentofthepowerof10theyrepresent.Thiswayyoucaneasilyvisualizerelationshipsbetweentheseunits.Forexample,ifyouknownanorepresents10-9andmicrorepresents10-6,youimmediatelyknowthatnanometersaresmallerthanmicrometersandthatthereare1000nanometersinamicrometer—theydifferbyafactorof103or1000.Likewise,ifyouknowthatpicorepresents10-12andkilorepresents103,youknowthatpicosecondsaresmallerthankilosecondandthatthereare1quadrillionpicosecondsinakilosecond-theydifferbyafactorof1015or1,000,000,000,000,000(1quadrillion).Ifyouknowtheserelationshipswell,youcanvisualizetheanswersgiveninyourmindasfollows,andyoushouldhavethecorrectanswerinlessthan15seconds.A)Correct10-910-310-2103.B)Incorrect10-310-910-2103C)Incorrect10310-210-910-3D)Incorrect10310-210-310-9E)Incorrect10-210-910310-3 AnswerA11. 5.5kilogram(s)containsthismanygrams.

5.5 kg x

1000g1 kg

= 5.5 x 103 g

Alwaysstartbywritingdownwhatyouaregiven,inthiscase,5.5kg.Thenthink—IwanttocreateaconversionfactorthatgetsridofkgandcreatesgsoIwillputkgonthebottomsoitcancelsandgonthetopsoitstays.Iknowthereare1000ginakgsoIaddthenumbersinandperformthemultiplication.Thereare2sigfigsintheanswer.Answer 5.5 x 103 g 12. Convert0.2974mtomm.

.2974 m x

1000mm1m

= 297.4mm

Alwaysstartbywritingdownwhatyouaregiven,inthiscase,.2974m.Thenthink—IwanttocreateaconversionfactorthatgetsridofmandcreatesmmsoIwillputmonthebottomsoitcancelsandmmonthetopsoitstays.Iknowthereare1000mmin1msoIaddthenumbersinandperformthemultiplication.Thereare4sigfigsintheanswer.Answer 297.4mm 13. 3.2secondscontainthismanypicoseconds.

3.2 s x

1012 pss

= 3.2 x 1012 ps or 3.2 s x

1ps10−12 s

= 3.2 x 1012 ps

Alwaysstartbywritingdownwhatyouaregiven,inthiscase,3.2s.Thenthink—IwanttocreateaconversionfactorthatgetsridofsandcreatespssoIwillputsonthebottomsoitcancelsandpsonthetopsoitstays.Iknowthereare1012psin1secondsoIaddthe

numbersinandperformthemultiplication.Conversely,Icouldalsosayin1ps,thereare10-12seconds.Thereare2sigfigsintheanswer.Answer= 3.2 x 1012 ps 14. 9.16secondscontainthismanynanoseconds.

9.16 s x

109 nss

= 9.16 x 109 ns or 9.16 s x

109 nss

= 9.16 x 109 ns

Alwaysstartbywritingdownwhatyouaregiven,inthiscase,9.16s.Thenthink—IwanttocreateaconversionfactorthatgetsridofsandcreatesnssoIwillputsonthebottomsoitcancelsandnsonthetopsoitstays.Iknowthereare109nsin1ssoIaddthenumbersinandperformthemultiplication.Conversely,Icouldalsosayin1ns,thereare10-9seconds.Thereare3sigfigsintheanswer.Answer= 9.16 x 109 ps

15. Thedistanceof54kmequals

54 km x

1000m1km

= 5.4x104 m (couldalsobe54,000m)

Alwaysstartbywritingdownwhatyouaregiven,inthiscase,54km.Thenthink—IwanttocreateaconversionfactorthatgetsridofkmandcreatesmsoIwillputkmonthebottomsoitcancelsandmonthetopsoitstays.Iknowthereare1000min1kmsoIaddthenumbersinandperformthemultiplication.Thereare2sigfigsintheanswer.Answer 5.4x104 m (Theanswer54000malsocontains2sigfigs).16. Whatisthemeasureofresistanceanobjecthastoachangeinitsstateofmotion?a) massb) weightc) volumed) lengthe) noneoftheseMakesureyouunderstandthedifferencebetweenmassandweight.Themassofanobjectistheamountofmatteritcontainsinit.Ifthemassofanobjectis15g,thisreflectstheamountofmattertheobjectcontainsandthisdoesnotchangenomatterwheretheobjectislocated—ontheearth,onthemoon,orinouterspace—themassisstill15g.However,iftheforceofgravityactsonthe15g,itwillhaveweight—weightisameasureoftheforceduetogravityonthe15g.Onearththisweightwouldbe.15Newtons(.033lb),onthemoonthisweightwouldbe.024Newtons(.0053lb),andinouterspacetheweightwouldbe0becausethereisnosignificantforceduetogravityactingonit.Asamoresubtlepoint,youareawarethatthegreaterthemassofanobject,themoredifficultitistomoveit.So,wealsosaythatthemassofanobjectisameasureofhowresistanceanobjectistobeingmoved.Thinkaboutthis—twoobjectsareinouterspaceandthereisnosignificantforceofgravityactingoneither.Thereforetheweightofbothis0.However,onehasamassof1000kg,whiletheotherhasamassof1kg.Eventhoughbothobjectsweigh0,theyhavedifferentmasses.Whichoneismoredifficulttomove?AnswerA

17. Thedegreeofagreementamongseveralmeasurementsofthesamequantityiscalled__________.Itreflectsthereproducibilityofagiventypeofmeasurement.a) accuracyb) errorc) precisiond) significancee) certaintyA)Incorrect.Accuracyisthedeterminationofhowcloseoneormoremeasurementsaretothecorrectanswer.B)Incorrect.Errorisadeterminationofhowfarawayfromtheexpectedvalueameasurementis.C)Correct.Foragroupofmeasurements,precisionisthetermusedtodescribehowclosetogetherthevaluesofthemeasurementsare.Iftheyareclosetogether,theyarepreciseandthemeasurementisreproducible.Thisisnottobeconfusedwiththetermprecisionasitisusedforasinglemeasurement.Inthatcontextitreferstothedegreeofuncertaintyforthemeasurement.D)Incorrect.Thetermsignificancereferstothenumberofcertainanduncertaindigitspresentinasinglemeasurement.E)Incorrect.Certaintyreferstothedegreetowhichdetermineameasurement—alldigitsinameasurementexceptthelastarecertain,whilethelastdigitisuncertain.AnswerC18. Aspartofthecalibrationofanewlaboratorybalance,a1.000-gmassisweighedwiththefollowingresults:

Trial Mass 1 1.201±0.001 2 1.202±0.001 3 1.200±0.001

Thebalanceis:a) Bothaccurateandprecise.b) Accuratebutimprecise.c) Precisebutinaccurate.d) Bothinaccurateandimprecise.e) Accuracyandprecisionareimpossibletodeterminewiththeavailableinformation.Allthreemeasurementsareveryclosetogetherandarethereforepreciseasagroupofmeasurements.However,asagroup,theyarenowhereneartheexpectedvalueandsoarenotaccurate.So,themeasurementsareprecisebutinaccurate.AnswerCUsethefollowingtoanswerquestions19-20:

Considerthefollowingthreearcherytargets:

I. II. III.

19. Whichofthefollowingfigure(s)representaresulthavinghighprecision?a) FigureIonlyb) FigureIIonlyc) FigureIIIonlyd) FigureIandFigureIIe) FigureIIandFigureIIIMeasurementsthatareclosetogetherhaveahighprecision,regardlessofwhetherornottheyareaccurate.Therefore,theresultsinbothdiagramsIIandIIIareprecise.AnswerE20. Whichofthefollowingstatementsconcerningthesefiguresiscorrect?a) FigureIrepresentssystematicerrorandFigureIIrepresentsrandomerror.b) FigureIrepresentsrandomerrorandFigureIIrepresentssystematicerror.c) FigureIandFigureIIrepresentrandomerror.d) FigureIandFigureIIrepresentsystematicerror.e) FigureIIIrepresentsnoerrors.Inagroupofmeasurementssystematicerroriserrorrelatedtoequipmentproblems,andsomeasuredvaluesdifferfromexpectedresultsbyaboutthesameamountandinthesamedirection,asshowninII.Randomerroriserrorrelatedtooperatorerrorandrandomvariationobservedinallprocesses,sovaluesdifferfromtheexpectedresultsinalldirectionsandbyvaryingamounts,asshowninI.InfigureIII,althoughallvaluesareclosetotheexpectedvalueandsoareaccurateandprecise,therewillalwaysstillbesmallamountsoferrorasshowninfigureIII.AsfigureIrepresentsrandomerror,figureIIrepresentssystematicerror,andfigureIIIstillrepresentssmallamountsoferror,Biscorrect.AnswerB21. Whichofthefollowingistheleastprobableconcerningfivemeasurementstakeninthelab?a) Themeasurementsareaccurateandprecise.b) Themeasurementsareaccuratebutnotprecise.c) Themeasurementsareprecisebutnotaccurate.d) Themeasurementsareneitheraccuratenorprecise.e) Alloftheseareequallyprobable.Ifmeasurementsareaccurate,astheywouldallclusteraroundtheexpectedvalue,theymustalsobeprecise.Therefore,whilemeasurementscanbepreciseandnotaccurate,asinsystematicerror,ortheycanbebothnotpreciseandinaccurate,ortheycanpreciseandaccurate,theycannotbeaccuratebutnotprecise.AnswerB.22. Youmeasurewaterintwocontainers:a10-mLgraduatedcylinderwithmarksateverymL,anda1-mLpipetmarkedatevery0.1mL.Ifyouhavesomewaterineachofthecontainersandaddthemtogether,towhatdecimalplacecouldyoureportthetotalvolumeofwater?a) 0.01mLb) 0.1mLc) 1mLd) 10mLe) noneoftheseForthe10mLcylinderwithmarksateverymL,youcouldestimatetothenearest0.1mL.Forthe1mLpipet,youcouldestimatetothenearest.01mL.Whenweaddmeasurements,welimitthenumberofsignificantfigurestothesmallestnumberofdecimalplaces—inthiscase,tothenearest0.1mL.AnswerB

23. Theagreementofaparticularvaluewiththetruevalueiscalleda) accuracyb) errorc) precisiond) significancee) certaintyWedeterminehowaccurateameasurementisbytryingtofigureouthowcloseitistoanexpectedvalue.AnswerA24. Theamountofuncertaintyinameasuredquantityisdeterminedby:a) boththeskilloftheobserverandthelimitationsofthemeasuringinstrumentb) neithertheskilloftheobservernorthelimitationsofthemeasuringinstrumentc) thelimitationsofthemeasuringinstrumentonlyd) theskilloftheobserveronlye) noneoftheseNotonlyisthereerrorintheabilityofahumantodeterminevaluesonascaleofmeasurement,thereisalsovariabilityinthemanufactureofsuchscales.Errorrelatedtobothmustbeconsidered.AnswerA25. Ascientistobtainsthenumber0.045006700onacalculator.Ifthisnumberactuallyhasfour(4)significantfigures,howshoulditbewritten?Rememberthatifanumberhasadecimalplace,leading0’sarenotsignificant,sothesignificantdigitsstartwiththenumberfourinthehundredthsplace.Onlythatandfollowingthreeplacesshouldbeused,andthe0beforethe6shouldberoundedup: Answer0.0450126. Expressthenumber0.000191inscientificnotation.Rememberthatwithscientificnotationthenumberyoumultiplyby10xshouldbeanumberbetween1.0000…and9.9999…..Thisnumbershouldonlyincludethesignificantdigitsinthevalue,withthedecimalpointafterthefirstdigit—sointhiscaseyouwouldstartwith1.91times10x.Youshouldvisuallybeableto“see”whattheexponentwillbeforthefirstseveraldecimalplaceswithouthavingtocount.Forexample,atleastbeabletoseeimmediatelythat: .1 tenths 10-1 .01 hundredths 10-2 .001 thousandths 10-3 .0001 tenthousandths 10-4Thenumberoftheexponentwillbetheplaceofthefirstnon-zerodigit—inthiscasethefirstnon-zerodigitisinthefourth(10,000ths)decimalplaceso10-4.Withinjustafewseconds(meaning2or3seconds)youshouldknowthattheansweris1.91x10-4. Answer1.91x10-427. Express165,000inexponentialnotation.Rememberthatinanumberwithnodecimalpointalltrailingzerosarenotsignificant(unlessyouhavewitnessedtheactualmeasurementtoknowthatatrailing0isactuallyanestimate).So,165,000onlyhasthreesignificantfiguresandthevalueyouaremultiplyingmustbe1.65.

Aswithdecimalnumberslessthan1,fornumbersgreaterthan1youshouldbeableto“see”whattheexponentwillbewithouthavingtocountdigits,foralmostanynumber.Thisishelpedbyknowingthefollowing:10 tens 101100 hundreds 1021000 thousands 10310000 tenthousands 104100,000 hundredthousands 1051,000,000 millions 1061,000,000,000billions 1091,000,000,000,000 trillions 1012Forthisproblem,youshouldimmediatelyknowthatbecausethenumberisinhundredthousands,theexponentwillbe105andyoushouldknowtheansweragainwithin2or3seconds. Answer1.65x10528. Expressthenumber0.0470inscientificnotation.Giventhediscussionofthe26and27,youcanimmediatelyseethattherearethreesignificantfiguresbeginningwiththe4inthehundredthsplacesothenumberyouaremultiplyingis4.70.Youcanalsoseethisnumberbeginsinthehundredthsplacesotheexponentvaluewillbe-2.Answer4.70x10-229. Expressthenumber6.04×10–3incommondecimalform.Rememberthattheforanumberexpressedinscientificnotationthenumberofdigitsofthemultipliednumberisalsothenumberofsignificantfigures—toexpressmeasurementsinunambiguousnumbersofsignificantfiguresisthepurposeofscientificnotation.Therefore,thegivennumberhasthreesignificantfigures—6.04.Becauseyoucanvisualizethenumberofdecimalplacesassociatedwithanegativeexponentyouknowthatfor10-3wouldbeanumberthatstartedinthethousandthsplace.Answer.0060430. Expressthenumber2.38×104incommondecimalform.Becauseyoucanvisualizethenumberofplacesassociatedwithapositiveexponentyouknowthat104wouldbeanumberthatstartedinthetenthousandthsplace.2.38x10,000wouldbe:Answer23,80031. Wegenerallyreportameasurementbyrecordingallofthecertaindigitsplus____uncertaindigit(s).a) nob) onec) twod) threee) fourYouareawareofourrulesforreportingmeasurements—wereportallcertaindigits,andthefirstestimatedoruncertaindigit.AnswerB32. Thebeakersshownbelowhavedifferentprecisionsasshown.

Supposeyoupourthewaterfromthesethreebeakersintoonecontainer.Whatwouldbethevolumeinthecontainerreportedtothecorrectnumberofsignificantfigures?Inthefirstdrawingleastcountmarksare1mLsowecanestimateto.1mL—Iwouldestimate26.4mL.Intheseconddrawingtheleastcountmarksare10mLsowecanestimateto1mL—Iwouldestimate27mL.Inthethirddrawingtheleastcountmarksare.1mLsowecanestimatetothenearest.01mL—Iwouldestimate26.42mL.Becauseweareaddingvaluestogetherthenumberofsigfigsisconstrainedbytheleastnumberofdecimalplaces,whichwillbe0decimalplaces.So,addingthethreevaluestogetherweget:Answer 26.4mL + 27mL + 26.42mL = 79.82 mL = 80 mL 33. Youareaskedtodeterminetheperimeterofthecoverofyourtextbook.Youmeasurethelengthas33.16cmandthewidthas24.83cm.Howmanysignificantfiguresshouldyoureportfortheperimeter?Becauseweareaddingvaluestogetherthenumberofsigfigsisconstrainedbytheleastnumberofdecimalplaces,whichwillbe2decimalplaces—rememberingthatforaperimeterweneedtoaddineachlengthandsidevaluetwice,weget 33.16cm+ 33.16cm+ 24.83cm+ 24.83cm = 115.98cm .So,ourfinalanswermusthavefivesignificantfigures.Rememberthatthiscorrespondstotheideathatwecanincreasetheprecisionofouranswersbyincreasingthevaluesoftheoverallmeasurements.Theuncertaintyoftheestimateddigitinthe0.01decimalplaceofthefinalansweraffectsthisvaluelessthanitdoesanyoftheindividualmeasurementswhicheachhaveonlyfoursignificantfigures.Answer5sigfigs34. Considerthenumbers23.68and4.12.Thesumofthesenumbershas____significantfigures,andtheproductofthesenumbershas____significantfigures.Forthesum,thenumberofsigfigsisconstrainedbythe2decimalplaces—theanswermustalsohave2decimalsplaces,soforthissum,thefinalanswerwillhavefoursigfigs.Multiplicationsareconstrainedbythevalueofthesmallestnumberofsignificantfigures.Inthiscase,4.12hasonly3sigfigssotheanswermusthave3sigfigs.Answer4,335. Usingtherulesofsignificantfigures,calculatethefollowing:6.167 815.10

Rememberformixedcalculations,forsigfigsweperformeachoperationaswecometoit.Thefirstoperationis6.167+81.Asthisisanaddition,wearelimitedbythenumberofdecimalplaces—thesmallestnumberofdecimalplacesinthiscalculationis0.Ifwedotheadditionweget87.167.Weusethisnumberfortherestofourcalculationtolimitroundingerrorbutinthefinalcalculation,thisnumberhasonlytwosigfigs.Itisusefultoindicate

thisinsomeinyourworksoyoudon’tforget—underliningorcirclingthenumberofsigfigsworksbest.Wethendivide87.167by5.10toget17.09157.Asthenumeratorisconstrainedto2sigfigs,andthislastoperationisadivision,thefinalanswermustalsohavetwosignificantfigures.Thus,thefinalansweris17.Answer17

6.167 +815.10

= 87 .1675.10

= 17.09157 = 17

36. Usingtherulesofsignificantfigures,calculatethefollowing:4.0021-0.247Asthisisasubtractionproblemwearelimitedtothesmallestnumberofdecimalplaces,whichisthree.Performthecalculationusingthenumbersgivenandlimittheresulttothreedecimalplaces.Theanswerendsupwithfoursignificantfiguresbutthisisokayeventhoughthesubtractedvalueonlyhasthreesigfigs—again,subtractionsareonlylimitedbydecimalplaces,notleastnumberofsigfigsinthevalues.Answer3.755 4.0021− .247 = 3.755 37. Howmanysignificantfiguresarethereinthenumber0.04560700?Rememberthatifadecimalnumberhasleadingzero’stheseARENOTsignificant.Foradecimalnumberwebegincountingatthefirstnon-zeronumber.Thisandallfollowingnumbers,including0’s,aresignificant.Beginningwiththefirstnon-zerodigit,therearesevensigfigs.Answer738. Howmanysignificantfiguresarethereinthenumber0.0006313?Rememberthatifadecimalnumberhasleadingzero’stheseARENOTsignificant.Foradecimalnumberwebegincountingatthefirstnon-zeronumber.Thisandallfollowingnumbers,including0’s,aresignificant.Beginningwiththefirstnon-zerodigit,thereare4sigfigs.Answer439. Howmanysignificantfiguresarethereinthenumber3.1400?Rememberthatifadecimalnumberhastrailingzero’stheseAREsignificant.Foradecimalnumberwebegincountingatthefirstnon-zeronumberafterthedecimalpoint.Thisandallfollowingnumbers,including0’s,aresignificant.Beginningwiththefirstnon-zerodigit,thereare5sigfigs.Answer540. Howmanysignificantfiguresshouldbereportedforthedifferencebetween18.6150mLand18.57mL?Asthisisasubtractionproblemwearelimitedtothesmallestnumberofdecimalplaces,whichistwo.Performthecalculationusingthenumbersgivenandlimittheresulttotwodecimalplaces.Beverycarefulhere—theanswerendsupwithonlyonesignificantfigure—whiletherearetwodecimalplacesoneoftheseisaleading0.Butthisisokayeventhoughbothgivenvalueshavemoresigfigs—again,subtractionsareonlylimitedbydecimalplaces,notleastnumberofsigfigsinthevalues.Answer1significantfigure.

18.6150mL−18.57mL = .045mL = .05mL 1 significant figure

41. Whatisthebestanswertoreportfor 3.478 g 1.164 g2.00 mL

× –0.349g/mL?

Rememberformixedcalculations,forsigfigsweperformeachoperationaswecometoit.

Thefirstoperationis3.478x1.164.Asthisisamultiplication,wearelimitedbythesmallestnumberofsigfigsinthevalues—inthiscase,bothhave4sigfigssotheanswerofthisoperationislimitedto4sigfigs.Ifwedothemultiplicationweget4.048392.Weusethisnumberfortherestofourcalculationtolimitroundingerrorbutinthefinalcalculation,thisnumberhasonly4sigfigs.Itisusefultoindicatethisinsomeinyourworksoyoudon’tforget—underliningorcirclingthenumberofsigfigsworksbest.Wethendivide4.048392gby2.00mLtoget2.024196.Asthisoperationisadivision,itwillbeconstrainedtothesmallestnumberofsigfigsineithervalue,inthiscase,3(from2.00mL).Noticethatthisgivesouranswer2decimalplaces.Wethensubtract.349from2.024196,rememberingthatthislastvaluereallyonlyhas2decimalplaces,soourfinalanswerisalsolimitedto2decimalplaces.Thesubtractiongives1.675196,which,limitedto2decimalplacesroundsto1.68g/mL.

3.478g x 1.164 g2.00mL

− .349g / mL = 4.048 3 92g**

2.00 mL− .349g / mL =

2.2 0 4196− .349 = 1.675196 = 1.68

Iwillgivethefirstpersonwhopointsouttheerrorintheaboveproblemacandybar.42. Whatisthebestanswertoreportfor(515×0.0025)+24.57?Rememberformixedcalculations,forsigfigsweperformeachoperationaswecometoit.Thefirstoperationis515x0.0025.Asthisisamultiplication,wearelimitedbythesmallestnumberofsigfigsinthevalues—inthiscase,2sigfigs(for.0025)sotheanswerofthisoperationislimitedto2sigfigs.Ifwedothemultiplicationweget1.2875.Weusethisnumberfortherestofourcalculationtolimitroundingerrorbutinthefinalcalculation,thisnumberhasonly2sigfigs.Itisusefultoindicatethisinsomeinyourworksoyoudon’tforget—underliningorcirclingthenumberofsigfigsworksbest.Wethenadd1.2875to24.57toget25.8575.Asthisisanadditionproblemsigfigsarelimitedtothesmallestnumberofdecimalplaces.Becausethepreviousnumberislimitedto2sigfigs,whichisthe.1place,thefirstnumberisonlysupposedtohave1decimalplace,sothislimitsthefinalanswertoonedecimalplace,or,25.9.

(515x.0025)+ 24.57 = 1. 2 875+ 24.57 = 25.8575= 25.9 43. Convert4450.4gtomg.Alwaysstartbywritingdownwhatyouaregiven,inthiscase,4450.4g.Thenthink—IwanttocreateaconversionfactorthatgetsridofgandcreatesmgsoIwillputgonthebottomsoitcancelsandmgonthetopsoitstays.Iknowthereare1000mgin1g(or1mgis.001g)soIaddthenumbersinandperformthemultiplication.Thereare5sigfigsintheanswer.Answer4.4504x106mg

4450.4 g x

1000mg1 g

= 4.4504x106 mg or 4450.4 g x1mg

.001 g= 4.4504x106 mg

44. Expressthevolume622.6cm3inliters.Alwaysstartbywritingdownwhatyouaregiven,inthiscase,622.6cm3.Rememberthatcm3isthesamethingasmL.Changecm3tomLandthenthink—I

wanttocreateaconversionfactorthatgetsridofmLandcreatesLsoIwillputmLonthebottomsoitcancels,andLontopsoitstays.Iknowthereare1000mLin1L(or1mLis.001L)soIaddthenumbersinandperformthecalculation.Thereare4sigfigsintheanswer.Answer.6626L

622.6cm3 = 622.6 mL x

1L1000 mL

= .6626L or 622.6cm3 = 622.6 mL x.001L1mL

= .6626L

45. Convert10.6m3tomm3.Alwaysstartbywritingdownwhatyouaregiven,inthiscase,10.6m3.Thenthink—Iwanttocreateaconversionfactorthatgetsridofm3andcreatesmm3soIputm3inthedenominatoroftheconversionfactorsoitcancels,andmm3soitstays.Iknowthatthereare1000mmin1msoIaddthesenumbersin.However,becausetheunitsarecubed,thevaluesalsoneedtobecubed.Thereare3sigfigsintheanswer.Answer1.06x1010mm3

10.6 m3 x

10003 mm3

1m3= 1.06x1010 mm3

46. Thepressureoftheearth'satmosphereatsealevelis14.7lb/in2.Whatisthepressurewhenexpresseding/m2?(2.54cm=1in.,2.205lb=1kg)Alwaysstartbywritingdownwhatyouaregiven,inthiscase,14.7lb/in2.Thenthink—Ineedasetofconversionfactorsthatisgoingtoconvertpoundstogramsandsquareinchestosquaremeters.Startbywritingdownthegivenvaluewithlbinthenumeratorandin2inthedenominator.Ialwaysbeginbycancellingoutthevalueinthenumerator,soweknowwewanttogetridoflbandcreateg.But,theconversionfactorwearegivenconvertslbtokg,sowewillhavetodothisfirstandthenconvertkgtog.Becausewewanttocancellbputthisinthedenominatoroftheconversionfactor;becausewewanttocreatekgputthisinthenumerator;thenaddinthecorrespondingvaluesoftheconversionfactorthataregiven.Then,inthenextfactorwewanttogetridofkgandcreategsoputkginthedenominatorandginthenumerator.Weknowthat1kgequals1000gsoputthesenumbersin.Nowthatwehavecreatedg,gobackandgetridofin2,andcreatem2.Unfortunately,theconversionfactorwearegivenonlychangescmtoin,sowewillfirsthavetochangein2tocm2,andthencm2tom2.Becausewewanttocancelin2,intheconversionfactorputin2inthenumerator,andthenputcm2inthedenominator.Thenputthevaluesgivenfortheconversionfactor,rememberingthatbecausetheunitsaresquared,thevaluesmustalsobesquared.Then,inthenextfactorwewanttogetridofthecm2,soputitinthenumerator,andwewanttocreatem2soputitinthedenominator.Weknowthatthereare100cmin1msoputthesevaluesin,rememberingtosquarethevalues.Finally,performthecalculation.Thereare3sigfigsintheanswer.Answer1.03x107g/m2

14.7 lb

2.205 lbx

1000g1kg

x1 in2

2.542 cm2x

1002 cm2

1 m2 = 1.03x107 gm2

47. Convert4315mLtoqts.(1L=1.06qt)Alwaysstartbywritingdownwhatyouaregiven,inthiscase,4315mL.Thenthink—IneedaconversionfactorthatgetsridofmLandcreatesquartssoIputmLinthedenominatorofconversionfactorsoitcancels.But,Ican’tputqtinthenumeratorbecauseIhaven’tbeen

toldhowmanymLisinaquart.ButIhavebeentoldhowmanyLareinaquartandIknowIcanconvertmLtoL.So,IputLinthenumerator.Iknowthereare1000mLin1LsoIaddthesenumberstotheconversionfactor.Then,IwanttogetridofLsoIputthisinthedenominatorofthenextconversionfactorsoitcancels.IwanttocreateqtsoIputthisinthenumerator.Ihavebeengiventhat1L=1.06qt,soIaddthesenumberstotheconversionfactor.ThenIperformthemultiplications.Thereare3sigfigsintheanswer.Answer4.57qt

4315 mL x

1 L1000 mL

x1.06qt

1 L= 4.57 qt

48. Convert24.3lbtog.(1lb=453.6g)Alwaysstartbywritingdownwhatyouaregiven,inthiscase,24.3lb.Thenthink—IneedaconversionfactorthatgetsridofpoundsandcreatesgsoIputlbinthedenominatoroftheconversionfactorsoitcancels.ThenIputginthenumeratorsoitstays.IhavebeengivenaconversionfactorthatconvertspoundsdirectlygramssoIjustaddthecorrespondingnumberstotheconversionfactor.Thenperformthemultiplicationtocompletetheproblem.Thereare3sigfigsintheanswer.Answer1.10x104g

24.3 lb x

453.6g1 lb

= 1.10x104 g

49. Convert72.3mitokm.(1m=1.094yds,1mi=1760yds)Alwaysstartbywritingdownwhatyouaregiven,inthiscase,72.3mi.Thenthink—IneedaconversionfactorthatgetsridofmiandcreateskmsoIputmiinthedenominatorofconversionfactorsoitcancels.But,Ican’tputkminthenumeratorbecauseIhaven’tbeentoldhowmanymiisinakm.ButIhavebeentoldhowmany1760ydarein1miandIhavealsobeentoldhowmanyydsarein1m.So,Iputydinthenumeratorandaddthecorrespondingnumberstotheconversionfactor.Then,IwanttogetridofydsoIputthisinthedenominatorofthenextconversionfactorsoitcancels.IwanttocreatemsoIputthisinthenumerator.Ihavebeengiventhat1m=1.094ydsoIaddthesenumberstotheconversionfactor.But,westillhavetogettokm.Forthelastconversionfactorweneedtocancelmsoweputthisinthedenominator.Wewanttocreatekmsoweputthisinthenumerator.Weknowthat1kmequals1000msoweaddthesenumberstotheconversionfactor.ThenIperformthecalculations.Thereare3sigfigsintheanswer.Answer1.16x102km

72.3 mi x

1760 yd

1m1.094 yd

1000 m= 1.16 x 102 km

50. Thedensityofliquidchloroformis1.48g/mL.Whatisitsdensityinunitsoflb/in3?(2.54cm=1in.,2.205lb=1kg)Alwaysstartbywritingdownwhatyouaregiven,inthiscase,1.48g/mL,ginthenumeratorandmLinthedenominator.Butwait—havingreadthroughtheproblemyourealizethattheunitsofvolumeyouneedwillbeincubicunits—so,rightfromthestart,becausemLandcm3arethesamething,changemLinthedenominatortocm3.Thenthink—IneedasetofconversionfactorsthatisgoingtoconvertgramstopoundsandmLtocubicinches.Ialwaysbeginbycancellingoutthevalueinthenumerator,soweknowwewanttogetridofgandcreatelb.But,theconversionfactorwearegivenconvertslbtokg,sowewillhavetoconvertgtokgfirst,andthenconvertkgtolb..Becausewewanttocancelgputthisinthe

denominatoroftheconversionfactor;becausewewanttocreatekgputthisinthenumerator.Weknowthat1kgequal1000gsoputthesenumbersintheconversionfactor.Then,inthenextfactorwewanttogetridofkgandcreatelbsoputkginthedenominatorandlbinthenumerator.Wearegiventhat1kgequals2.205lb,soputthesenumbersin.Nowthatwehavecreatedlb,gobackandgetridofcm3,andcreatein3.Becausewewanttocancelcm3,intheconversionfactorputcm3inthenumerator,andthenputin3inthedenominator.Thenputthevaluesgivenfortheconversionfactor,rememberingthatbecausetheunitsarecubed,thevaluesmustalsobecubed.Finally,performthecalculation.Thereare3sigfigsintheanswer.Answer5.35x10-2lb/in3

1.48 g

1000 gx

2.205lb1kg

x2.543 cm3

1 in3 = 5.35x10−2 lbin3

51. Convert0.0898ft3toL.(2.54cm=1in.,1L=1dm3)Alwaysstartbywritingdownwhatyouaregiven,inthiscase,.0898ft3.Thenthink—Ineedaconversionfactorthatgetsridofft3andcreatesLsoIputft3inthedenominatorofconversionfactorsoitcancels.But,Ican’tputLinthenumeratorbecauseIhaven’tbeentoldhowmanyLisinaft3.ButIhavebeentoldhowmanycmarein1in,soIcanstartbyconvertingft3toin3,andthenconvertthattocubiccm.So,createcubicinchesinthenumerator.Then,youknowthatthereare12inchesin1ft—addthesenumbersinrememberingyouhavetocubethevaluesbecausetheunitsarecubed.Then,convertcubicinchestocubiccm—putcubicinchesinthedenominatorofthenextconversionfactorsoitcancelsandcubiccminthenumeratortocreateit.Addinthegivennumbersforeach—2.54cmand1in—again,rememberingtocubethesevalues.Nownotice,wearenotgivenhowmanycubiccmarein1L,wearegiventhat1dm3equals1L.So,wehavetoconvertcubiccmtocubicdm.So,inthenextconversionfactorputcubiccminthedenominatortocancelitandcubicdminthenumerator.Thereare10cmin1dmsoputthesenumbersin,rememberingtocubethem.Finally,convertcubicdmtoLbycancellingdmandcreatingLinaconversionfactor.Completethecalculations.Thereare3sigfigsintheanswer.Answer2.54L

.0898 ft3 x

123 in3

1 ft3x

2.543 cm3

1 in3x

103 cm3x

1dm3= 2.54L

52. InMarch2008,goldreachedamilestonevalueof$1000pertroyounce.Atthatprice,whatwasthecostofagramofgold?(1troyounce=31.10g)Youwerenotrequiredtoshowyourworktofindthisanswer.Ifyoudidnot,hopefullyyouwereabletoquicklyrealizethatyousimplyneededtoestimatetheanswerbydividing$1000byabout30g,whichwouldgiveyouanestimateofabout$33pergram,oranswer(c).Ifyoucalculatedthisanswer,alwaysstartbywritingdownwhatyouaregiven,inthiscase$1000pertroyounce,withdollarsinthenumeratorandtroyouncesinthedenominator.Thenthink—IneedtokeepdollarssoIdon’twanttocreateaconversionfactorforit.But,Ineedaconversionfactorthatgetsridoftroyouncesandcreatesgrams.So,becausetroyouncesisinthedenominator,tocancelitputtroyouncesinthenumeratorintheconversionfactor.Then,tocreategramsputgramsinthedenominator.Finally,addthecorrespondinggivennumbersfortheconversionfactor.Completethecalculations.Togettherightanswer

youonlyneededanestimate.However,forthiscalculation,therewouldbe4sigfigs.AnswerC

$10001 troy ounce

x1 troy ounce

31.10 g= $32.15

53. Itisestimatedthaturaniumisrelativelycommonintheearth'scrust,occurringinamountsof4g/metricton.Ametrictonis1000kg.Atthisconcentration,whatmassofuraniumispresentin1.2mgoftheearth'scrust?a) 5ngb) 5µgc) 5mgd) 5×10–5ge) 5cgAlwaysstartbywritingdownwhatyouaregiven.Inthiscasefiguringoutwhatyouaregivenmaybedifficult.Whatyouarereallygiventostartwithisthatthereare1.2mgofcrust.Thenthink—IYouwantaconversionfactorthatgetsridofmgofcrust,andcreatesgramsofUranium.So,putmgcrustinthedenominatortomakeitcancel.But,youcannotputgramsofUraniuminthenumeratorbecauseyouhavenotbeentoldhowmanygramsofuraniumare1mgofcrust.ButyouhavebeentoldhowmanygramsofUraniumarein1metricton.Therefore,youshouldputmetrictonsofcrustinthenumeratoroftheconversionfactor.Nowyouneedtoreasonouthowmanymgarein1metrictonsoyoucancompletetheconversionfactor.Youhavebeengiventhat1metrictonis1000kg.Youalsoknowthat1kgis1000g.Eachoftheseisafactorof103.Therefore,1metrictonisequalto106g.But,wewantmg.Youknowthatthereare1000mgin1g,sothisisanotherfactorof103.1metrictonisequivalentto109mg.Sonowwecanputthesenumbersintoourconversionfactor.NowwecancreateaconversionfactorthatcancelsmetrictonsandcreatesgramsofUranium.Theconversionfactoryouhavebeengivenstatesthatthereare4gUforevery1metrictonofcrust—so,putmetrictonsinthedenominatorsoitcancels,and4gUinthenumeratorsowekeepthis.Thencompletethecalculations..Thereis1sigfigintheanswer.Answer5ng

1.2 mg crust x

1 metric ton crust

109 mg crustx

4 g U1 metric ton crust

= 5 ng

54. A20.0mLsampleofglycerolhasamassof25.4grams.Whatisthedensityofglycerolinounces/quart?(1.00ounce=28.4grams,and1.00liter=1.06quarts)Alwaysstartbywritingdownwhatitisthatyouaregiven.Inthiscase,determiningthiscouldbeconfusing.Youaregiventhatthereisatotalmassof25.4gramsandyouaretoldthatitiscontainedinavolumeof20.0mL.Thisbecomesyourgivenvaluewiththemassinthenumeratorandvolumeinthedenominator—valuesarealreadyintheformofdensity!Thenthink—IneedasetofconversionfactorsthatisgoingtoconvertgramstoouncesandmLtoquarts.Ialwaysbeginbycancellingoutthevalueinthenumerator,soweknowwewanttogetridofgandcreateounces.Beginbyplacingginthedenominatorsothatitcancelsandozinthenumerator.Wehavebeentoldthatthereare28.gin1.00ouncessoplacethesenumbersintheconversionfactor.TocreatetheotherconversionweknowwewanttogetridofmLsoweputmLinthenumeratoroftheconversionfactortocancel—but,wearenottoldhowmanymLareinaquartintheconversionwearegiven,sowewillfirst

havetochangemLtoL.So,putLinthedenominatortocreateit.Weknowthereare1000mLin1Lsoaddthesenumbersin.ThenwecanconvertLtoquartswithafinalconversionfactor.WewanttocancelLsoputLinthenumerator.Wewanttocreatequartssoweputquartsinthedenominator.Wehavebeentoldthat1.06qtisequalto1.00L.Addthesenumbersin.Completethecalculations.Thereare3sigfigsintheanswer.Answer42.2oz/qt

25.4 g

20.0 mLx

1.00 oz28.4 g

x1000 mL

1.00 L1.06 qt

= 42.2ozqt

55. Duringaphysicsexperiment,anelectronisacceleratedto87percentofthespeedoflight.Whatisthespeedoftheelectroninmilesperhour?(speedoflight=3.00×108m/s,1km=0.6214mi)Alwaysstartbywritingdownwhatitisthatyouaregiven.Inthiscase,determiningthiscouldbeconfusing.Youaregiventhatanelectronisacceleratedto87%ofthespeedoflight,whichisthesamethingas.87xthespeedoflight,or,withtheactualvalue,.87x(3.00x108m/s)Thisbecomesyourgivenvaluewithmetersinthenumeratorandsecondsinthedenominator.Thenthink—Ineedtocreateasetofconversionfactorsthatconvertmetersintomilesandsecondsintohours..Ialwaysbeginbycancellingoutthevalueinthenumerator,soweknowwewanttogetridofmandcreatemi.Beginbyplacingminthedenominatorsothatitcancels.However,wecannotputmiinthenumeratorbecausewehavenotbeentoldhowmanymetersareinamile—but,wehavebeentoldhowmanykmareinamile,sowecanfirstconvertfrommtokm.So,createkmbyputtingitinthenumerator.Iknowthereare1000minakmsoIaddthesenumberstotheconversionfactor.Thenconvertkmtomi—putkminthedenominatoroftheconversionfactorsoitcancels.Nextputmiinthenumeratortocreateit.Aswearegiventhat1kmequals.6214mi,addthesenumbersin.Nowthatmetershasbeenconvertedtomiles,nextconvertsecondstohours.Youshouldalreadyknowaconversionfactorforthis—1hris3600seconds.Placesecondsinthenumeratoroftheconversionfactortocancelthis.Thenplacehrinthedenominatortocreateit.Finally,placethecorrespondingnumbers.Completethecalculations.Thereare2sigfigsinthefinalanswer.Answer5.8x108mi/hr

.87x(3.00x108 m)s

1000mx

.6214mi1km

x3600s1hr

= 5.8x108 mihr

56. Inthespringof2008,petrolcost£1.029perlitreinLondon.Onthesameday,theexchangeratewas$1=£0.497.WhatwasthepriceofLondonpetrolindollars($)pergallon?(1gal=3.7854L)Alwaysstartbywritingdownwhatitisthatyouaregiven,inthiscase,£1.029perL,with£1.029inthenumeratorand1Linthedenominator.Thenthink-Ineedasetofconversionfactorsthatconverts£to$andLtogal.Ialwaysbeginbycancellingoutthevalueinthenumerator,soweknowwewanttogetridof£andcreate$.Beginbyplacing£inthedenominatorsothatitcancelsand$inthenumerator.Wehavebeentoldthatthereare.497£per1dollarsoplacethesenumbersintheconversionfactor.Then,toconvertLtogallonswewanttogetridofLsowewillplacethisinthenumeratoroftheconversionfactorsoitcancels.Then,tocreategallonswewillplacethisinthedenominator.Wearetoldthat1gallonequal3.7854Lsoaddthesenumbersin.Completethecalculations.Thereare3sigfigsinthefinalanswer.Answer$7.84/gal

£ 1.0291 L

£ .497x

3.7854 L1 gal

= $7.84gal

57. Forspringbreakyouandsomefriendsplanaroadtriptoasunnydestinationthatis1705milesaway.Ifyoudriveacarthatgets32milespergallonandgascosts$3.399/gal,abouthowmuchwillitcosttogettoyourdestination?Alwaysstartbywritingdownwhatitisthatyouaregiven,inthiscase,1705mi.Thenthink—Ineedaconversionfactorthatgetsridofmilesandcreatesgallons,andthenIneedaconversionfactorthatgetsridofthegallonsandcreatesdollars.Tocreatethefirstconversionfactorplacemilesinthedenominatorsoitcancels.Thenplacegallonsinthenumeratortokeepit.Wearegiventhateachgallonisequivalentto32miles,soaddthesenumbersintotheconversionfactor.Tocreatethesecondconversionfactorwewanttogetridofgallonssoplacethisinthedenominatorsoitcancels.Thenplacedollarsinthenumeratortokeepit.Wearegiventhat1gallonisequivalentto3.399dollars,soaddthesenumbersintotheconversionfactor.Completethecalculations.Thereare2sigfigsinthefinalcalculation.Answer$180

1705 mi x

1 gal32 mi

x$3.3991 gal

= $180

58. Convert7.1kgtolb.(1kg=2.205lb)Alwaysstartbywritingdownwhatitisthatyouaregiven,inthiscase,7.1kg.Thenthink—Ineedaconversionfactorthatgetsridofkgandcreatespounds,soIputkginthedenominatorsoitcancels.Thentocreatepounds,Iputthisinthenumerator.Ihavebeengiventhat1kgequals2.205poundssoIplacethesenumbersintheconversionfactor.Completethecalculations.Thereare2sigfigsinthefinalcalculation.Answer16lb

7.1 kg x

2.205lb1kg

= 16 lb

59. Manganesemakesup1.3×10–4percentbymassoftheelementsfoundinanormalhealthybody.Howmanygramsofmanganesewouldbefoundinthebodyofapersonweighing261lb?(2.205lb=1kg)Alwaysstartbywritingdownwhatitisthatyouaregiven.Inthiscase,determiningthiscouldbeconfusing.Youaregiventhat1.3x10-4%ofa261lbbodyismanganese.Becauseapercentisafraction,thisisthesameassaying(1.3x10-6)x261lbismanganese,andthisbecomesyourgivenvalue.Thenthink—startingwiththisnumberofpounds,Ineedaconversionfactorthatgetsridoflbandgivesmegrams.Togetridofpounds,placeitinthedenominatorsoitcancels.But,wecannotyetputgramsinthenumeratortocreatethisbecausewehavenotbeentoldhowmanygramsarein1pound.Wehavebeentold,however,thatthereare2.205lbin1kg,sowewillstartbyputtingkginthenumerator.Thenwecanconverttograms.Placekginthedenominatorofthenextconversionfactorsoitcancels.Thenplaceginthenumerator.Weknowthatthereare1000gin1kgsoaddthesenumbersintothisconversionfactor.Answer.15g

(1.3 x 10−6 )(261 lb) x

1kg2.205lb

x1000g

1kg= .15g

60. In1928,44.9gofanewelementwasisolatedfrom660kgoftheoremolybdenite.Thepercentbymassofthiselementintheorewas:

Thepercentmassofasubstancewouldsimplybethefractionofanelementinasampleoforex100togiveapercent.Thefractionofthenewelementwouldbe44.9goutof660kg,or44.9goutof660,000g.Thereare2sigfigsinthefinalcalculation.Answer6.8x10-3%

44.9g660,000g

x100 = 6.8 x 10−3 %

61. 424KelvinequalswhatindegreesCBecausethesamenumberofdegreesexistsbetweenthefreezingandboilingpointinboththeKelvinandCelsiusscale(1000),thereisnoneedtouseaconversionfactor.However,theCelsiusscaleisoffsetfromtheKelvinscaleby2730.0intheKelvinscaleisabsolute0andso,thetheoreticalpointatwhichnokineticenergyremainsinparticlesatthattemperature.The00pointoftheCelsiusscaleis2730abovethis(actually,273.150).Therefore,toconvertfromKdegreestoCdegrees,oneneedstosubtract2730(0C=K–2730).ToconvertfromCdegreestoKdegrees,oneneedstoadd2730(K=0C+273).So,forthisproblem,toget0Cweneedtosubtract2730fromthegiventemperature.Answer1510C

0C = K − 2730 = 424 K − 2730 = 1510C 62. Themeltingpointofacertainelementis346°C.WhatisthisontheFahrenheitscale? (T°F=T°C×(9°F/5°C)+32°F)Inthisproblemyouarealreadygivenaformulatoconvert0Cto0F.Simplyplugthenumbersinandapplytheformula.Answer6550C

x90 F50C

⎛⎝⎜

⎞⎠⎟+ 320 = 3460C x

90 F50C

⎛⎝⎜

⎞⎠⎟+ 320 = 6550C

a) 465°Fb) 224°Fc) 896°Fd) 655°Fe) 591°F63. Convert:–46.9°C=____________°F.(T°F=T°C×(9°F/5°C)+32°F)Inthisproblemyouarealreadygivenaformulatoconvert0Cto0F.Simplyplugthenumbersinandapplytheformula.Answer-52.40C

x90 F50C

⎛⎝⎜

⎞⎠⎟+ 320 = −46.90C x

90 F50C

⎛⎝⎜

⎞⎠⎟+ 320 = −52.40C

64. Aswarmwatersitsinacoolroom,youmeasurethetemperaturechange(ΔT=Tfinal–Tinitial).Whichofthefollowingistrue?a) Thetemperaturechange(ΔT)isbiggerifyouaremeasuringin°F.b) Thetemperaturechange(ΔT)isbiggerifyouaremeasuringin°C.c) Thetemperaturechange(ΔT)willbethesameregardlessofthescaleyouuse.d) AnswerAorBiscorrect,dependingonthedifferenceintemperaturebetweenthewaterandtheroom.e) Noneoftheabove.Notethatthewaterisbetweenthefreezingandboilingpoints.Thetemperaturedifferencebetweenthefinalandinitialtemperatureswilldependonthenumberofdegreesthatseparatethefreezingandboilingtemperaturesintheparticularscale.Thegreaterthenumberofdegreesseparatingthefreezingandboilingpoints,thegreaterthedifferencebetweenthefinalandinitialtemperatures.Becausetherearemoredegreesseparatingthe

thefreezingandboilingpointintheFscalecomparedtotheCscale,thetemperaturechangewillbebiggerifmeasuringintheFscale.AnswerA65. Themeltingpointofpicolinicacidis136.5°C.WhatisthemeltingpointofpicolinicacidontheFahrenheitscale?Inthisproblemyouarealreadygivenaformulatoconvert0Cto0F.Simplyplugthenumbersinandapplytheformula.Answer277.70C

x90 F50C

⎛⎝⎜

⎞⎠⎟+ 320 = 136.50C x

90 F50C

⎛⎝⎜

⎞⎠⎟+ 320 = 277.70C

(T°F=T°C×(9°F/5°C)+32°F)66. In1984,somedrumsofuraniumhexafluoridewerelostintheEnglishChannel,whichisknownforitscoldwater(about15°C).Themeltingpointofuraniumhexafluorideis148°F.Inwhatphysicalstateistheuraniumhexafluorideinthesedrums?(T°F=T°C×(9°F/5°C)+32°F)Thisproblemrequiresthatyoufirstmakeacalculationtoconvert0Cinto0F,thencomparetheFtemperatureofthewaterwiththatofthemeltingpointofuraniumhexafluoridetoseewhetheritisaboveorbelowthemeltingpoint,andthussolidorliquid.Youplug150Cintothegivenformulatofindoutthatthetemperatureofthewateris590C,whichissignificantlylessthanthemeltingpointofuraniumhexafluoride.Thismeansthissubstanceissolid.Becausethisisamultiplechoicequestionyoudidnotneedtoshowthecalculation.AnswerA

x90 F50C

⎛⎝⎜

⎞⎠⎟+ 320 = 150C x

90 F50C

⎛⎝⎜

⎞⎠⎟+ 320 = 590C

a) solidb) liquidc) gasd) amixtureofsolidandliquide) notenoughinformation67. Themeltingpointofindiumis156.2°C.At323°F,whatisthephysicalstateofindium?(T°F=T°C×(9°F/5°C)+32°F)Thisproblemrequiresthatyoufirstmakeacalculationtoconvert0Finto0C,thencomparetheCtemperatureofthewaterwiththatofthemeltingpointofindiumtoseewhetheritisaboveorbelowthemeltingpoint,andthussolidorliquid.Youplug3230FintothegivenformulaAFTERmanipulatingtheformulatoisolate0C,tofindoutthatthetemperatureofthetemperaturein0Cis1620.Thisisslightlygreaterthanthemeltingpointofindium,suggestingitisaliquid.However,wearen’tgiventhissubstance’sboilingpointandsodon’tknowatthistemperatureifindiumisaliquidoragas,sothereisnotenoughinformationtobecertainwhatthephysicalstateis.Becausethisisamultiplechoicequestionyoudidnotneedtoshowthecalculation.AnswerD

x90 F50C

⎛⎝⎜

⎞⎠⎟+ 320 → T

= (T0F

− 320 F ) x50 F90C

⎛⎝⎜

⎞⎠⎟= (3230 F − 320 F )x

50 F90C

⎛⎝⎜

⎞⎠⎟= 1620C

a) Solid.b) Liquid.c) Gas.d) Notenoughinformation.e) At323°F,theindiumispartiallysolidandpartiallyliquid;thereisanequilibriumbetweenthetwostates.

68. Onanewtemperaturescale(°Z),waterboilsat120.0°Zandfreezesat40.0°Z.Calculatethenormalhumanbodytemperatureusingthistemperaturescale.OntheCelsiusscale,normalhumanbodytemperaturecouldtypicallybe37.1°C,andwaterboilsat100.0°Candfreezesat0.00°C.Torelateonetemperaturescaletoanotheryouknowthattheamountofenergyrepresentedbythetemperaturesbetweenthefreezingandboilingpointofwateristhesameregardlessofthetemperaturescale.So,thenumberofdegreesbetweenfreezingandboilinginonescaleisequivalenttothenumberofdegreesbetweenfreezingandboilinginanotherscale,andweshouldbeabletocreateaconversionfactortoallowustoconvertbetweenthescales.Todothissimplyputthedifferenceindegreesbetweenboilingandfreezinginonescaleinthenumeratorofaconversionfactorandthedifferenceindegreesbetweenboilingandfreezingintheotherscaleinthedenominator.Initially,itdoesnotmatterwhichisthenumeratorordenominator.Youcanuseavisualaidtohelpifnecessary.Iwriteouttheendsofeachscaleandthedifferenceindegreesbetweenthesenumbersinthemiddle.Forthisexamplethishelpsmeseethatforevery80.0degreesoftheZscale,Ihave100.0degreesoftheCscale,andthisbecomesmyconversionfactor,whichIthenwriteouttotheside.So,ifIamgivenZdegrees(say500Z)andIwantCdegrees,IwouldwanttocancelZandcreateC,soIwouldmultiply:

50.00 Z x

100.00C80.00 Z

IfIamgivenCdegrees(say500C)andIwantZdegrees,IwouldwanttocancelCandcreateZ,soIwouldmultiply:

50.00 C x

80.00 Z100.00 C

Thereisoneotherthingtodealwith—thescalesdonothavethesame“zero”point—ifeither(orboth)scale(s)donotbeginatzero,wehavetoadjustforthis.Tomakethisadjustment,whereverthevalueforthescalethatdoesnotstartatazeropointappearsintheequation,subtractthedifferencefrom0fromthisvalue.Usingtheaboveexample,astheCscalestartswith0,itdoesn’tneedadjustment.But,anytimeavalueforZisstated,youwillhavetosubtract40.00.Fortheaboveexample,ifIamgivenCdegrees(say500C)andIwantZdegrees:

0 Z − 40.00 Z = 500C x

80.00 Z100.00C

→ 0Z = 500C x80.00 Z

100.00C+ 40.00 Z = 800 Z

Fortheexample,ifIamgivenZdegrees(say500Z)andIwantCdegrees:

0C = (50.00 Z − 40.00 Z ) x

100.00C80.00 Z

= 12.50C

Forthecurrentproblem,combiningtheuseoftheconversionfactorandadjustmenttothe0point,youaregiventheCtemperatureof37.1andaskedfortheZtemperature.Soyou

createanequationthatsays“degreesZ(-40.00toadjustto00)equalstheCtemperaturetimestheconversionfactorthatcancelsCandcreatesZ.”Answer69.70Z

0 Z − 40.00 Z = 37.10C x

80.00 Z100.00C

→ 0Z = 37.10C x80.00 Z

100.00C+ 40.00 Z = 69.70 Z

69. ThecalibrationpointsforthelinearReaumurscalearetheusualmeltingpointoficeandboilingpointofwater,whichareassignedthevalues0°Rand80°R,respectively.Theboilingpointofbromineis58.8°F.Whatisthistemperaturein°R?Torelateonetemperaturescaletoanotheryouknowthattheamountofenergyrepresentedbythetemperaturesbetweenthefreezingandboilingpointofwateristhesameregardlessofthetemperaturescale.So,thenumberofdegreesbetweenfreezingandboilinginonescaleisequivalenttothenumberofdegreesbetweenfreezingandboilinginanotherscale,andweshouldbeabletocreateaconversionfactortoallowustoconvertbetweenthescales.Todothissimplyputthedifferenceindegreesbetweenboilingandfreezinginonescaleinthenumeratorofaconversionfactorandthedifferenceindegreesbetweenboilingandfreezingintheotherscaleinthedenominator.Initially,itdoesnotmatterwhichisthenumeratorordenominator.Youcanuseavisualaidtohelpifnecessary.Iwriteouttheendsofeachscaleandthedifferenceindegreesbetweenthesenumbersinthemiddle.Forthisexamplethishelpsmeseethatforevery80degreesoftheRscale,Ihave180degreesoftheFscale,andthisbecomesmyconversionfactor,whichIthenwriteouttotheside.So,ifIamgivenRdegrees(say500R)andIwantFdegrees,IwouldwanttocancelRandcreateF,soIwouldmultiply:

50.00 R x

180.00 F80.00 R

IfIamgivenFdegrees(say500F)andIwantRdegrees,IwouldwanttocancelFandcreateR,soIwouldmultiply:

50.00 F x

80.00 R180.00 F

Thereisoneotherthingtodealwith—thescalesdonothavethesame“zero”point—ifeither(orboth)scale(s)donotbeginatzero,wehavetoadjustforthis.Tomakethisadjustment,whereverthevalueforthescalethatdoesnotstartatazeropointappearsintheequation,subtractthedifferencefrom0fromthisvalue.Usingtheaboveexample,astheRscalestartswith0,itdoesn’tneedadjustment.But,anytimeavalueforFisstated,youwillhavetosubtract32.00.Fortheaboveexample,ifIamgivenRdegrees(say500R)andIwantFdegrees:

0 F − 32.00 Z = 500 R x

180.00 F80.00 R

→ 0F = 500 R x180.00 F80.00 R

+ 32.00 F = 1450 F

Fortheexample,ifIamgivenFdegrees(say500F)andIwantRdegrees:

0 R = (50.00 F − 32.00 F ) x

80.00 R180.00 F

= 80 R

Forthecurrentproblem,combiningtheuseoftheconversionfactorandadjustmenttothe0point,youaregiventheFtemperatureof58.8andaskedfortheRtemperature.Soyoucreateanequationthatsays“degreesRequalstheF(-32.00toadjustto00)temperaturetimestheconversionfactorthatcancelsFandcreatesR.”Answer11.90R

0 R = (58.80 F − 32.00 F ) x

80.00 R180.00 F

= 11.90 R

70. Amonolayercontaining3.23×10–6gofoleicacidhasanareaof20.0cm2.Thedensityofoleicacidis0.895g/mL.Whatisthethicknessofthemonolayer(thelengthofanoleicacidmolecule)?Youneedtounderstandbyreadingthroughthequestionthatthegiveninformationisaskingyoutovisualizeasinglelayer(amonolayer)ofmoleculesofoleicacid.Theyhavegivenyoutheareaofonesurfaceofthelayer,andareaskingyoutofindthethickness—inotherwords,thethirddimensionofthevolumetakenupbythismonolayer.Aquickdrawingofthesituationwouldprobablybehelpful: Youaregiventheabilitytofindthetotalvolumeusingthedensity.If thedensityis.895g/ml,andthereare3.23x10-6ginthesample,you canfindthetotalvolume:

Vso V = m

D= 3.23x10−6 g

.895 g / mL= 3.61x10−6 mL

Sinceyouhavethetotalvolumeofthemonolayer,andyouaregiventhearea,youcandividethevolumebytheareatogetthethirddimension,thethickness.Tomaketheunitsmatch,youwillneedtoturnmLintocubiccm.Answer1.81x10-7cm

Thickness = V

A= 3.61x10−6 cm3

20cm2 = 1.81x10−7 cm

71. Thedensityofgasolineis0.7025g/mLat20°C.Whengasolineisaddedtowater:a) Itwillfloatontop.b) Itwillsinktothebottom.c) Itwillmixso,youcan'tseeit.d) Themixturewillimprovetherunningofthemotor.e) Noneofthesethingswillhappen.Itshouldbeintuitivethatifyouhaveliquidsofdifferentdensitiesthelessdensefluidwillfloatonthesurfaceofthemoredensefluid.Inthiscase,aswaterhasanapproximatedensityof1g/mL,itismoredensethangasoline,sogasolinewillfloatonthesurfaceofthewater.AnswerA72. Apieceofantimonywithamassof17.41gissubmergedin46.3cm3ofwaterinagraduatedcylinder.Thewaterlevelincreasesto48.9cm3.Thecorrectvalueforthedensityofantimonyfromthesedatais:Pleasemakesureyouunderstandthatthisisaverysimplewaytodetermineaveryimportantintensivepropertyofasubstance(anintensivepropertyisonewhichdoesnotchangenomatterhowmuchofthesubstanceyouhave).Measurethemassofthesubstance—thendeterminethesamplevolumebywaterdisplacement.Massdividedby

volumethenequalsthedensity.Wewentthroughawholelabexercisedesignedtogiveyoutheskillstounderstandtheuncertaintyandprecisioninvolvedinsuchameasurement.Inthiscase,tofindthevolumeofthesampleofantimony,thedifferenceinfinalandinitialvolumesisdetermined.Thenthemassisdividedbythisvalue:

V= 17.41g

48.9mL− 46.3mL= 6.7 g / mL

Atrickypointtothisquestionisthatthesubtractiononthebottomislimitedto1decimalplaceintermsofsignificantfigures.Becausetheresultofthissubtraction(2.6)onlyhas2significantfigures,thefinalansweralsohasonly2sigfigs.Usethefollowingtoanswerquestions73-74:Thedensityofaliquidisdeterminedbysuccessivelyweighing25,50,75,100,and125mLoftheliquidina250-mLbeaker.73. Ifvolumeofliquidisplottedalongthehorizontalaxis,andtotalmassofbeakerplusliquidisplottedontheverticalaxis:a) Thex,orhorizontal,interceptisthenegativevalueoftheweightofthebeaker.b) They,orvertical,interceptistheweightoftheemptybeaker.c) Theslopeofthelineis1.0.d) Thelinewillpassthroughtheorigin.e) Theslopeofthelineisindependentoftheidentityoftheliquid.ItmighthelptomakeaquicksketchoftheplotofthisinvestigationTheplotshouldbelinearbecausewewouldexpectthesameincrementofmassforevery25mLofliquidaddedtothebeaker.Becausetheslope(riseoverrun)wouldbeequivalenttomassovervolume,theslopeshouldreflectthedensity—butitisunlikelythatthiswouldbeavalueofone,unlesstheliquidiswater(eliminate(c)).So,theslopeisdependentontheidentityoftheliquid(eliminate(e)).Wherethelineinterceptsthexaxiswouldindicatebotha0massandnegativevolume(eliminate(a).Itshouldmakesensethatifthereare0mLofliquidinthebeakerthatthelinecreatedbytheplotwillintercepttheyaxisatthemassofthebeaker,sothelinecannotpassthroughtheorigin(eliminate(d),confirm(b)).AnswerB74. Consideringtheplotoftotalmass(y-axis)versusvolume(x-axis),whichofthefollowingistrue?a) Theplotshouldberatherlinearbecausetheslopemeasuresthedensityofaliquid.b) Theplotshouldbecurvedupwardbecausetheslopemeasuresthedensityofaliquid.c) Theplotshouldbecurvedupwardbecausethemassoftheliquidishigherinsuccessivetrials.d) Theplotshouldbelinearbecausethemassofthebeakerstaysconstant.e) Noneoftheabove.(a)True.Themeasurementofmassovervolumeisthedensityofasubstance.Asweareincrementingthevolumebythesameamountwitheachmeasurement(increasing25mL)wewouldpredictthatthemasswouldalsoincreasebythesameamountwitheachincrement,meaningthatthevalueofriseoverrun(theslope),whichwouldbeequivalenttothemassovervolume(density)wouldremainconstant—thismeansthattheplotshouldbelinear.(b)False.Becausethelineshouldbelinearbecausetheslopemeasuresthedensity,(see(a))thiswouldbeinconsistentwithstatement(b).(c)False.Whilethemassoftheliquidishigherinsuccessivetrials,itisincreasedbythesameamountwitheachtrial,andso,

therelationshipislinear.(d)False.Whilethemassofthebeakerstaysconstant,themassoftheliquidinthebeakerincreases,sowhilethemassofthebeakerisconstant,thisisnotwhytheplotislinear.(e)False.AnswerA75. A20.0mLsampleofglycerolhasamassof25.2grams.Whatisthemassofa51-mLsampleofglycerol?Thisproblemrequiresyoutofirstdeterminethedensityofglycerolbasedonthemassandvolumeofglycerolgiventoyou,andthen,usingthatdensity,findthemassofadifferentsampleofglycerolofadifferentvolume.Tofindthedensityofglycerolsimplydividethemassgivenbythevolumegiven:

Dglycerol =

= 25.2g20.0mL

= 1.26g / mL

Thenrearrangethedensityformulatoisolatemassandpluginthevaluefordensityandthenewvolumeofglyceroltofindthemass:

m =VDglycerol = (51mL) 1.26g

mL⎛⎝⎜

⎞⎠⎟= 64g

Thefinalanswerislimitedto2sigfigs.Answer64g76. Supposethatyoupurchasedawaterbedwiththedimensions2.55m×2.53dm×225cm.Whatmassofwaterdoesthisbedcontain?a) 1.45×103gb) 1.45×104gc) 1.45×105gd) 1.45×108ge) 1.45×106gTheproblemgivesyoutheinformationtofindthealargevolumeofwaterandyoushouldknowthatthestateddensityofwateris1.00g/cm3.Youcanfindthevolumeofthewaterbedincm3andthenmultiplythisbythedensityofwatertogetthenumberofgramsofwater.Firstconvertallofthedimensionvaluesintocm—youshouldbeabletodothisinyourhead.Thenmultiplybydensity.Answer1.45x107g

2.55m x 2.53 dm x 225 cm = 255 cm x 25.3 cm x 225cm = 1.452 x 106 cm3

Dwater =

mwater

Vso mwater =VDwater = (1.452 x 106 cm3) x 1.00 g / mL = 1.45x106 g

77. AfreightercarryingacargoofuraniumhexafluoridesankintheEnglishChannelinlateAugust1984.Thecargoofuraniumhexafluorideweighed2.251×108kgandwascontainedin30drums,eachcontaining1.47×106LofUF6.Whatisthedensity(g/mL)ofuraniumhexafluoride?ItisasimplemattertocalculatethedensityofUF6inkg/Lfromtheinformationgiven—justdividethegivenmassbythegivenvolume(rememberthatthereare30drumat1.47x106).However,thequestionwantstheanswering/mLsothisbecomesaconversionproblem.Itwouldbebettertonotdeterminedensitytobeginwith—justhaveitasthestartingvaluefortheconversion.Thestartingvaluewouldbe2.251x108kg(inthenumerator)per(30)(1.47x106L)(inthedenominator).Thenthink—IneedasetofconversionfactorsthatisgoingtoconvertkilogramstogramsandLtomL.Ialwaysbeginbycancellingoutthevalueinthenumerator,soweknowwewanttogetridofkgandcreateg.Beginbyplacingkginthedenominatorsothatitcancelsandginthenumerator.Iknowthatthereare1000gin1kgsoplacethesenumbersintheconversionfactor.Tocreatetheotherconversionweknowwe

wanttogetridofLsoweputLinthenumeratoroftheconversionfactortocancel,andthencreatemLbyputtingitinthedenominator.Weknowthatthereare1000mLin1Lsoweaddthesevaluesintothisconversionfactor.Completethecalculations.Thereare3sigfigsinthefinalvalue.Answer5.10g/mL

2.251 x 108 kg(30)1.47 x 106 L

x1000g

1L1000mL

= 5.10 g / mL

78. Theboilingofwaterisaa) physicalchangebecausethewatermerelydisappearsb) physicalchangebecausethegaseouswaterischemicallythesameastheliquidc) chemicalchangebecauseheatisneededfortheprocesstooccurd) chemicalchangebecauseagas(steam)isgivenoffe) chemicalandphysicaldamageAphysicalchangechangesthephysicalnatureofasubstancebutdoesnotchangethechemicalnatureofthesubstance.Whenweboilwaterwechangeitfromthephysicalstateofbeingaliquidtothatofbeingasolid,butthesubstanceisstillH2O,sothechemicalnatureisnotchanged.Therefore,(c),(d),and(e)areeliminated.(a)and(b)arebothtruestatements,but(b)offersamuchmorethoroughexplanationastowhythisisaphysicalchange.AnswerB79. Thestateofmatterforanobjectthathasadefinitevolumebutnotadefiniteshapeisa) solidstateb) liquidstatec) gaseousstated) elementalstatee) mixedstateRememberthatwecandefinehowdifferentstatesofmatterappearintermsofvolumeandshape.Gasestaketheshapeandtakeupthefullvolumeofspacetheyarecontainedinandsohavenodefinitevolumeorshape.Liquidstaketheshapeofthespacetheyarecontainedinandsohavenodefiniteshape,buttakeuponlythevolumethatcorrespondstotheirdensityandsohavedefinitevolume.Solidsdonottaketheshapeofthespacetheyarecontainedandsohavedefiniteshape,andtheytakeuponlythevolumethatcorrespondstotheirdensityandsohavedefinitevolume.AnswerB80. Thestateofmatterforanobjectthathasbothdefinitevolumeanddefiniteshapeisa) solidstateb) liquidstatec) gaseousstated) elementalstatee) mixedstateRememberthatwecandefinehowdifferentstatesofmatterappearintermsofvolumeandshape.Gasestaketheshapeandtakeupthefullvolumeofspacetheyarecontainedinandsohavenodefinitevolumeorshape.Liquidstaketheshapeofthespacetheyarecontainedinandsohavenodefiniteshape,buttakeuponlythevolumethatcorrespondstotheirdensityandsohavedefinitevolume.Solidsdonottaketheshapeofthespacetheyarecontainedandsohavedefiniteshape,andtheytakeuponlythevolumethatcorrespondstotheirdensityandsohavedefinitevolume.AnswerA

81. _________aresubstanceswithconstantcompositionthatcanbebrokendownintoelementsbychemicalprocesses.a) Solutionsb) Mixturesc) Compoundsd) Quarkse) HeterogeneousmixturesAnysubstancewithaconstantcompositionisnotamixture.Therefore,solutions,mixturesandheterogeneousmixturescanbeimmediatelyeliminated.Quarksaresubatomicparticlesthatmakeupothersubatomicparticlesandcannotbebrokendown.Thisleaves(c)astheonlyvalidchoice.However,thedefinitionshouldalsotellyouthisimmediatelywithouthavingtoeliminateanyanswers.Ifasubstancecanbebrokendownintoelementsitmusteitherbeacompoundoramixture.Ifthesubstancehasaconstantcomposition(definiteproportions)itmustbeacompoundbydefinition.AnswerC82. Amethodofseparationthatemploysasystemwithtwophasesofmatter,amobilephaseandastationaryphase,iscalleda) filtrationb) chromatographyc) distillationd) vaporizatione) homogenizationChromatographyisamethodofseparationinwhichcomponentsofafluidmixture,eachofwhichhasauniqueabilitytodevelopinterparticleforcesofattractionareplacedonastationarymedium(thestationaryphase).Thestationarymediumwiththeappliedmixtureisthenplacedinasolvent,ormobilephase,whichbydiffusion,cantravelthroughthestationaryphase.Becauseparticlesofthecomponentsofthemixturewilllikelybeabletodevelopdifferentforcesofattractionforparticlesofthestationaryphasevsparticlesofthemobilephase,thecomponentsofthemixturewillbecarriedthroughthestationaryphasebythesolvent,andsowillbecomeseparated.Filtrationisamethodofseparationthatseparatescomponentsofamixturebasedonwhetherthecomponentisasolidoraliquid.Distillationisamethodofseparationofliquidsthatisbasedondifferencesofboilingpoints.Vaporizationissimplytheprocessofcreatingavapor,notaseparationmethod.Homogenizationissimplytheprocessofmakingamixturemorehomogeneousbyblending,notaprocessofseparation.AnswerB83. Whichofthefollowingstatementsisfalse?a) Solutionsarealwayshomogeneousmixtures.b) Theterms“atom”and“element”canhavedifferentmeanings.c) Elementscanexistasatomsormolecules.d) Compoundscanexistasatomsormolecules.e) Atleasttwooftheabovestatements(A-D)arefalse.A)True—solutionsarestatedtobehomogeneousmixturesbecausetheparticlesofthecomponentsofthemixturesareuniformlymixturegivingthemixtureahomogeneousappearance.B)True—atomsisageneraltermreferringtothesmallestparticlesofanyelement,whilethetermelementreferstoasubstancecontainingatomsofaspecifictype.C)True—samplesofmostelementsconsistofcollectionsofindividualatomsthatareheldtogetherthroughforcesthatwewilllearnarecalled“metallic”bonds.However,samplesof

someelementsarecontaincollectionsoftwo(ormore)atomsheldtogetherwithcovalentbonds(throughthesharingofpairsofelectrons),inwhichcasetheparticlesoftheelementsareconsideredtobemolecules.D)False—bydefinition,theparticlesofcompoundsaremadeupofatomsoftwoormoretypesthatarechemicallybondedtogetherindefiniteproportions—thatis,theyaremadeofmolecules.E)Becauseonly(d)isfalse,thisstatementisalsofalse.AnswerD84. Anexampleofapuresubstanceisa) elementsb) compoundsc) purewaterd) carbondioxidee) alloftheseYouareawarethatmatterisdividedintotwotypesofsubstances—puresubstancesandimpuresubstances,whichwealsocallmixtures.Puresubstancesarefurtherdividedintotwocategories—thosethatcontainonlyatomsofonetype(elements)orpuresubstancesthatcontainatomsofmorethanonetype(compounds).Purewater(H2O)andcarbondioxideareexamplesofcompounds.Therefore,allfouranswers(a-d)areexamplesofpuresubstances.AnswerE85. Asolutionisalsocalledaa) homogeneousmixtureb) heterogeneousmixturec) puremixtured) compounde) distilledmixtureAsolutionisamixtureinwhichtheparticlesaresouniformlydistributedthatyoucannottellthatitisamixture—theonlysituationinwhichthiscanhappeniswhentheparticlesofthecomponentsofthemixtureareatomic/molecular/ionicsized.Becausetheparticlesareuniformlymixedwecallthisahomogeneousmixture.AnswerAUsethefollowingtoanswerquestions86-89:Considerthefollowingchoiceswhenansweringquestions86-89.

86. Whichbestrepresentsahomogeneousmixtureofanelementandacompound?

Amixtureofsubstanceswouldrequiretwoormoredistinctsubstances.Thiswouldallowustoeliminate(b)and(c)because(b)containsonlyatomsofonetype(singlegraycircles)and(c)containsonlyacompoundofonetype(eachparticlemadeofalargewhiteatomandasmallergrayatom).(a),(d),and(e)allhaveparticlesofmorethanonetype.Saying“homogeneousmixture”wouldallowustoeliminate(d)because,whilethereelementsoftwodifferenttypes(smallgrayatomsandsmallwhiteatoms),theyarenotuniformlymixed.Theparticlesin(a)and(e)ontheotherhand,appeartobeuniformlymixed.Saying“amixtureofanelementandacompound”wouldallowustoeliminate(a)(andselect(e))becausetheparticlesin(a)arebothelements,whiletheparticlesin(e)areandelement(whitesquares)andacompound(whitecircularatombondedtoagraycircularatom).AnswerE87. Whichbestrepresentsagaseouscompound?a) optionab) optionbc) optioncd) optionde) optioneSayingthat“acompoundisgaseous”impliesthattheparticlesarehaphazardlyarrangedinspacefarapartfromeachother.Thisallowsustoeliminate(b),(d),and(e)whicharealluniformlyarranged.Saying“compound”allowsustoeliminate(a)(andselect(c))because(a)containstwogaseouselements(smallwhitecircularatomsandsmallgraycircularatoms)while(c)containsgaseousmoleculesofasinglecompound(smallwhitecircularatomsbondedtolargergraycircularatoms).AnswerC88. Whichbestrepresentsasolidelement?a) optionab) optionbc) optioncd) optionde) optioneSayingelementimpliesthatallparticlesareofthesametype.Thisimmediatelyeliminates(a),(c),(d)and(e),regardlessofanythingelse,andselect(b).Additionally,theparticlesinasolidpuresubstanceshouldappearasbeinglockedinarigidrepeatingpattern,asin(b)and(e)—however,theparticlesin(e)representasolidmixtureoftwodifferentparticles,anelementandacompound.AnswerB89. Whichbestrepresentsaheterogeneousmixtureoftwoelements?a) optionab) optionbc) optioncd) optionde) optioneAmixtureofsubstanceswouldrequiretwoormoredistinctsubstances.Thiswouldallowustoeliminate(b)and(c)because(b)containsonlyatomsofonetype(singlegraycircles)and(c)containsonlyacompoundofonetype(eachparticlemadeofalargewhiteatomandasmallergrayatom).(a),(d),and(e)allhaveparticlesofmorethanonetype.Saying

“mixtureoftwoelements”wouldallowustoeliminate(e)becausethisisamixturebetweenandelement(smallwhitecircularatoms)andacompound(chemicallycombinedparticlesofsmallwhitecircularatomsandlargergraycircularatoms)while(a)and(d)aremixturesoftwoelements.Saying“heterogeneous”mixtureallowsustoeliminate(a)becauseitappearstobeauniformorhomogeneousmixturewhiletheparticlesin(d)appearnottobemixedinauniformfashion(alsonotethattheparticlesinthemixturein(e)aremixedinauniformfashion).AnswerD90. TFAllphysicalchangesareaccompaniedbychemicalchanges.Probablythebestexampleofthisbeingafalsestatementischangeofstate.Whenwaterchangesfromgastoliquidtosolid,thisrepresentsaphysicalchange,butthereisnochemicalchangeoccurringduringthesephysicalchanges—thesubstanceundergoingthechangeisstillH2Oinallcircumstances.AnswerFalse91. TFColorchangesalwaysindicateachemicalchange.Whilecolorchangesoftenresultwhensubstancesinteractchemical,changesincoloralsoaccompanymanyphysicalchanges.Itisnotuncommonforformationofsolutions(whichisaphysicalchange—acreationofamixture)toresultinacolorchange.Ifametalisheatedtoahigherandhighertemperature,thecolorofthemetalchangesfromadeepredcolortoanorange,thenyellow,thenbluetowhitecolor.Thesearejustacoupleofexamplesofphysicalchangesoccurringthatresultincolorchanges.Socolorchangesarenotlimitedtochemicalchanges.AnswerFalse92. Whatarethecomponentsofthescientificmethod?MakeobservationsaboutnaturalphenomenaObservationsmakeyouaskquestionsaboutwhythenaturalphenomenonworks thewayitdoesQuestionsleadtoyouproposingahypothesisaboutwhythenaturalphenomenon worksthewayitdoesHypothesisleadstoproposingtestsofthehypothesisResultsoftestleadyoutodrawconclusionsaboutthevalidityofthehypothesis (Claimandevidence)93. Garfield(weighing24lbs)tookaflighttothemoononthespaceshuttle.Asusual,hestuffedhimselfwithlasagnaduringtheentireflightandnappedwhenhewasn'teating.Muchtohisdelightwhenhegottothemoonhefoundheweighedonly6lbs.Heimmediatelyproclaimedaquickweightlossdiet.Explainthefallacyinhisreasoning.Assumegravityonthemoontobeaboutone-sixththatofEarth.Ifaquestionasksyoutoexplainthefallacyinsomeone’sreasoning,youreallyneedtoexplicitlystatethefallacy.InthiscaseyouneedtomakethepointinsomewaythatthefallacyinGarfield’sreasoningisthatweightisanindicatoroftheamountofmatterapersoncontains.Inreality,weightdependsontheforceduetogravity,andasgravityisdifferentindifferentlocations,itcannotbeusedtoindicatetheamountofmatter.Thisdependsonthemass,andGarfield’smasswillhaveatleaststayedthesame,ifnotincreased.94. Contrastthetermsprecisionandaccuracy.Precisionisthedegreetowhichagroupofmeasurementsofthesamequantityareinagreement—howclosetheyaretoeachother.Accuracyisthedegreetowhichameasurementagreeswiththetruevalueofthequantity.Measurementscanbeprecise(closetogether)withoutbeingaccurate.Measurementscanalsobepreciseandaccurate.However,

itisimpliedthatagroupofmeasurementsthatisaccuratearealsoclosetogether—therefore,agroupofmeasurementscannotbeaccuratebutnotprecise.95. WhatdatawouldyouneedtoestimatethemoneyyouwouldspendongastodriveyourcarfromLosAngelestoChicago?Provideasamplecalculation.YouwouldneedthenumberofmilesbetweenLAandChicago,thenumberofgallonsrequiredforeachmile,andthepricepergallon.

Cost = 2615 mi x

23.3 mix

$ 2.65gal

= $297

96. Onanewtemperaturescale(°Y),waterboilsat155.0°Yandfreezesat0.00°Y.Calculatethenormalhumanbodytemperatureusingthistemperaturescale.OntheFahrenheitscale,normalhumanbodytemperatureis98.6°F,andwaterboilsat212.0°Fandfreezesat32.0°F.Torelateonetemperaturescaletoanotheryouknowthattheamountofenergyrepresentedbythetemperaturesbetweenthefreezingandboilingpointofwateristhesameregardlessofthetemperaturescale.So,thenumberofdegreesbetweenfreezingandboilinginonescaleisequivalenttothenumberofdegreesbetweenfreezingandboilinginanotherscale,andweshouldbeabletocreateaconversionfactortoallowustoconvertbetweenthescales.Todothissimplyputthedifferenceindegreesbetweenboilingandfreezinginonescaleinthenumeratorofaconversionfactorandthedifferenceindegreesbetweenboilingandfreezingintheotherscaleinthedenominator.Initially,itdoesnotmatterwhichisthenumeratorordenominator.Youcanuseavisualaidtohelpifnecessary.Iwriteouttheendsofeachscaleandthedifferenceindegreesbetweenthesenumbersinthemiddle.Forthisexamplethishelpsmeseethatforevery155degreesoftheYscale,Ihave180degreesoftheFscale,andthisbecomesmyconversionfactor,whichIthenwriteouttotheside.So,ifIamgivenYdegrees(say500Y)andIwantFdegrees,IwouldwanttocancelYandcreateF,soIwouldmultiply:

500Y 1800 F

1550Y⎛⎝⎜

⎞⎠⎟

IfIamgivenFdegrees(say500F)andIwantYdegrees,IwouldwanttocancelFandcreateY,soIwouldmultiply:

500 F 1550Y

1800 F⎛⎝⎜

⎞⎠⎟

Thereisoneotherthingtodealwith—thescalesdonothavethesame“zero”point—ifeither(orboth)scale(s)donotbeginatzero,wehavetoadjustforthis.Tomakethisadjustment,whereverthevalueforthescalethatdoesnotstartatazeropointappearsintheequation,subtractdifferencefrom0fromthisvalue.Usingtheaboveexample,astheYscalestartswith0,itdoesn’tneedadjustment.But,anytimeavalueforFisstated,youwillhavetosubtract32.Fortheaboveexample,ifIamgivenYdegrees(say500Y)andIwantFdegrees:

0 F − 320 = 500Y 1800 F1550Y

⎛⎝⎜

⎞⎠⎟

0F = 500Y 1800 F1550Y

⎛⎝⎜

⎞⎠⎟+ 320 = 900 F

Fortheexample,ifIamgivenFdegrees(say500F)andIwantYdegrees:

0Y = (500 F − 320 ) 1550Y1800 F

⎛⎝⎜

⎞⎠⎟= 15.50Y

Forthecurrentproblem,combiningtheuseoftheconversionfactorandadjustmenttothe0point,youaregiventheFtemperatureof98.6andaskedfortheYtemperature.Soyoucreateanequationthatsays“degreesYequalstheFtemperature(fromwhichyouwillhavetosubtract32degrees)timestheconversionfactorthatcancelsFandcreatesY.”Answer57.30Y

0Y = (98.60 F − 320 ) 1550Y1800 F

⎛⎝⎜

⎞⎠⎟= 57.30Y

97. ExplainhowArchimedesmighthaveusedtheconceptofdensitytodeterminewhethertheking'scrownwaspuregold.(densityofgold=19.32g/cm3)Knowingthedensityofgold,Archimedeswouldhaveknownthatthecrownmusthaveacertainvolume.Hecouldhavemeasuredthevolumebywaterdisplacementsoseeifthemasspervolumeofthecrowncorrespondedtotheknowndensity.98. Explainthemaindifferencesbetweenacompoundandamixture.Thecomponentsofacompoundarejoinedtogetherchemicallythroughchemicalbondsindefiniteproportionswhilethecomponentsofamixturearenotjoinedtogetherchemicallyandarenotpresentinthemixtureindefiniteproportions.Therefore,acompoundwillrequirechemicalmeanstoseparatethecomponentswhilemixtureswillrequirephysicalmeanstoseparatethecomponents.99. Givethreephysicalmethodsusedbychemiststoseparatemixturesandidentifythetypeofmixturebestsuitedforeachprocess.Filtrationisbestforseparationofsolidsfromliquids.Ittakesadvantageofthedifferenceinphysicalstructureofsolidsandliquids,aphysicalpropertydependentonthestrengthoftheinterparticleforceswithinthesubstances,byplacingabarrierthatallowtheliquidtopassthroughwhilepreventingthesolidfrompassingthrough.Distillationisbestforseparationofmixturesofvolatileliquids.Ittakesadvantageofdifferencesinboilingpoint,aphysicalpropertydependentonthestrengthoftheinterparticleforceswithintheliquids.Chromatographyisbestforseparatingmixturesofsolublesubstances.Ittakesadvantageofthefactthatbecauseofdifferencesininterparticleforces,componentsofthemixturewillhavedifferentaffinitiesforthestationaryphasevsthemobilephaseofthechromatographysetupandwillmovethroughthesystematdifferentrates,thusbecomingseparated.100. Namethreemethodsfortheseparationofmixtures.

Answer: Distillation Chromatography Filtration

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