高中化学计算题High school chemistry calculation problems a.docx
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高中化学计算题High school chemistry calculation problems a.docx
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高中化学计算题Highschoolchemistrycalculationproblemsa
高中化学计算题
(一)(Highschoolchemistrycalculationproblems(a))
Thereisaclassofcomputationalproblems,theamountofreactants,reactionchemicalequationsaredifferent,suchascarbondioxidegasesintotheNaOHsolution,dependingonhowmuchaccessamount,therearetwoequationsCO2+NaOH=NaHCO3,CO2+2NaOH=Na2CO3+H2O
Whenfacedwithsuchproblems,wecanusetwoyuanforasetofequationssolutionofsodiumbicarbonateCO2xmol,sodiumbicarbonateCO2ymol
Theapplicationofconservationlawcanbecalculated
Namely:
conservationandconservationofcarbonatomsofsodiumion
TheproblemsolvingmethodsinComputationalChemistry
Thechemicalcalculationproblemisakindofproblemstudentsheadacheinchemistrylearning,akindoftopicintheirtestsandexamsinthemostdifficulttoscore,canchoosethemostsuitablemethodforquicklyandaccuratelysolvethecalculationproblems,toimprovelearning,reinforcementlearningefficiency,isofgreatsignificance.
Choosetheappropriatemethodtosolvethecalculationproblems,cannotonlyshortenthesolvingtime,alsohelpstoreducethecomputationalcomplexityintheprocess,toreducetheerrorintheprocessofoperationopportunityasmuchaspossible.Forexample,therearetwokindsofdifferentmethods,incontrast,isnotdifficulttoseetheimportanceofselectingtheappropriatemethod:
Thereactionofnitricacidsolutionwith1]30mLconcentrationand5.12gramsofcopper,copperiscompletedwhenallreaction,collectedatotalof2.24litersofgas(S.T.P),whiletheconcentrationofnitratesolutionmaterialatleast
A.9mol/LB.8mol/LC.5mol/LD.10mol/L
Asolution:
becausethesubjectinthatnitrateisthickorthin,sotheproductcannotbedetermined,accordingtothetwoequationofreactionofcopperandnitricacid:
(1)3Cu+8HNO3(rare)=3Cu(NO3)2+2NO=+4H2O
(2)Cu+4HNO3(strong)=Cu(NO3)2+2NO2=+2H2O,canbesetinthereaction
(1)Cuxmol,NOgasgeneratedbythereactionof2/3xmol,theconsumptionofnitricacidreactionfor8/3xmol,andthensetupinresponse(2Cu)ymol,NO2gasgeneratedbythereactionof2ymol,theconsumptionofnitricacidreactionis4ymol,andtheequationsarelisted:
(x+y)*64=5.12,[(2/3)x+2y]*22.4=2.24,x=0.045moly=0.035mol,isobtained,theconsumptionofnitrateis8/3x+4y=0.26mol,theconcentrationofmol/L(0.26/0.03),between8-9,canchooseA.
Methodtwo:
accordingtothelawofconservationofmass,becauseonlycoppernitrateandcompletereactionofCu2+,itshouldbetheproductofcoppernitrate,andtheamountofsubstancewiththeoriginalcopper,are5.12/64=0.08friction,chemicalproductsfromCu(NO3)2canbeseeninNO3-HNO32*0.08=0.16frictionmetathesisreactionthegaseousreactionproducts;andwhetheritisNOorNO2,everymoleculecontainsaNatom,thenumberofmolesofgasisequivalenttothetotalnumberofmoleculesinvolvedintheredoxreactionofHNO3,sothefrictionconsumptionperoneHNO3generates22.4Lgas(eitherNOorNO2oramixtureofboth).Theexistinggas2.24L,namely0.1frictionHNO3isinvolvedinredoxreaction,sotheconsumptionofnitratefor0.16+0.1=0.26friction,theconcentrationofmol/L(0.26/0.03),between8-9,canchooseA.
Fromtheabovetwomethodscanbeseen,thisisthechoice,aslongasthecalculatedresultscan,regardlessofthewayandthekeypointsandsolvingstandard,theproblemiswhethertheskilledapplicationofthelawofconservationofmass,
Secondmethodsofusingconservationmethod,sothecomputationismuchless,alsodoesnotneedthefirstchemicalformulalist,trim,itcangreatlyshortenthesolvingtime,avoidsbecauseIdonotknowwhichonetogettheequationaccordingtothenitratecausedpanic.Lookunderthetitle:
The2]inaclosedcontainerof6liters,3litersinX(gas)and2literY(gas),thereactionundercertainconditions:
4X(gas)+3Y(gas)2Q(gas)+nR(gas)reachedequilibriumaftertheconstanttemperaturecontainer,thepressureofmixedgastheincreaseof5%thantheoriginal,theconcentrationofXdecreased1/3,thereactionequationisthenvalueof
A.3B.4C.5D.6
Solution:
toseizethe"concentrationofXreduced1/3,combinedwiththechemicalequationcoefficientisequaltothevolumeratio,theinitialstateofthematerialcanbelistedseparately,andthefinalstatevariables:
4X3Y2QnR
Theinitialstate3L2L00
-1/3*3L=1L-3/4*1L=3/4L+2/4*1L=1/2L+n/4*1L=n/4Lvariables
3-1=2L2-3/4==5/4L0+1/2=1/2L0+n/4=n/4Lfinalstate
Fromtheaboveformulashowsthatthebalance(finalstate)mixedgasvolume(2+5/4+1/2+n/4)L(15+n)/4L,accordingtothe"mixedgaspressurethantheoriginal5%"(15+n)/4-5=5*5%,obtainedn=6.
Methodtwo:
usethedifferencemethod,accordingtothe"mixedgaspressurethantheoriginal5%increasedbythe"mixedgaspressurethantheoriginal5%",namelythemixedgasvolumeincreasedby*5%=0.25L(2+3),accordingtotheformula,4X+3Ycangenerate2Q+nR,namelyevery4volumeofXreaction,thetotalvolumechangefor(2+n)-(4+3)=n-5,Xreactionofexisting1/3*3L=1L,namelythetotalvolumechangequantityis1L*[(N-5)/4]=0.25L,tocalculaten=6.
Methodthree:
seizethepressureofmixedgasof5%thantheoriginal,wasobtainedbyX+Yatthebeginning,thebalancewillbeshiftedtotheright,aftertheformationofQandRpressureincreases,thatitisdefinitelyincreasedvolumereaction,reactioncoefficientofXandYintheequationandhavelessthanQandthecoefficientofRand4+3<2+n,sothatn>5,infour,onlytheDn=6optiontomeettherequirements,shouldbeselectedfortheanswer.
Thetestisaboutthechemicalequilibrium.Oneistofollowthelawofchemicalequilibriumsolution,thenormwentthroughchannelsalthoughcertainlycancalculate,thecorrectanswer,butdidnotgraspthe"choice,don'task,aslongasthecharacteristicsoftheresult",asacalculationtodo,ordinarystudentsatleast5minutesthatspendingmoretime.Usingthesolutionoftwodifferencemethodwithvariablevolume,containingn(differentialequation)toestablish,calculatethevalueofWenXialanran,butstillfailedtomakefulluseofchoice"choice"characteristics,timeto1minutes.Threetounderstandtherelationbetweensolutionandequilibriumthevolumechangethoroughly,nothalfaminutecanbeobtainedonlythecorrectanswer.
Thus,intheprocessofcalculationforthetopicselectionofcharacteristicsofdifferentmethods,oftenhelpstoreducethetimeconsumptionofoperationandtheerrorintheprocessoftheopportunitytoachieverapid,accurateproblem-solvingeffect,andtheuseofmoreproblem-solvingmethodsusuallyhavethefollowing:
Takingmorethan1.method:
thismethodismainlyappliedtothesolutionoforganicmatter(especiallyhydrocarbon)knowthemolecularweightdeterminedthemolecularformulaofaclassofproblemsforhydrocarbon,
ThealkanegeneralformulaCnH2n+2,molecularweightis14n+2,withthecorrespondingCnH2n+1alkanetype,molecularweightof14n+1,olefinandnaphthenegeneralformulaCnH2n,molecularweightis14N,alkylformulacorrespondingtotheCnH2n-1,themolecularweightof14n-1,alkynesanddienesformulaCnH2n-2,molecularweightis14n-2,alkylthecorrespondingformulawasCnH2n-3,molecularweightis14n-3,soitcanbeknownmolecularorganicmatterminustheamountamountofoxygencontainingfunctionalgroups,thedifferencedividedby14(14,inadditiontodirecthydrocarbon)isthelargestmanufacturerforcarbonatomnumber(nvalue),theremainderbythemolecularweightwiththeformula.Itisthecategory.
[3]astraightchainalcohols14gramscanreactwithsodiummetal,0.2gramsofhydrogengeneration,whilethenumberofisomersofthealcoholas
A.6B.7C.8D.9
Becauseofapolyolcontainingonlyone-OHpermol,alcoholcanonlyconvert1/2molH2,generatedby0.2gramsof14gramsofalcoholshouldbeinferredfromH20.2mol,sothemolarmassof72grams/mole,amolecularweightof72,afterdeductinghydroxyltypequantity17,theremaining55,dividedby14,thebiggestbusinessis3.Morethan13,isnotreasonable,shouldbetaking4,morethan-1,thenthemolecularweightformulaforalkenyloralkylringof4carbonatoms,with"straightchain",soastodeducethenumberofisomersof6.
2.theaveragecompositionofthismethodismostsuitableforqualitativesolutionmixture,whichisonlypossiblemixtures,regardlessofthecontentofeachcomponent.Accordingtovariousphysicalquantity(suchasmixturedensity,volume,molarmass,concentration,massfraction)ofthedefinitionoftopicoracombinationoftheconditionsfor.Youcancalculatetheaveragevalueofthemixtureofcertainphysicalquantities,andtheaveragevaluemustbethesamephysicalquantityofeachcomponentvaluebetweenmixtures,inotherwords,thephysicalquantityoftwocomponentsinamixturemustbelargerthantheaveragevalue,alowerthanaverage,inordertomeettherequirements.Whichmaydeterminewhetherthemixturemayform.
[casesof4]willbetwokindsofmetalelementsaddedtothemixtureof13g,asufficientamountofdilutesulfuricacid,totalreleaseof11.2Lgasunderstandardconditions,thetwokindsofmetalmaybe
A.ZnandFeB.AlandZnC.AlandMgD.MgandCu
将混合物当作一种金属来看,因为是足量稀硫酸,13克金属全部反应生成的11.2L(0.5摩尔)气体全部是氢气,也就是说,这种金属每放出1摩尔氢气需26克,如果全部是+2价的金属,其平均原子量为26,则组成混合物的+2价金属,其原子量一个大于26,一个小于26.代入选项,在置换出氢气的反应中,显+2价的有Zn,原子量为65,Fe原子量为56,Mg原子量为24,但对于Al,由于在反应中显+3价,要置换出1mol氢气,只要18克Al便够,可看作+2价时其原子量为27/(3/2)=18,同样假如有+1价的Na参与反应时,将它看作+2价时其原子量为23*2=46,对于Cu,因为它不能置换出H2,所以可看作原子量为无穷大,从而得到A中两种金属原子量Aremorethan26,twokindsofmetalatomsinCwerelessthan26,soA,CdoesnotmeettherequirementsofBAlatomicweighti
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