answer for chemistry
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8/6/2019 Answer for chemistry
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8/6/2019 Answer for chemistry
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No.of molesFe= 2.80/56= 0.05molNo.of molesCl2 (0.05 3)12= 0.07QmolVofume f cl2 ; o.ozs 24 = 1'8dm3 1,800 m3
11 (a) Formulahat shorvshe simplestatioof henumber f atomsor eacfielementin hemmpound. 1"'1
Element c HMass %) 92.3 7. 7Number fmoles 92'3= z.zt2 7'7 = z:tIRatioof moles 1 1
4.5.6.111 .. .6EI
111
(b)
Empiricalormula:CHRMMof GH)n= 76[12+1]n = 7813n = 78n =6
mass s obtainedRecordallthemass10. Results:
(d) - Cannot eparateoppertommagnesiumxide- CannotweighoPPer
11.. .5Molecularormula C5H5(c) Procedure:1. Cleanmagnesiumibbonwithsandpaper 12. Weighcru-ciblend ts id 13. put ragnesiumibbonnto hecnrcible ndweigh hecrucible ith ts id 'l4. Heatstronglyhe cruciblewithout b lid 1S. Coverhe &uciOle hen he magnesiumtartso burnand ift/raisehe id 1a litdeat intervals6. Removehe idwhen hemagnesiumumtcompletely 17. Heatstronglyhecrucibleor a fanrminutes 18. Coolandweigh hecnrciblewithts idand he content 1
g. Repeathepricessesof heating, ooling ndweighing ntila constant 1
111114...Max12
I1m15 (a ) (i ) AtomY:2.8.7Answer ChemistryPerfect ScoreModule 2009
Element Mg oMass g) y-x z-yI . t-- | ,- yNur Etno;6olformula: g"oo/MgO -6Simolestatioof moles a b
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8/6/2019 Answer for chemistry
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Atom :2.8.8.1Group 7Because tomY has7 valence lecfronperiod3Because tomY has hreeshellsoccupied itheledrons
(ia) 2Na+ Ye* 2NaYConect ormulaof reactantsandproductconedBalance quationYe+2NaOH + NaY+NaOY+HzOConect ormulaof reactants ndproduc{s orrec{Balance quation
15 (b) lonicomoound/bond1- Theeledron nangementf atomP= 2, 8, 1 2.8. 1 llTheelectonanangementf atomQ= 2, 8, 7 I 2.8.72- to achievehe stableeleclronanangement$ atomof P donates givesone eledron o theatomof Qto form P*// halfequation4 - atomof Q receives acceptsone elecfon ftomatomof Pto formQ'l/ hatfequation5 -theP* ion andQ- ionsare atraded by a strong electrostaticorce oform onicbond&withformula PQ coneddiagramCovalent omooundbond7-Theelectron rangement fatom R= 2, 6 / 2.6&to acfiievehe stableelecfon anangement$[atom R and atomQ shareelectrons]10 atomRcontributes eledronsandatomQcontributesne. eledron11- one atomR and4 atomQshare4 pairsof eledrons-12- o formcovalent ompound ith he ormulaRQr diagram
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AnswerChemistry erfectScoreModule2009f*
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8/6/2019 Answer for chemistry
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Numberofmole 0.48t'12ilo.M 0.05/1//0.05 0.28t14ilo.o2 0.16/16tn.o1Thesimplestratio 0.04/0.01i l4 0.05/0.01i l5 0.020.01i l2 0.01/0.01l l1
13 (a)
Empiricalormula CaH5N2O[CrHsNzOn = 194[97]n=194n= 19487l2Molecularormula CeHioNrOzAble o calculatehemolarmassand hepercentagef nitrogen ymass neachof the three eftilisersand cftoose he best ertiliser.1 molarmassof ammoniumulphate 1329/mol2 percentagef nibogennammoniumulphate= 28l132x1OOo/ol21.2o/o3 molarmassof urea= 60g/mol4 percentage f nitrogen n urea= 28/ 60 x 100% l 46.70/05 molarmassof hydrazine 329/mol6 percentagef nitogen n hydrazine281132 100% l 87.5o/o7 Hydrazineas herichest ource f nitrogen ompares ithotherfertilizers.8 The armer hould ftoose ydrazine
1. Theproton umbers 11 l Number fprotons 112. Nucleonumbers23llAtomicmass s 233. Number fnzutron 2.g-^11124. Nucfeusontains 1pand12n5. Position f electron irculatinghe nucleus6. Conednumber hellconsists f electron7. Symbol f sodiums Na any6
11111. . . . .8(c)
11111111... . .8a.111111... . .6
Formulahatshowsimplest ratio number f atomsof eachelementncompound 11. Refativemolecularmassforn(CHzO)180 l 112n+2n+16n= 1802. n =63. CoHrzOe
Element FE clMass/o 2.80 5.32No.of moles 2.80/56= 0.05 5.3235.5 0.15Ratioof moles/Simolestatio 0.05/0.05 1 0.15/0.05 34. Empiricalormula FeCls
1. Formula f the reactants2. Formula f produc{s3. Balance qudion//, . /ZFe+3Clz+2Fe0ls
(b) (i)
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Ans'werChemistry erfectScoreModule2009#
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8/6/2019 Answer for chemistry
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8/6/2019 Answer for chemistry
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STRUCTURE APER
SuqqestedAn$rver(a )0 s = 95.0' C,30s = 85.0' C,60 s = 82.0 C,90s = 80.0'Q,120s 80.0'C,150s 80.0'C,180s = 78.O C. 21O :7O.0 '
(c) ( i )&( l l ) Score3&31
An-rwerChemistryPerfectScoreModule2009!* a
10
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8/6/2019 Answer for chemistry
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d(i) SuoqestednswerTheconstantemperaturetwhich iquidbecomes solid 3d(ii) tiuooestedAnstverTheheat eleased hen heparticlesn he iquidarangeto formsolid balanced v the heat oss o the sunoundinos. 3e Suggested nsrer
Theairtapped n the conicaliask s a poormnductorof heat.The air helpedo minimize eat oss o thesunoundinqs.to ensure niform oolino. 3f Fuss$d 3ns4cr 3
(ii) Whitesolid s formed/The assof crucible nditscontent ncreases-Magnesiumxide s formedMagnesiumeactswithoxygen(i)(iD
Themassof crucible nd id= 25.35g.Themassof cnrcible,id and magnesiumibbon=27.75 .Themassof crucible,idandmagnesiumxidewhencooled= 29.35g(i) ThemassofMg= 27.75-25.35)9=2AOg(iD Themass f02=(29.3$27.75\=1.69(iiD Thenumber f molesMg=0.1moleThenumber f molesO= 0.1moleTherat ioof g:O=1:1Theempiriealormulas MgO.0.1mde of Mg eaclswith0.1moleof O/1moleof Mg reactswith 1 moleof O
(b)
(c)
(d)
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l8 (a ) Able opredid themanipulatedvariable,he respondingariableandthe mndant variable ompletely.Manipulatedariablemetals f Group1elements /sodium,ithium, otassium.Respondingariable:thereactivity f fte reactionwithwater / the speedof movement ntrewater surtaceConstant ariable:sizer/massf metals.Volume f water
Able o date how o controlthemanipulatedariablesconecilyRepeathe experimenty usinghe metals f sodium,llthium ndpotassiumAble o $:ate onectly he way o nntrol themanipulated ariableToobserve ow ast he metalsmoveon he surfaceof water;Able o use he metalgranuleswiththesamesizeUse he metalgranules ith hesamesize.
18 (b) Able o state fie rela0bnsht:petween he manipulatedariableand he esponding ariableconectly..suggested nsrirer: he eactivity f Group1elementsncreasesoingdorn thegroup.18(c) Able o anangeconec{y he reactivtty eriesof the metalsaccordingo descending rder.Ansurec otassium, odium, ithium18 (d ) Able o classiff he ionscorrectly.[tonameor wtte all he ormula f the onscorectlyat the cations ndanions roup.lAnswer: ositiveon/cation sodiumon/ Na+,hydrogenon/H+Negativeon/anionhydroxideon/OH-
AnswerChemistry erfectScoreModule2009a*, \(u*'\t ,$de'tr\
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