stpm trials 2009 chemistry answer scheme kelantan

8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan

1/11

Answer

1. B 26. D

2. D 27. C

3. D 28. A4. A 29. D

5. A 30. C

6. B 31. D

7. C 32. D

8. D 33. A

9. C 34. C

10. D 35. B

11. A 36. D

12. B 37. A

13. D 38. A

14. C 39. B

15. D 40. C

16. C 41. A17. A 42. D

18. D 43. A

19. C 44. D

20. D 45. A

21. B 46. D

22. B 47. A

23. C 48. B

24. C 49. D

25. B 50. C


2/11

(b)

SECfIONA

I. (a) (i)mi.154156

158(ii)

mi.154

156

158

,STPM Trial exam2009PAPER 2Kelantan

IonsC H,"CI" Br ,C, H,"sr"C I+c,Hl 'CI"Br, C,H} 'Br"CI+c,Hl 'C I" Br , C,H." Br"CI +c)Hl7CI8I Br"" C,H." Br"CI'

Relative abundance"CI"sr : or 3

: ) "CI" Br : (lx l )2 - 1-8 or 4"CI" Br : ( lxl)2 2 or I

(i) I , ' 2,' 2p'3, ' 3p'(ii) PCI , - ,p ' and PCI, - ,p' d(iii) Nitrogen cannot form a penthachloride, Nels becausenitrogen do not have d orbital.

I

(1+1+1]

[1 +1+ 1](I](1+I][I + ]

Total 11Muimum 10

2. (a) The power to which the concentration ofa substance is raised in an experimentallydetermined the rate equation . [2J(b) Rate k[H,O,](c)(i) Rate (3.42 X 10 ' X1.25 X 10.2 )


3/11

3.

In 2(e) H a l f l i f e k0.693

- 3.42xI0 )

(f)[H,o,)moldm -)

- 202.6 s

1.0

0.5

0.25

(a) (i) Boron oxide(ii) Nickel oxide

(b) In glass , there is no regular arrangement of particles.The arrangement is just like in liquid.

Time I s

However, in crystalline ionic so lids, the ions are arranged in an ordered andclole;lyoacked manner.

(e) (i)

PbCI. +

i l lTotal 10

[I]

[ I)

[I )[I)[I)

[11

[4)(ii) S i C ~ and b C ~ have empty d orbitals which are snacked by lone pair electrons of water molecules.

But in eeL.. the carbon atom does not contain d orbitals. [ 1+1]Total: 10


4/11

... .4 (a)

C H BrNo . ofmoles 135 .2112 = 2.9 I 6.5 /1= 6.5 I 58 .3/80 = 0.73II Mole . . io 4 9 1 1m

Molar mass : 136 .9Molecular fonnula : C.H,Br lUI

:!:@ I (i) I I .. . ICH ,I 1mCH, - CH - CH,B,

(U) Nucleophilic substitution SN2 (primary haloi:t'n alkanes) 1m(e) (Q H HI I 1mCH.J - C = C- CH, or CH, CH = CHCH,

(n) CH ,I 1mCH, - C - CH,IBr(iU) I SOMer . CH ) CH2 - CH "" CH2 or CH,- C - CH, 1mICH,

EquIJrlo,. :CH, CH, - CH - CFh + 5(OJ CH, CH2 - COOH +C0z+ HP lIDOr C(CH,lFCH, + 4[0] (CH,),C-o + CO , + H2O

(d) -Wann wit h aqueous acidified KMnO . 1m.Alcohol P does oot decolourisc the purple solution.-2-metbyl- l- propanol decolour1ses the purpk solution. (oxidation of 1mLE ' alcohols10m


5/11

SECTION B5(a)(i)

NH,

HCI

co,

bonds between> r e

dcr Waals forces orbetween Co, HCI + H,O 7 H,O' + crmolecules arestronger than van derWaals forces betweenHCI molecules

H +

are nonwhereas water molecules arepolar

(a)(ii) H 'til HH

N 5e4H 4eCharge +1 8e 4 bonding pair.;Shape : Tetrahedron

5(c) (i) Zn (s) +. 2W(aq) 7 Zn" (aq) + H,(g)-+1).54 =

+0 .54 =

[z '- ]?E' 0.059 I n H ,"" - 2 og [H ' ]'(0-(-0.76 _ 0.059 10 1. 0% 1.02 g [W ] '+0. 76 + 0.059 x 210g[W]2

log [H1 = -3.73pH - -log [H']- 3.73(ii) If the pH of acid increases, IWI decreases

emf of the cell decreases

[I+ I]

[1 +1+1]

-[ I]

[I][I ]

[I]

[ I]

[ I ]

[I][ I][I][I]-1Total IS


6/11

5

6.(a) Le Chlltelier 's principle states that if asys tem in eq uilibrium is disturbed, the equilibriumwill shift to ca .Dcel out the effect of the change. [2)

(b) (i) Increasing the pressu", will couse the equilibriwn shift to the left [I]

(ii)

because the foward reaction involves an increase in the number ormolcs of gas .or I mol? 2 mol [I](ii) The reaction to the ripht is endothermic . [I]Thus heating will cause the equilibrium position ti l move to the right. [IJ(iii) The equilibrium will shift to the right because more cr04' ions react r ]with W ions to produce CC20,"and H20.Thus the concentration ofW ions can be reduced. [I]

(c) (i) Total gas ratio - 2+ I =3Initial paroal press"", of So, = 2/3 x J = 2 atmInitial partial press"", of0, - 113 x J - I atm2So,(g)+ O,(g) +:! 2S0,(g)

1.9atm

[ I]

From the equation above, to produce 1.9 atm o[SO), 1.9 aIm S02and 0.95 atm 0[02 mustreact.Psm = 2 - 1.9 - 0.1 atmPo, = I - 0.95 - 0.05 atmPlOW "" PSOl + POl + POOl= 0.1 + 0.05 + 1.9

[I][I)

= 2.05 atm [I]Percentage of cbange from So, to SO, .JQitial pressure of S(h is 2 atm. 1.9 atm of S02 has reacted to become SOJ.Therefore, percentage of change - 1.9 / 2.0 x 100 - 95% [I]2S0,(g) + 0, 2S0, (g)Kp =

( P ~ ) (Pm)- iH;'(OO5)= 7220 atm-!

[I )

[I]7

Total : 15


7/11

6

7 (a) 0) The enthalpy change when one mole of an ionic compound is fonned from its element, alstandard condition. [I J

1Na (,) + 2" Ch (J) - NaCI (,)[I + IJ

(ii) The energy released when one mole of an ionic comoound is fonned from itsgaseous ion at standard condition.

(b) (i)

(ii)

-969.0

---'O!MFW+731.0 322.0+284 .6 +79 .0

M (s) +

6 H((M) = + 284.6 +731.0 + 79 .0 - 322.0 - 969 .06 H((MF) = -196 .4 icJmor '

[lJ[I + IJ

[5J[lJ[lJ

(iii) MF because its enthalpy of fannalian is more exothermic . This show s thatMF is more stable since its enthaJpy is lower. [2JTotal: IS


8/11

8.(a) Transition elements are defined as d-block elements that can form at least one stable ion withpartially-filled d orbital. [I]

[Notes:

3d 4s(b) Cu ' in [CuC],]" 1 1 ~ 1 1 ~ 1 1 j j 1 ~ 1 1 ~ 1 Q

CU"in[Cu(H,Ol6]" 1 1 ~ 1 1 v 1 1 0 1 0 1 I 0Splitting of the energy of the 3d-orbitals occur.

[I][I]

[I]Electrons in the low energy 3d-level of Cu2+ absorbs (red) light and move into the high energy3d- level. I dod transition occurs. [I]No dod transition occurs. [I]

(c) (i) The basicity of the oxides increases down the group.CO is neutral .Si02 and C02 are acidic.Oxides ofGe, Sn and Pb are amphoteric.

[I][I]

[I + ][ IJ

(ii) Lead (IV) oxide I PbO, is an oxidizing agent.I Pb4+ is reduced to Pb 2+.! +2 oxidation state of lead is more stable than +4 oxidation state.Mn 2+ is oxidized to Mn 7+ MnO,,' .! Oxidation state for Mn change from +2 to +7.A redox reaction occurs.Solution turns purple ! violet ! pink at the end of reaction.! dark-brown solid dissolves. ! solid dissolves.

[IJ[I][I J[I]

Total: 15

Transition complexes are coloured because of electronic transition between non-degenerate d orbitals.The transition .(Oetal ions have incomplete 3d sub-shells, d l to d9. As a result, d-d electronic transition ispossible resulting in pair of the visible light spectrum being absorbed. Other non -transition metal ions haveeither completely filled 3d orbital or no 3d electron at all. Hence, dod electronic transition is not possible,they are colourless or white. ]


9/11

9(a)In G, the CI atom is bonded to the side chain hence it can be replaced by hydroxideduring hydrolysis.

Gis 0 -CH,CIo CH ,CI + NaOH 7 o _CH ,OH + NaCIIn J. the Cl atom is attached to the aromatic rin2. The C- CI bond in H is less polar and strong .

J s or CI - CH ,CICI

9(b)K is o NH,With aqueous bromine. 2,4,6-tribromophenylamine is fonned as a white precipitate.

o -NH , + 3Br, 7 Br- o Br. -NH ,Br

With etbanoyl chloride. a substituted amide is fonned.Sr Sr

+ 3HBr

-NH , + CH,COCI 7 B.r- .-NHCOCH,Br (L ) Br + HCI

K with ethanoyl chlorideo NH, + CH,COCI 7 NHCOCH, + HCIWith bromine J substitution occurs at benzene ringo _NHCOCH, + Br, 7

[I][I ]

[I]

[I ]

[I]5

[I]

[I]

[11

[I]

[I ]


10/11

9(c)

N has a ch iral center because it exists as a pair of enantiomers .It is an a1cohol as it reaclS wi th PCI,: to give He) gas.N is CH,CH(OH)CH,CH,

[I][I ][I]C!i,CH(OH)CH,CH, + PCI, CH, CHCICH,CH, + HCI + POC!, [I]

CH,IHif'J "'-CH,CH,H

,,,,,,

No n- superimpossable[I]

5Total : 15

9 (a) He l or H2S0"or H+ .. acidconc(ifHCIoDlyVdilute/aqueo us+ heal1m1m

(b) two ri.ngs only ( I ring around the (I-C ofryrosinc & I around me (Z -e of lysine) 1m(e) C H ~ ~ (or disp layed ronnuta) 1m(d) (i) NH 2CH2C02. (Ns (e ither -C02"Na"'0' -C02Na but NOT -CO-O-Na) 1m

(ii) (N.,) '-Q-C6H4-CH2CH(NH,)CO'- (No ') Iw+lm(iii) (C I)+NH )(CH,).CH(NH)' )CO , H (Cl) Im+lm

(6m )


11/11

(el

(I)(Q(Q.g.

(i)

'0

CH CO.H

stnktWt [IJIII least one peptide group identified [I J

anorOl -01 - 00 'If"CI\- CI\- H1!-

IIIIII

[I)

orClCOI---

(iii) Condensat ion polyrnerisationMaximum :

2m

3m

15m

fllTOIaI : 16

15

stpm trials 2009 chemistry answer scheme kelantan

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