stpm trials 2009 chemistry answer scheme kelantan
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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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
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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(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 )
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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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
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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... .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
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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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
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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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
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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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
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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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. ]
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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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 ]
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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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 )
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8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Kelantan
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(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
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