8/8/2019 STPM Trials 2009 Chemistry Answer Scheme (Johor)
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Question 1(a) (i)
00
(ii) (8X2X1/8) + (6X2X112) =8(1 )(1 )
(b) (i) A set of temperature and pressure values where solid, liquid and gas exist inequilibrium.
(ii)
liquid
.cUd5 , . . ______ __ _______L---
0",
!r,_ ._.._- ---.--- :r, '. ...._- --," .._..._-_ . ...- 31 "" - -
General sketch of graph .The correct axes.The labellings.
(1 )
(1 )(1 )(1 )
(iii) because its triple pOint occurs at a pressure above the normal atmosphericpressure. (1 )
8/8/2019 STPM Trials 2009 Chemistry Answer Scheme (Johor)
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p
p =5.50 x 8.31 x (24.0+273) x 10-3 kPa10.0 x10-3 X 44.0
= 30.9 kPaPI = 30.9 + 93.7 = 124.6 kPa
Question 2
(1 )
(1 )(1 )
Total: 10 marks
(a) Ca2+(g) + 2e- + 2CI(g) (This is the only answer for the top line) (1)Ca2+(g) + 2e- + CI2(g) (1)Ca+ (g) + e- + CI2 (g) (1)Ca(g) + CI2(g) (state symbols and electrons essential) (1)Note: CI2to 2CI can be in any order but Ca must be in sequence)
Max. 3 marks(b) Enthalpy of formation = 178 + 590 + 1145 + (2 x 121) - (2 x 364) - 2237 (1)= - 810 kJ mor1 (1)(c) flH = flH(lattice formation) + :LflH(hydration) (1)
(or cycle with state symbols, numbers or labels)=2237 -1650 - (2 x 364)= 141 kJ mor1 (1)(d) Moles of NH4CI = 2153.5 = 0.0374
Heat absorbed = 15 x 0.0374 = 0.561 (1)Q=mcf lTf lT= Qlmc = 0.561x1000)/(50x4.2) =2.6 (0C) (1)(allow 2.5 to 2.7; can use 52; ignore units, answer must be at least 2 sig figs)
Final temperature = 20 - 2.6 = 17.4 c (1)(Answer is for 20 - previous ans; must be < 20)Note; may not use moles (loses first 2 marks)so f lT= 15x1000)/(50x4.2)So answers of 71.4 and 68.7 score last 2 out of first 4.(allow no units for temperature, penalise wrong units)
Total: 10 marks
8/8/2019 STPM Trials 2009 Chemistry Answer Scheme (Johor)
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Question 3(a) 3s2 3p1(b) (i) aluminium hydroxide
(ii) AI(OH)4 -(c) (i) white precipitate
Effervescence(ii) AI(H20)6 ]3+ + H20 _ AI(H20)s(OH)f+
(d) (i) AICh + CH3CI(ii) as lewis acid
(1 )(1 )(1 )(1)(1)(1 )(1 )(1 )
In AICI3 ,AI has an incomplete octet and accepts a pair of electrons from the CIatom in CH 3CI @ AICh is an electron-deficient compound. and acts as anelectron-pair acceptor.(e) light
Question 4(a) (i) Ag+/silver
(1 )(1 )
Total = 10 marks(1 )
(ii) Carbonyl or C =0 (1)(iii) Helps to speed up the reaction/slow at room temperaturelflammable (1)(iv) Propanal: silver mirror formed
Propanone: no change/solution remains colourless(b) (i) optical isomerism
(ii)(a)
(13)
H rN a ( } - - Hb H cH,
(c) (i) nucleophilic addition(ii) dry ether, room temperature
(1 )(1 )(1 )
(1 )
(1 )(1 )(1 )
Total = 10 marks
8/8/2019 STPM Trials 2009 Chemistry Answer Scheme (Johor)
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rQuestion 5(a) Lone pair donated / both electrons supplied by one atom
from N (to W) [ignore missing charge or hydrogen atom)dative/dative covalent/coordinate bonding
(b) Van der Waals ( or VrNV) forces between methane molecules(or VdW forces in methane)
Hydrogen bonding in ammonia and waterHydrogen bonds are stronger than van der Waals forces( orVdW forces are the weakest)
(1 )(1)(1 )
(1)
(1 )(1)
Hydrogen bonds in water more extensive than ammonia because 0 has two lonepairs (N one) (1 )[or Water forms more H bonds per molecule (than ammonia)(or H bonds in water stronger because 0 more electronegative than N)(or H bonds in water stronger because 0 - H bond more polar than N - H)
(c) (i) The change in concentration per unit of time (1)Both axes must be labelled to gain marks for graph. y axis cone N02and x axis time (1)Curve starts at origin and levels off. (1)
Note: If candidates graph does not level off then second mark can be scored for aCUNe with a continuously decreasing gradient.
Initial rate can be found by finding the gradient at t = 0 (1)Note: Candidates may score this mark if they have shown this on their graph
NO is a catalystit is regenerated at the end of the reactionprovides an alternative route of lower activation energy
(1 )(1)(1 )(1)
Total = 5 marks
8/8/2019 STPM Trials 2009 Chemistry Answer Scheme (Johor)
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- -_ . -.- - . --Question 6(a) (i) A buffer solution has the ability to resist changes in pH when a little acid or base isadded .
When a little acid is added, it reacts with NH3 . The acid is used up, pH remainsunchanged
NH3 + H+ NH/
(1 )
(1 )(1 )
When a little base is added, it reacts with OW The base is used up, pH remainsunchanged
NH/ + OH-
(ii) pOH = 14.0 - 9.0 = 5.0QIU .v I [NH;1p L1 =Pf tb + og [NH,J
5.0 = log1.7x10-5 + log [NH/] 10.10[NH/] =0.17 mol dm-3
mass of NH4CI in 450 cm3=0.17 x 53.5 x 450/1000 = 4.1 g(b) (i) acid strength increases from no. 1 to no. 3 or
down the table or
(1)(1)
(1 )(1 )
(1)(1 )
as number of CI increase (1)due to the electron-withdrawing effect orelectronegativity of chlorine (atoms) (1)stabilising the anion or weakening the O-H bond (1)NOT H+ more available
(ii) chlorine atom is further away (from O-H) in no. 4, so has less influence (1)(iii) either: pH = pKa - IOg10 [acid) or Ka = 10 - pKa = 1.259 x 10-3= 4.9 + 2) [H+) = (Ka. c) = .55 x 10"" (1)
'"' 3.4 (allow 3.5) pH =3.4 (1)Note: [1) for correct expression & values;
[1] for correct workingTotal = 5 marks
8/8/2019 STPM Trials 2009 Chemistry Answer Scheme (Johor)
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Question 7(a) (i) Increase (1)
(ii) Colourless solution tums to a brown solutionl black ppt(need both colours) (1)CI2 + 2KI 12 + 2 KCI (1)CI 2 is an oxidising agent (1)
(iii) Add silver nitrate solution (1 )KBr forms creamy ppt (1)KI forms yellow ppt (1)AgN03+ KBr AgBr + KN03 orAgN03 + KI Agi + KN03 (1)or ionic equationsThen add (dilute or conc) ammonia (1)AgBr/ cream ppt dissolves in conc NH3 orslightly dissolves in dilute NH3 (1)Agi is insoluble in dilute or conc NH3 (1)
(b) SiCl, reacts/hydrolyses, CCl, does not (1)[This must be clearly stated and not just implied](lone) pair of electrons (from the oxygen atom) in a water molecule (1)cannot form a bond with/be donated to the C atomOr cannot be accepted by C atom (1)because C has no available orbitalOR no 3d orbitals in COR C is a small atom surrounded by CI atomsOR CI atoms are large and surround C atom (so attack is sterically hindered) (1)Rejects CCI4 has no d orbitalsSi has (available) 3d orbitals (1 )Rejects SiCI4 has available 3d orbitals
8/8/2019 STPM Trials 2009 Chemistry Answer Scheme (Johor)
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Question 8(a) white precipitate formed fo r barium nitrate
no visible change for magnesium nitrateH2S04(aq) + Ba(N03h --. ..Ba2+(aq) + SO/-(aq) BaS04(s) + 2HN03(aq) @BaS04(s)
(1)(1 )
(1 )
(b) ionic size of Mg2+ < Ba2+ (1)magnitude of hydration energy of Mg2+ 8a2+ (1)ionic size of sol- size of cations Mg2+ and 8a2+ (1)
magnitude of lattice energy for MgS04 slightly more than 8aS04 (1)LiHsolution= LiHhydration -LiH lattice (1 )
For MgS04 LiHsolution= xothermic; For BaS04 LiHsolution= ndothennic ; (1)barium sulphate - inscluble in water; magnesium sulphate - soluble (1)
(c) (i) (Ksp =) [Mg2+][OH12 (1)units are mol3dm-9 (1)
(ii) Let [Mg(OHh(aq)) =Mg2+] =Ksp = X 10-11 = x3 (1)x =1_71 X 10-4 mol dm-3 (1)
(iii) less soluble because of the common ion effector the equilibrium Mg(OHh(s) --> Mg2+(aq) + 20W(aq) is moved to the left
(1 )Total = 15 marks
to
8/8/2019 STPM Trials 2009 Chemistry Answer Scheme (Johor)
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Question 9(a) (i)
(b)
CH3CH 2He- + H-) --:- CI -H
CH 3CH 21
HO - / ,, - CIH H(1)
bimolecular nucleophilic substitution
" ; :H 2CH 3_ HO -C - H
""H(1 )
+ C ~
(1 )(ii) rate increases. (1)
(iii)
(i)
atomic size of Br>CI . (1)Hence the C-Br bond is weaker and )s easier to break than C-CI bond (1 )! "
8/8/2019 STPM Trials 2009 Chemistry Answer Scheme (Johor)
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Question 10They don't react with Fehling's solution - they are ketones (1)A and B gives positive iodoform test - they have the structureCH3 C = 0 or methylketone (1)IBoiling point of A> B - A is less branched compared to B (1)
A is CH3COCH2CH2CH3 (1)
o and E gives a positive test with Fehling's solution - they are aldehydes (1)o s optically active. Hence isomers of 0 are
Reduction of E gives a long chain compound, FE is CH3CH 2CH 2CH 2CHOF is CH3CH2CH2CH2CH20HPis copper(l) oxide (reject CU20)Examples of equations:
H3CH 3CH2CH 2C=O + H2NNH- - \Q,>-- N02 --_
12-
(2)(1 )(1 )(1 )(3)
Total = 15 marks
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