Chemistry 12 Semester 2 Revision
1
Coghlan
Semester 2 Revision - Answers
These are currently incomplete.
1.
(a) Cl2 (aq) + 2Br - (aq) → 2Cl - (aq) + Br2 (aq)
The brown solution becomes more red.
(b) 2Na(s) + 2C2H5 OH(l) → 2C2 H5 ONa(l) + H2 (g)
The sodium melts into a spherical shape and moves around on the surface of the liquid
accompanied by a hissing sound. The reaction appears to stop once all the sodium has
disappeared.
(c) SO3 (g) + H2 O(l) → H+(aq) + HSO4 -(aq)
The yellow solution gradually turns red as the gas is bubbled through.
(d) Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s)
The copper becomes coated at first with a black solid and then progressively with silvery white
crystals. The colourless solution gradually turns a pale blue colour.
2.
[Co(CN)6 ]3A ligand is a polar molecule or an anion that may attach itself to a transition metal ion. By
sharing a lone pair with the central metal ion, a coordinate covalent bond is formed.
CNCNCN
-
CNCo3+
CN-
CN-
3.
A catalyst provides a reaction surface so that an alternative lower energy reaction pathway is
enabled. The activated complex for a catalysed reaction is different than that for an uncatalysed
reaction. Because the activation energy is lower, it follows that more collisions will have
sufficient energy to form the activated complex.
Reaction path without
catalyst
Potential
Energy
Reaction path without
catalyst
Progress of reaction
Chemistry 12 Semester 2 Revision
4.
2
Coghlan
Consider the four test tubes set up below.
Iron
Iron
A
Zinc
Iron
B
Copper
Iron
C
Magnesium Iron
D
A contains a Fe(NO3 )2 solution. A mild amount of corrosion due to presence of H2O and
dissolved oxygen. No redox reaction.
B contains a Fe(NO3 )2 solution.
Zn(s) → Zn2+(aq) + 2e0.76 V
Fe2+(aq) + 2e- → Fe(s)
-0.44 V
The overall Eo is positive so the above redox reaction will proceed and a significant amount of
corrosion would occur.
C contains a CuSO4 solution.
Cu2+(aq) + 2e- → Cu(s)
0.34 V
Fe(s) → Fe2+(aq) + 2e0.44 V
The overall Eo is positive so the above redox reaction will proceed and a significant amount of
corrosion would occur. The amount of corrosion will probably be greater than B because of the
greater Eo value.
D contains a Mg(NO3 )2 solution.
Mg(s) → Mg2+(aq) + 2e2.37 V
Fe2+(aq) + 2e- → Fe(s)
-0.44 V
There would be very little corrosion of the iron as the Mg is acting as a sacrificial anode.
However, once all the Mg was used up, normal rusting would then commence due to the presence
of water and dissolved oxygen.
5.
(a) HCl is a strong acid and so fully ionises.
[H+] = 1 x 10-3 mol L-1
(b) [OH -][H+] = 10-14
[OH-]
= 10-14 /10-3
= 1 x 10-11 mol L-1
(c) [Cl-]
= [H+]
= 1 x 10-3 mol L-1
Chemistry 12 Semester 2 Revision
6.
3
Coghlan
Bleaching is the conversion of coloured dyes into colourless molecules – this can be achieved
over a long period of time by exposing the material to the sun. The UV radiation initiates
oxidation reactions which accomplish this. A solution of sodium hypochlorite also wil act as a
bleaching agent. The material to be bleached is first passed through a solution of NaClO and
then through a dilute solution of H2 SO4 . This generates hypochlorous acid.
ClO- (aq) + H+(aq) → HClO (aq)
This then undergoes reduction while the dyes undergo oxidation and bleaching occurs.
HClO (aq) + H+(aq) + 2e- → Cl- (aq) + H2 O(l)
7. (a) the mass of copper deposited
Cu2+(aq) + 2e- → Cu(s)
Q
=It
= 2.00 x 5.40 x 60 x 60
= 3.888 x 104 C
n(e ) = 3.888 x 104
9.65 x 104
= 0.403 mol
n(Cu) = ½ x 0.403
m(Cu) = ½ x 0.403 x 63.55
= 12.8 g
(b) the concentration of the copper (II) ions in solution immediately after the current was turned
off.
Moles Cu initially
= 0.250 x 1.00
= 0.250 mol
Moles Cu deposited
= ½ x 0.403
= 0.2015 mol
Moles Cu remaining
= 0.250 – 0.2015
=0.0485 mol
c(Cu2+) = 0.0485
0.250
= 0.194 mol L-1
8.
(a) Write an equation for dissolving aluminium hydroxide in a NaOH solution.
Al(OH)3 (s) + OH-(aq) → [Al(OH)4 ]-(aq)
(b) The impurities, among them being Fe 2O3 , settle out of solution as insoluble material and this
is then pumped out of the reaction vessel.
(c) No because the acid would merely produce both Al3+ and Fe3+ in solution and further
chemical reactions would have to proceed to separate them.
N2 O4 (g)
2NO2 (g)
∆ H = + 58 kJ mol-1
(a) The left to right reaction is endothermic, therefore the addition of more heat would cause the
forward reaction to speed up initially to minimise this change. The gas mixture would become
more brown.
Chemistry 12 Semester 2 Revision
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Coghlan
(b) The concentration of the N2 O4 (g) would decrease and the concentration of the NO2 (g) would
increase resulting in more NO2 (g) molecules causing a darker brown colour.
(c) See (a) above.
(d) If the volume of the container were suddenly doubled, the pressure would decrease to one
half of the original value. That reaction would proceed which would make more molecules
so that the change in pressure could be minimised. Therefore the forward reaction would
speed up initially and the colour would become more brown.
10.
O
O
C
This is the functional group of
the esters.
CH3
Benzene ring
OH
C
O
This is the functional group of
the alkanoic acids.
11.
(a) Balance this equation.
2C8 H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2 O(g)
Petrol has a density of 702.5 g L-1. For each litre of petrol which is consumed, calculate:
(b) the mass of oxygen required
m(C8 H18 ) =702.5 g
n(C8 H18 ) = 702.5
M (C 8 H18) = (8 x 12.01) + (18 x 1.008)
114.224
= 114.224 g mol-1
= 6.15 mol
n(O2 )
= 12.5 x 6.15
m(O2 )
= 12.5 x 6.15 x 32.00
= 2460 g
(c) the volume of carbon dioxide emitted, measured at 100.7 kPa pressure and 22.0o C.
n(O2 )
= 12.5 x 6.15
n(O2 )STP = 12.5 x 6.15 x 22.4
= 1722 L
P1
V1
T1
= 101.3 kPa
= 1722 L
= 273 K
P1 V1
T1
= P2 V2
T2
P2
V2
T2
= 100.7 kPa
=?
= 295 K
Chemistry 12 Semester 2 Revision
V2
5
Coghlan
= P1 V1 T2
P2T1
= (101.3 x 1722 x 295)
(100.7 x 273)
= 1872 L
12.
(a) CH3 CH2 CH2 CH2 OH
1 - butanol
(b) CH3  C  CH2 CH3
O
CH3
(c)
butanone
H
HCCH
(d) (CH3 )2 C =CHCH3
2-methyl-2-butene
CH3 CH3
methylbutane
13.
14.
(a) sodium nitrate solution and sodium sulfate solution
Add some colourless barium nitrate solution to both. The sodium nitrate solution would show
not reaction whereas the sodium sulfate solution would produce a white precipitate of barium
sulfate.
(b) carbon dioxide gas and hydrogen gas
Bubble both gases through lime water. The gas which produces a milky colouration is the carbon
dioxide.
(c) solid magnesium hydroxide and solid lead (II) sulfate
Add some dilute hydroxide to both solids. The solid which dissolves is the magnesium
hydroxide.
(d) silver nitrate and sodium nitrate
Dissolve both in distilled water. Add some sodium chloride solution to both. The one which
produces the white precipitate is the silver nitrate.
The diagram below indicates the structure of an addition polymer.
CH3
CH3
C1
CH3
CH3
 C  CH2  C  CH2  C  CH2  CH2  C  CH2  C 
C1
Cl
CH3
C1
Which of the following compounds could polymerise to form this chain?
(a) CH3  C = CH2
Cl
(b) CH3  CH = CH
Cl
C1
Chemistry 12 Semester 2 Revision
6
(c) CH2 = C  CH = CH2
Cl
15.
(d) CH2 = CH  CH2
Cl
What is the oxidation number of Xe in:
(a)
(b)
(c)
(d)
16.
Coghlan
XeO4 2XeF4
PtXeF6 (Pt2+)
Na2 XeO 4
+6
+4
+4
+6
Draw the structures of the following species, representing all valence shell electron pairs as
either • or x.
(a)
trichloromethane (chloroform)
H
xx
x
xx x
H C Cl
H
(b)
carbonate ion
(c)
sulphite ion
(d)
ethanol
Chemistry 12 Semester 2 Revision
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Coghlan
17.
(a) a halogen which is a liquid at room temperature and pressure - bromine.
(b) a common corrosion resistant alloy of iron – stainless steel.
(c) a salt whose water solution can be used as an oxidizing agent – sodium hypochlorite.
(d) a positive ion which is normally green in solution – Cr3+.
(e) a compound which is a gas at room temperature and pressure and whose molecules have a
dipole moment – HCl, NH3 .
(f) a substance, other than water, which is a liquid at room temperature and pressure and which
has H-bonding between its molecules. – ethanol.
18.
Cu(s) → Cu2+(aq) + 2e-0.34 V
+
2H (aq) + 2e → H2 (g) 0.00 V
Overall E0 value is negative so the reaction will tend not to happen. For the metals which
produce hydrogen from dilute acids, the E0 values are positive.
19.
22.40 mL of 0.1000 mol L-1 KMnO4 solution was required to oxidise 20.00 mL of an acidified
FeSO 4 solution. Determine:
(a) the concentration of the FeSO4 solution.
MnO 4 - (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (l)
x2
Fe2+ (aq) → Fe3+ (aq) + eX5
2MnO4- (aq) + 16H+ (aq) + 5Fe2+ (aq) → 2Mn2+ (aq) + 8H2 O (l) + 5Fe3+ (aq)
n(MnO4 - ) = 0 .02240 x 0.1000
n(Fe2+)
= 5/2 x 0 .02240 x 0.1000
= 0.005600 mol
c(FeSO4 ) = 0.005600
0.02000
= 0.2800 mol L-1
(b)
the mass of FeSO4 in 20.00 mL of solution.
m(FeSO4 ) = 0.005600 x 319.46
= 1.789 g
(c) the mass of Fe in 1.250 L of the FeSO4 solution.
m(FeSO4 ) = 0.005600 x 319.46 x 1250/20
= 111.8 g
20.
(a) What are the two types of electrochemical cell?
Cells that cannot be recharged (the common dry cell) and those which can be recharged (lead
accumulator and NiCd batteries).
(b) Outline the relationship between the chemical reaction and the electrical energy which is
involved in the two types of electrochemical cell.
In a normal dry cell when it is discharging, Chemical Potential Energy → Electrical Energy.
In a rechargeable cell in its recharging mode, Electrical Energy → Chemical Potential Energy.
Chemistry 12 Semester 2 Revision
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Coghlan
21.
(a)
NH4 +
(b)
BeC12
(c)
H2 Te
(d)
SiH3 F
Tetrahedral – every bond is equivalent and so repel one another
equally.
The cation is of course polar because it is charged.
Linear – the individual bonds are polar but the molecule is non-polar
since the dipoles are in opposite directions and cancel.
Bent – there are two non-bonding electron pairs on the Te which force
the two H-Te bonds to be bent.
The molecule will be basically tetrahedral but not quite symmetrical
due to the strongly polar nature of the Si- F bond.
22. To produce a soap, fats and oils are hydrolysed with concentrated NaOH solution. The
hydrolysis of glyceryl tristerate produces the soap, sodium stearate. Detergents are
alkylbenzene sulfonates produced from the reactions of benzene and alkanes from the
petroleum refining industry.
23.
The rusting of iron involves an electrochemical process. Explain the chemistry of rusting (using
chemical equations). Comment on how it is possible to reduce the rate at which corrosion occurs.
In the absence of other metals;
Fe(s) → Fe 2+(aq) + 2eO2 (g) + 2H2 O(l) + 4e- → 4OH-(aq)
Iron (II) hydroxide forms and then oxidises further to rust.
4Fe(s) + 3O2(g) + 2H2 O(l) → 2Fe2 O3 .H2O(s)
It is possible to reduce the rate of reaction by modifying the environment, modification of the
iron, cathodic protection and protective metallic and non-metallic coatings.
24.
(a) Ba2+(aq) + SO42-(aq) → BaSO4 (s)
m(BaSO4)
= 1.70 g
n(BaSO4)
= 1.70
233.36
= 7.285 x 10-3 mol
n(MgSO4)
= 7.285 x 10-3 mol
c(MgSO4)
= 7.285 x 10-3
5.00
= 1.457 x 10-3 mol L-1
c(MgSO4)
= 1.457 x 10-3 x 120.37 x 1000
= 175 mg L-1
(b)
Mg2+(aq) + CO3 2-(aq) → MgCO3 (s)
n(MgSO4)
= 1.457 x 10-3 x 100 mol
n(Na2 CO3 .10H2 0)
= 1.457 x 10-3 x 100 mol
m(Na2 CO3 .10H2 0)
= 1.457 x 10-3 x 100 x 286.15
= 41.7 g
Chemistry 12 Semester 2 Revision
25.
26.
9
Coghlan
MnO4 - (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2 O (l)
x2
2Cl (aq) → Cl2 (g) + 2e
X5
2MnO4 - (aq) + 16H+ (aq) + 5Cl- (aq) → 2Mn2+ (aq) + 8H2 O (l) + 5Cl2 (g)
n(KMnO4 )
= 5.50
158.04
= 3.48 x 10-2 mol
n(Cl2 )
= 5/2 x 3.48 x 10-2 mol
V(Cl2 ) STP = 5/2 x 3.48 x 10-2 x 22.4
= 1.949 L
P1
V1
T1
= 101.3 kPa
= 1.949 L
= 273 K
P1 V1
T1
V2
= P2 V2
T2
= P1 V1 T2
P2T1
= (101.3 x 1.949 x 300)
(109.5 x 273)
= 1.98 L
Oxidation numbers
+7 –2
WO 4 - (aq)
→
P2
V2
T2
= 109.5kPa
=?
= 300 K
+2
W2+(aq)
Because the oxidation number of the W has decreased it has been reduced. Therefore electrons
have been gained.
WO 4 - (aq) + 8H+(aq) + 5e- →
W2+(aq) + 4H2 O(aq)
27.
(a) CH3  CH2  C = O
O  CH2  CH3
Ethyl propanoate
28.
(b) CH2 = CH  CH  CH2  CH3
CH2  CH2  CH3
3-ethyl-1-pentene
Electron configuration
F 1s2 , 2s2, 2p5
Cl 1s2 , 2s2 , 2p6, 3s2, 3p5
The bonding electrons in fluorine come from energy level 2 whereas in chlorine they come from
the higher energy level 3. Because of this the physical size of the Cl2 molecule and bond length
is bigger than that of F2 .
Chemistry 12 Semester 2 Revision
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Coghlan
29.
OH - (aq) + H+(aq) → H2 O(l)
n(HCl)
= 0.0159 x 0.100
= 0.00159 mol
n(OH-) in 25 mL
= 0.00159 mol
n(OH ) in 100 mL
= 0.00159 x 4
= 0.00636 mol
n(NaOH) in original solution
= 0.0400 x 0.5
= 0.0200mol
n(OH-) difference
= 0.0200 – 0.00636
= 0.0136 mol
This must have been the number of moles of OH- that originally reacted with the ammonium salt
on a one to one mole ratio.
So the original compound must have contained 0.0136 moles of NH3 .
m(NH3 )
= 0.0136 x 17.034
= 0.232 g
% (NH3 )
= 0.232 x 100
1.00
= 23.2%
30.
The cost of obtaining a sufficiently large amount of Al by far exceeds the value of the Fe
produced.
31.
(a) In Step 1, what gas was evolved? CO2
(b)Write a balanced equation for the reaction of this gas with the calcium hydroxide solution.
Ca(OH)2(aq) + CO2 (g) → CaCO3 (s) + H2 O(l)
(c) In Step 2, what was the formula of the black powder? CuO
(d) Write a balanced equation for the reaction of the black powder with the sulphuric acid.
CuO (s) + 2H+ (aq) → Cu2+ (aq) + H2 O(l)
(e) Write the equation for the cathodic reaction in Step 3.
Reduction at the cathode – gain of electrons.
Cu2+(aq) + 2e- → Cu (s)
(f) Write equations for the two reactions in Step 4.
Cu2+(aq) + 2OH- (aq) → Cu(OH)2 (s)
Cu(OH)2(s) + 4NH3 (aq) → [Cu(NH3 )4 ]2+(aq) + 2OH- (aq)
(g) Identify the original green powder. CuCO3
32.
4Au (s) + 8CN- (aq) + 2H2O(l) + O2 (g) →
4[Au(CN)2 ]-(aq) + 4OH-(aq)
(a) The oxidation state of the gold in the complex ions is +1.
(b) The species that is reduced is the oxidising agent so it must be oxygen gas.
(c) Gold is not normally oxidised by oxygen but the cyanide ion helps in this process by forming
the complex ion.
Chemistry 12 Semester 2 Revision
11
Coghlan
33. Draw the structures of each of the following molecules.
(a) 1,1,2-trichloro-4-methyloctane
( b) propanoic acid
(c) methyl ethanoate
(d) ethanal
34.
The aluminium wheels of jumbo jets are fitted with tyres which are inflated with nitrogen.
Why would nitrogen be preferred to air?
M(air)
= 0.21(32.00) + 0 .79(28.02)
= 28.86 g mol-1
This means that the weight of fully inflated tyres with nitrogen would be less than if inflated with
air so this is less weight the aircraft has to lift.
There also may be a problem with the direct contact between air and aluminium. Even though
aluminium forms an impervious oxide layer when reacting with air, this slight oxidation may
compromise the integrity of a highly stressed aircraft part.
Chemistry 12 Semester 2 Revision
35.
12
Coghlan
1-chloropropane and 2-chloropropane are structural isomers. One way to assign the correct
structure to each molecule is to chlorinate them further and count the number of products derived
from them. Predict the number of dichloropropanes (draw the structures) formed from each
compound.
On further chlorination, 1-chloropropane forms 3 possible isomers while 2-chloropropane only
forms 2.
36.
(a) MnO4- (aq) + 8H+ (aq) + 5e- → Mn2+ (aq) + 4H2O (l)
x2
H2 C2 O4 (aq) → 2H+ (aq) + 2CO2 (g) + 2eX5
2MnO4 - (aq) + 6H+ (aq) + 5H2 C2 O4 (aq) → 2Mn2+ (aq) + 8H2 O (l) + 10CO2 (g)
(b) Calculate the number of moles of KMnO4 used in the titration.
n(KMnO4 ) = 0.03440 x 0.0216
= 7.430 x 10-4 mol
(c) Determine the number of moles of oxalic acid which would react with the KMnO4 .
n(H2 C2 O4 ) = 5/2 x 7.430 x 10-4
= 1.858 x 10-3 mol
(d) Use the result from (c) and the mass of oxalic acid crystals to determine the molecular mass
of the H2 C2 O4 . xH2 O.
m(H2 C2 O4 ) = 1.858 x 10-3 x (90.036 + 18.016 X)
0.234
= 1.858 x 10-3 x (90.036 + 18.016 X)
X
= 1.99
=2
M(H2 C2 O4 .2H2 O)
= 126.07
(e) Determine the value of x in the oxalic acid.
Chemistry 12 Semester 2 Revision
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Coghlan
See (d) above.
(f) Sketch the structure of oxalic acid.
(g) Give the equation for the reaction of oxalic acid with excess ethanol.
HOOCCOOH(l) + 2C2H5 OH(l) → C2 H5OOCCOOC2 H5 (l) + 2H2O(l)
(h) Is oxalic acid a polyprotic acid or a monoprotic acid? Explain.
It is a polyprotic acid because it has two active hydrogens, one on either end of the structure,
which are able to react with the ethanol.
37.
Zn(s) → Zn2+(aq) + 2e0.76 V
2+
Fe(s) → Fe (aq) + 2e
0.44 V
Because Zn has a more positive Eo value than Fe, it will preferentially oxidise and thus act as a
sacrificial anode.
38.
A 1.0 mol L-1 KI solution is electrolysed using inert Pt electrodes.
(a) Species present – K+, I-, H2 O (very small concentrations of OH- and H+)
Reduction occurs at the cathode so possible reduction half-equations are;
K+ + e- → K
-2.93 V
2H2O + 2e- → H2 + 2OH-0.83 V This reduction will tend to occur as its Eo value is
the more positive.
Oxidation occurs at the anode so possible oxidation half-equations are;
2I- → I2 + 2e-0.54 V
+
2H2O → O2 + 4H + 4e -0.82 V This reduction will tend to occur as its Eo value is the
more positive.
(b) EMF = 0.83 + 0.82
= 1.65 V
+ (c) K , I , H+, OH-
39.
(a) Copper – excellent electrical conductor – electrical wiring.
(b) Hydrochloric acid – reacts with carbonates and hydroxides – used to clean waste cement off
newly constructed brick walls.
(c) sodium hypochlorite – produces hypochlorous acid, HClO – used as the oxidising agent in
bleaches.
(d) sodium stearate – a long chain molecule one end of which is hydrophillic and the other is
hydrophobic – used as a cleaning agent, soap, the hydrophobic end dissolves non-polar molecules
such as fat and grease while the hydrophillic end attaches itself to water and is removed with the
whole chain when water drains off.
Chemistry 12 Semester 2 Revision
14
40.
41.
Which of the following is a characteristic property of a covalent molecular compound?
(a) dissolves in polar solvents – polar substance, ionic compound.
(b) relatively low melting point – covalent molecular compound.
(c) malleable and ductile- metal.
(d) conducts electricity when molten but not when solid – ionic compound.
42.
A dilute acid is one in which the concentration of the substance causing the acidity is low.
e.g. 0.20 mol L-1 HCl.
A weak acid is one which is only partially ionised in solution.
e.g. ethanoic acid.
43.
(a) Molten sulfur
(b) Sulfur dioxide.
(c) Divanadium pentoxide.
(d) 2S(s) + 2O2 (g) → 2SO 2 (g)
2SO 2 (g) + O 2(g)
2SO 3 (g)
2SO 3 (g) + H2 SO4 (l) → H2 S2 O7 (l)
H2 S2 O7 (l) + H2 O(l) → 2H2SO4 (l)
n(H2 SO4)
n(O2 )
m(O2 )
n(S)
m(S)
= 2500 x 12 x 0.98
= 2.94 x 106 mol
= 2.94 x 106 mol
= 2.94 x 106 x 32.00
= 9.41 x 107 g
= 9.41 tonne
= 2.94 x 106 mol
= 2.94 x 106 x 32.06
= 9.43 x 107 g
= 9.43 tonne
Coghlan
Chemistry 12 Semester 2 Revision
44.
15
Coghlan
Many of the transition metals form coloured solutions. Write down the colour of aqueous
solutions of the following pure substances. State the species that is responsible for the colour.
(a) K2 Cr2 07 orange, Cr2 07 2(b) CuSO4 blue, Cu2+
(c) NiCl2 green, Ni2+
(d) FeCl3 Fe3+, brown
45.
Sodium
98
Argon
-189
Melting point
o
C
Heat of fusion
2.5
1.3
kJ mol-1
Boiling point
892
-186
o
C
Heat of
89
6.7
vaporisation
KJ mol-1
Melting point and boiling point – sodium is held together by metallic bonding which is the
simultaneous electrostatic attraction between the sodium ions and the delocalised electrons. This
strong metallic bonding causes the mp/bp to be reasonably high. There are only weak dispersion
forces between argon atoms and consequently it has a low mp/bp.
Both the latent heats of fusion and vaporisation involve changing the potential energies or the
spacing between the particles. The fact that both latent heats are similar for both species
indicates that the spacing between the particles in both the solid, liquid and gaseous phases are
similar so that the energies required to separate the species is similar.
46.
A metallic object, to be plated with Cu, is one electrode in an electrolytic cell. A bar of Cu
(pure) is the other electrode.
(a) It must be the cathode because the Cu2+ will migrate to the cathode to be reduced.
(b) anode Cu(s) → Cu2+(aq) + 2ecathode Cu2+(aq) + 2e- → Cu(s)
(c) The electrolyte in this cell would be an aqueous solution of Cu2+ together with cyanide
complexes to ensure a low concentration of free copper ions so that the electroplating produces
an acceptable product.
(d) Q
= It
= 10.0 x 12.5 x 60
= 7500 C
n(e-)
= 7500
96500
= 0.0777 mol
n(Cu)
= ½ x 0.0777
m(Cu)
= ½ x 0.0777 x 63.55
= 2.47 g
Chemistry 12 Semester 2 Revision
47.
16
Coghlan
Draw complete structural formulae, including all hydrogen atoms, and give the IUPAC names for
the following.
(a)
propanal
(b)
2-methyl-2-butanol
(c)
(d)
methyl propanoate
(e)
48.
butanal
1,1-dibromoethene
(a) Determine the empirical formula of the compound.
n(CO2 )
= 0.0220/44.01
= 5.00 x 10-4 mol
n(C)
= 5.00 x 10-4 mol
m(C)
= 5.00 x 10-4 x 12.01
= 0.006005 g
n(H2 O)
= 0.0135/18.016
= 7.49 x 10-4 mol
n(H)
= 1.498 x 10-3 mol
m(H)
= 7.49 x 10-4 x 2 x 1.008
= 0.00151 g
m(O)
= m(sample) - m(C) - m(H)
= 0.0155 – 0.006005 – 0.00151
= 0.007985 g
n(O)
= 0.007985/16.00
= 4.99 x 10-4 mol
Empirical formula
C
H
-4
n
5.00 x 10
1.498 x 10-3
÷ 4.99 x 10-4
1
3
CH3O
(b) Determine the relative molecular mass of the compound.
P1
= 101.3 kPa
P2
= 101.3 kPa
V1
= 0.306 L
V2
=?
T1
= 473 K
T2
= 273 K
P1 V1
T1
= P2 V2
T2
O
4.99 x 10-4
1
Chemistry 12 Semester 2 Revision
17
Coghlan
V2
= P1 V1 T2
P2T1
= (101.3 x 0.306 x 273)
(101.3 x 473)
= 0.177 L
n(sample) = 0.177/22.4
= 0.00790
MR
= 0.490/0.00790
= 62.0
(c) Molecular formula of the compound.
Empirical formula CH3 O
MR
= 12.01 + 3 x 1.008 + 16.00
= 31.034
Molecular formula = 2 x Empirical formula
= C2H6 O2
Analysis of the compound revealed that it reacted with Na metal to produce hydrogen gas. It was
also found to decolourise bromine water and produce a 1,2-substituted product. The compound
was also known to be a hydrate.
(d) The compound is an alcohol.
(e) The production of a 1,2-addition product indicates that the compound is unsaturated.
(f) With one water of hydration the molecular formula becomes C2H4 O.H2 O and the structural
formula
49.
Carbon disulphide can be manufactured by the endothermic reaction of sulphur trioxide and
carbon dioxide according to the equation:
2SO 3 (g) + CO2 (g) → CS2 (g) + 4O2 (g)
If this reaction is allowed to come to equilibrium at 100o C and a pressure of 400 kPa, what would
be the effect on the equilibrium amount of carbon disulpfide of each of the following changes?
(a) Raising the temperature to 200oC whilst maintaining the pressure of 400 kPa would cause the
equilibrium amount of the carbon disulfide to increase.
(b) Reducing the pressure to 200 kPa whilst maintaining the temperature at 100o C kPa would
cause the equilibrium amount of the carbon disulfide to increase.
(c) Removing oxygen from the equilibrium mixture kPa would cause the equilibrium amount of
the carbon disulfide to increase.
50.
(a) chlorine atom 1s2 2s2 2p6 3s2 3p5
(b) sodium ion 1s2 2s2 2p6
(c) aluminium ion 1s2 2s2 2p6
(d) bromide ion 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5