Copyright reserved / Kopiereg voorbehou
DEPARTMENT OF CHEMISTRY
DEPARTEMENT CHEMIE
CHM 172
Final Examination / Finale Eksamen
Pages / Bladsye:
15
Examiners / Mrs BA Castleman
Eksaminatore: Mrs AC Botha
Mr NJ de Beer
External Examiner /
Mrs N Conradie
Eksterne eksaminator :
ia
17 November 2014
3 hours / ure
100
or
Date /Datum:
Time /Tyd:
Marks / Punte:
Initials
Voorletters
Pr
et
Surname
Van
Student Number
Studente Nommer
Signature
Handtekening
Section /
Afdeling
Question / Vraag
©
B
Question 3 / Vraag 3
13
Question 4 / Vraag 4
16
Subtotal A / Subtotaal A
62
rs
17
38
Subtotal B / Subtotaal B
TOTAL / TOTAAL:






INSTRUCTIONS
All answers (calculations, sketches, and diagrams) must
be given in black or blue ink.
All calculations must be shown in full.
Information pages are attached (pages 16 - 18).
Only the information pages may be carefully removed.
Use the back of the page if you need more space to
answer a question. Clearly indicate this action in your
answer.
You may use the molecular model kit.
Examiner /
Eksaminator
16
Question 2 / Vraag 2
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A
Marks/ Punte
ity
Question 1 / Vraag 1
of
RESULTS / RESULTATE
100






%
INSTRUKSIES
Alle berekeninge, sketse en diagramme moet in swart of
blou ink gegee word.
Alle berekeninge moet volledig vertoon word.
Inligtingsblaaie is aangeheg (bladsye 16 - 18 ).
Slegs die inligtingsblaaie mag versigtig verwyder word.
Gebruik die agterkant van die bladsy indien jy meer
spasie benodig vir ‘n antwoord. Dui hierdie aksie duidelik
aan in jou antwoord.
Jy mag die molekulêre modelstel gebruik.
STUDENT NUMBER
STUDENTENOMMER
SECTION A
[62]
AFDELING A
Question 1 / Vraag 1
1.1
Die element lood, Pb, bestaan uit vier natuurlike
isotope, met atoommassas 203.97302 𝑢,
205.97444 𝑢, 206.97587 𝑢 en 207.97663 𝑢. Die
persentasievoorkoms van die twee ligste isotope
is: 204𝑃𝑏= 1.40% and 206𝑃𝑏= 24.1%.
Bepaal die persentasievoorkoms van die ander
twee isotope.
[3]
ia
The element lead, Pb, consists of four naturally
occurring isotopes with atomic masses
203.97302 𝑢, 205.97444 𝑢, 206.97587 𝑢 and
207.97663 𝑢. The percentage abundances of the
two lightest isotopes are:
204
𝑃𝑏= 1.40% and 206𝑃𝑏= 24.1%.
Determine the percentage abundance of the
other two isotopes.
[3]
or
1.1
[16]
of
Pr
et
[Answer: 23.79% & 50.71%]
Consider a neutral hydrogen atom.
1.2.1
Calculate the energy of the electron in the
hydrogen atom if the electron has been excited
to the third excited state.
[2]
-19
J]
Beskou ‘n neutrale waterstofatoom.
1.2.1
Bereken die energie van die elektron in die
waterstofatoom indien die elektron tot die derde
opgewekte toestand opgewek is.
[2]
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[Answer: -1.3620 × 10
1.2
ity
1.2
1.2.2
How many discrete spectral lines could be
observed as the electron returns to the ground
state? Use a diagram to illustrate your answer.
[3]
1.2.2
Hoeveel diskrete spektrale lyne kan
waargeneem word as die elektron terugkeer na
die grondtoestand? Gebruik ‘n diagram om jou
antwoord te illustreer.
[3]
©
[Answer: 6]
CHM 172
Final Examination
2/15
© University of Pretoria
2014
STUDENT NUMBER
1.3.1
A halogen compound consists of carbon, oxygen
and chlorine only. Combustion of a 1.380 𝑔
sample of the pure compound produces
0.6139 𝑔 of CO2 and 0.9893 𝑔 of Cl2. The mass
percentage oxygen is 16.174%.
The value of 𝑛 in �𝐶𝑥 𝐶𝑙𝑦 𝑂𝑧 �𝑛 is 1. Determine
the molecular formula of this compound.
[4]
1.3
1.3.1
‘n Halogeenverbinding bestaan slegs uit
koolstof, suurstof en chloor. Verbranding van ‘n
1.380 𝑔 monster van die suiwer verbinding
produseer 0.6139 𝑔 CO2 en 0.9893 𝑔 Cl2. Die
massapersentasie van suurstof is 16.174%
Die waarde van 𝑛 in �𝐶𝑥 𝐶𝑙𝑦 𝑂𝑧 �𝑛 is 1. Bepaal
die molekulêre formule van die verbinding. [4]
ia
1.3
STUDENTENOMMER
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[Answer: CCℓ2O]
1.3.2
Draw a Lewis dot structure and the line
structure to illustrate the geometry of this
halogen compound. Clearly show all bond
angles and lone pairs. Indicate polarity of the
molecule.
[4]
1.3.2
Line structure / Lynstruktuur
©
Dot structure / kolstruktuur
Teken ‘n Lewis kolstruktuur en ‘n lynstruktuur
om die geometrie van hierdie halogeenverbinding te illustreer. Wys alle
bindingshoeke en alleenpare duidelik in die
lynstruktuur. Dui die polariteit van die molekule
aan.
[4]
CHM 172
Finale Eksamen
3/15
© Universiteit van Pretoria
2014
STUDENT NUMBER
STUDENTENOMMER
Question 2 / Vraag 2
CH2 C
H2 N
OH
2.1.1
+
O2
Gebruik bindingsdissosiasie-energieë om die
energie wat vrygestel word deur die verbranding
van een mol glisien te bereken, gegee die
ongebalanseerde vergelyking en bindingsentalpieë:
CO2 + H2O + N
O
Bond / Binding
C−C
C−H
C−O
C=O
C−N
346
413
358
745
305
O−O
O=O
N−H
N=O
H−O
Bond dissociation energy /
Bindingsdissosiasie-energie
(kJ/mol)
146
498
391
607
463
Pr
et
Bond / Binding
Bond dissociation energy /
Bindingsdissosiasie-energie
(kJ/mol)
O
ia
Using bond dissociation energies, calculate the
energy released by the combustion of one mole
of glycine, given the unbalanced equation and
bond enthalpies:
or
2.1.1
[17]
Balanced equation / Gebalanseerde vergelyking:
[2]
___ C2H5O2N + ___ O2 → ___ CO2 + ___ H2O + ___ NO
of
Calculation / Berekening
[4]
2.454 𝑔 of glycine is combusted in a bomb
calorimeter which contains 750.0 𝑔 of water.
The temperature changes from 22.150 ℃ to
24.540 ℃. Determine the heat capacity of the
calorimeter.
[Hint: use the answer from 2.1.1]
[5]
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2.1.2
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[Answer: -708 kJ/mol]
2.1.2
2.454 𝑔 glisien word in ‘n bomkalorimeter wat
750.0 𝑔 water bevat verbrand. Die temperatuur
verander van 22.150 ℃ tot 24.540 ℃. Bepaal
die hittekapasiteit van die kalorimeter.
[Wenk: gebruik die antwoord van 2.1.1]
[5]
©
[Answer: 6.539 kJ ℃−1 ]
CHM 172
Final Examination
4/15
© University of Pretoria
2014
STUDENT NUMBER
2.2
STUDENTENOMMER
Consider the reaction:
2.2
Beskou die reaksie:
𝟐𝐍𝐎(𝐠) + 𝐂𝐥𝟐 (𝐠) → 𝟐𝐍𝐎𝐂𝐥(𝐠)
𝑺° (𝑱/𝒎𝒐𝒍 ∙ 𝑲)
210.8
223.1
261.7
∆𝒇 𝑯° (𝒌𝑱/𝒎𝒐𝒍)
90.29
51.71
or
Substance / Stof
NO (g)
Cℓ2 (g)
NOCℓ(g)
Bereken naastenby die ewewigskonstant vir
hierdie reaksie by 431 𝐾.
[6]
Gegee:
ia
Calculate the approximate equilibrium constant
for this reaction at 431 𝐾.
[6]
Given:
Question 3 / Vraag 3
One half-cell in a voltaic cell at 25℃ contains an
acidified 𝑉𝑂2+ / 𝑉𝑂2+ solution.
The [ 𝑉𝑂2+ ] = 2.00 𝑀 and the
[ 𝑉𝑂2+ ] = 1.00 × 10−2 𝑀.
This is combined with a zinc electrode
immersed in a 1.00 𝑀 Zn(NO3)2 solution.
Given:
3.1
Write down the fully balanced cell reaction. [2]
3.1.1
U
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3.1
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[Answer: 1036]
Een halfsel in ‘n voltaïese sel by 25℃ bevat ‘n
2+
aangesuurde VO+
oplossing.
2 / VO
+
Die [ VO2 ] = 2.00 𝑀 en die
[ VO2+ ] = 1.00 × 10−2 𝑀.
Dit word gekombineer met ‘n sink-elektrode
wat gedompel is in ‘n 1.00 𝑀 Zn(NO3)2
oplossing. Gegee:
+
−
𝟐+ (𝐚𝐪)
𝐕𝐎+
+ 𝐇𝟐 𝐎 (𝓵)
𝟐 (𝐚𝐪) + 𝟐𝐇 (𝐚𝐪) + 𝐞 → 𝐕𝐎
𝐄° = 𝟏. 𝟎𝟎 𝐕
Skryf die volledig-gebalanseerde selreaksie
neer.
[2]
©
3.1.1
[13]
CHM 172
Finale Eksamen
5/15
© Universiteit van Pretoria
2014
STUDENT NUMBER
3.1.2
STUDENTENOMMER
Under these conditions the measured cell
potential is 1.89 𝑉. Calculate the pH of the
acidified solution.
3.1.2
[5]
3.1.3
Calculate ∆𝐺° for this redox reaction.
[2]
3.1.3
Bereken ∆𝐺° vir hierdie redoksreaksie.
[2]
of
[Answer: -340 kJ]
Pr
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or
ia
[Answer: 0.078]
Onder hierdie toestande is die gemete
selpotensiaal 1.89 𝑉. Bereken die pH van die
aangesuurde oplossing.
[5]
3.2
ity
An electroplating apparatus is used to coat a ring
with 6.10 𝑔 of gold. It is set up as the cathode in
a bath containing basic [Au(CN)4] - ions.
The balanced half reaction is
rs
3.2
‘n Elektroplaterings-apparaat word gebruik om
‘n ring met 6.10 𝑔 goud te plateer. Dit word
opgestel as die katode in ‘n bad wat basiese
[Au(CN)4]- ione bevat. Die gebalanseerde
halfreaksie is
[𝐀𝐮(𝐂𝐍)𝟒 ]− (𝐚𝐪) + 𝟑𝐞− → 𝐀𝐮(𝐬) + 𝟒𝐂𝐍 − (𝐚𝐪)
U
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If a steady current of 5.00 𝐴 is applied to the
system, how many minutes will it take to plate
the ring?
[4]
As ‘n gelykmatige stroom van 5.00 𝐴 toegepas
word op die sisteem, hoeveel minute sal dit
neem om die ring te plateer?
[4]
©
[Answer: 29.9 min]
CHM 172
Final Examination
6/15
© University of Pretoria
2014
STUDENT NUMBER
STUDENTENOMMER
Question 4 / Vraag 4
ia
Die massapersentasie van chloriedione in ‘n
20.00 𝑚𝐿 monster van seewater is bepaal deur
die monster met silwernitraat te reageer.
 Die silwernitraatoplossing is voorberei
deur 110.36 𝑔 silwer nitraat in genoeg
gedistilleerde water op te los om 250.0 𝑚𝐿
oplossing te maak.
 25.0 𝑚𝐿 van hierdie oplossing is verdun
tot 200.0 𝑚𝐿.
38.25 𝑚𝐿 van die verdunde silwernitraatoplossing word benodig om volledig met die
chloride in die monster seewater te reageer.
or
The mass percentage of chloride ions in a 4.1
20.00 𝑚𝐿 sample of seawater was determined by
reaction of the sample with silver nitrate.
 The silver nitrate solution was prepared by
dissolving 110.36 𝑔 silver nitrate in
enough distilled water to make 250.0 𝑚𝐿
solution.
 25.0 𝑚𝐿 of this solution was diluted to
200.0 𝑚𝐿.
38.25 𝑚𝐿 of the diluted silver nitrate solution
was needed to completely react with the chloride
in the sample of seawater
Pr
et
4.1
[16]
Cℓ¯(aq) + Ag+(aq) + NO3¯(aq) → AgCℓ(s) + NO3¯(aq)
4.1.1
What is the chloride ion concentration in 4.1.1
seawater?
[4]
©
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[Answer: 0.621 M]
Wat is die chloriedioonkonsentrasie in die
seewater?
[4]
CHM 172
Finale Eksamen
7/15
© Universiteit van Pretoria
2014
STUDENT NUMBER
4.1.2
STUDENTENOMMER
What is the mass percentage of chloride ion in 4.1.2
seawater if the density of seawater is
1.025 𝑔/𝑚𝐿?
[3]
Pr
et
or
ia
[Answer: 2.15%]
Wat is die massapersentasie van chloriedione in
seewater indien die digtheid van seewater
[3]
1.025 𝑔/𝑚𝐿 is?
The silver was recovered from the precipitate
with a yield of 83.6%. What is the mass of silver
that was recovered?
[3]
4.1.3
Die silwer word herwin uit die neerslag met ‘n
opbrengs van 83.6%. Wat is die massa silwer
wat herwin is?
[3]
of
4.1.3
©
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[Answer: 1.12 g]
CHM 172
Final Examination
8/15
© University of Pretoria
2014
STUDENT NUMBER
The following two solutions are mixed:
4.2
3
1) 45.0 cm of 0.258 M CdCℓ2
2) 75.0 cm3 of 0.443 M NaOH
Calculate [OH−] and [Na+] in the final solution
after the precipitation is complete.
[6]
Die volgende twee oplossings word gemeng:
1) 45.0 cm3 van 0.258 M CdCℓ2
2) 75.0 cm3 van 0.443 M NaOH
Bereken [OH−] en [Na+] in die finale oplossing
nadat die neerslag volledig gevorm het.
[6]
CdCℓ2(aq) + 2NaOH(aq) → Cd(OH)2(s) + 2NaCℓ(aq)
ity
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or
[Answer: 0.0834 M]
ia
4.2
STUDENTENOMMER
SECTION B
©
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2.
3.
4.
INSTRUCTIONS
Answer the following questions on the computer answer
sheet.
Use only side 2 of the answer sheet.
Only one answer per question is allowed.
No marks are considered for unclear answers. It is your
responsibility to ensure that the answer sheet is readable
by the optical mark reader. All instructions are provided
on the answer sheet.
The allocation of marks per question may vary, but is
indicated at each question.
Answers are not marked negatively.
rs
1.
[38]
5.
6.
Question 1
Mark option J of Question 1 on your computer answer sheet.
This is for control purposes only and ensures that you use
1.
2.
3.
4.
5.
6.
AFDELING B
INSTRUKSIES
Beantwoord die volgende vrae op die rekenaarantwoordblad.
Gebruik slegs kant 2 van die antwoordblad.
Slegs een antwoord per vraag is toelaatbaar.
Geen punte word oorweeg vir onduidelike antwoorde nie.
Dit is u verantwoordelikheid op te sorg dat die
antwoordblad leesbaar is vir die optiese merkleser. Alle
instruksies is op die antwoordblad aangebring.
Die puntetoekenning per vraag mag varieer en word by
elke vraag aangedui.
Antwoorde word nie negatief nagesien nie.
Vraag 1
Merk opsie J van Vraag 1 op u rekenaarantwoordblad.
Dit is slegs vir kontroledoeleindes en verseker dat u kant 2
van die antwoordblad gebruik.
side 2 of the answer sheet.
Question 2
Mark option H of Question 2 if you are certain that you have
written your student number in the horizontal row of boxes at
the top of the answer sheet and that you have correctly coded
each digit vertically.
Vraag 2
Merk opsie H van Vraag 2 indien u seker is dat u u
studentenommer op die rekenaarantwoordblad in die boonste
horisontale ry blokkies geskryf het, en dat u elke syfer daarvan
vertikaal korrek gekodeer het.
The last column here must be blank.
Die laaste kolom hier moet oop wees.
CHM 172
Finale Eksamen
9/15
© Universiteit van Pretoria
2014
STUDENT NUMBER
STUDENTENOMMER
Vraag 3
[2]
3+
Die nukliede simbool vir ‘n element is 51
23𝑋 .
Waaruit bestaan hierdie katioon?
A 28 protone, 23 neutrone, 31 elektrone
B 23 protone, 28 neutrone, 25 elektrone
C 51 protone, 28 neutrone, 48 elektrone
D 28 protone, 23 neutrone, 25 elektrone
E 23 protone, 28 neutrone, 20 elektrone
F
51 protone, 23 neutrone, 48 elektrone
Question 4
The formula for barium bromite is
Vraag 4
[2]
Die formule vir bariumbromiet is
D
E
F
or
BaBr2
Ba2Br
Ba(BrO2)2
Ba(BrO3)2
BaBrO2
BaBrO3
Pr
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A
B
C
ia
Question 3
3+
The nuclide symbol for an element is 51
23𝑋 .
What does this cation consist of?
A 28 protons, 23 neutrons, 31 electrons
B 23 protons, 28 neutrons, 25 electrons
C 51 protons, 28 neutrons, 48 electrons
D 28 protons, 23 neutrons, 25 electrons
E 23 protons, 28 neutrons, 20 electrons
F
51 protons, 23 neutrons, 48 electrons
B
C
E
n = 4,  = 2, m = 1, ms = +½
n = 4,  = 2, m = 2, ms = –½
F
n = 5,  = 1, m = 0, ms = +½
n = 5,  = 1, m = 1, ms = +½
n = 5,  = 2, m = 0, ms = +½
n = 5,  = 2, m = 1, ms = +½
None of the above / Geen van bogenoemde nie
G
n = 4,  = 2, m =– 1, ms = +½
n = 4,  = 2, m =– 2, ms = +½
n = 5,  = 0, m = 0, ms = +½
n = 5,  = 0, m = 0, ms = –½
rs
D
n = 4,  = 1, m = 1, ms = +½
n = 4,  = 2, m = 2, ms = +½
ity
A
Vraag 5
[2]
Watter van die volgende stelle kwantumgetalle vir enige twee
ongepaarde elektrone in ‘n neutrale rodium (Rh) atoom in die
grondtoestand is geldig?:
of
Question 5
Which of the following sets of quantum numbers for any two
of the unpaired electrons in a neutral rhodium (Rh) atom in
the ground state is valid?
©
U
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Question 6
Which of the following atoms, designated by their electron
configurations, has the highest first ionisation energy?
A
B
C
D
[Kr] 4d105s25p3
[Xe] 4f145d106s26p3
[Ne] 3s2 3p2
[Ne] 3s2 3p3
E
F
G
Vraag 6
[3]
Watter van die volgende atome, aangewys deur hulle elektronkonfigurasies, het die hoogste eerste ionisie-energie?
[Ar] 3d104s24p3
[Xe] 6s1
[Ne] 3s1
Question 7
Vraag 7
[3]
The air in a room is found to have toxic levels of carbon
Daar word bevind dat die lug in ‘n kamer toksiese vlakke
koolstofdioksied bevat. Die digtheid van die lug is
dioxide. The density of the air is 1.425 𝑚𝑔/𝑐𝑚3 in a room of
1.425 𝑚𝑔/𝑐𝑚3 in ‘n kamer van 5.27 𝑚 × 7.92 𝑚 × 2.85 𝑚.
5.27 𝑚 × 7.92 𝑚 × 2.85 𝑚. Analysis shows that there is
Analise wys dat daar 12.76 𝑘𝑔 CO2 in die kamer is. Wat is die
12.76 𝑘𝑔 CO2 in the room. What is the mass percentage of
massapersentasie van CO2 in die lug?
CO2 in the air?
A 0.654%
E 65.4%
B 7.53%
F 75.3%
C 6.54%
G None of the above /
D 0.753%
Geen van bogenoemde nie
CHM 172
Final Examination
10/15
© University of Pretoria
2014
STUDENT NUMBER
STUDENTENOMMER
Questions 9 to 11 relate to the following redox reaction
which takes place in aqueous basic medium. Balance the
reaction on the back of the previous page and then answer
the questions.
A
B
C
D
E
F
G
H
I
J
ia
lead nitrate
lead nitrate and ammonium perchlorate
ammonium perchlorate
lead bromide
lead bromide and ammonium bromide
sodium perchlorate
sodium perchlorate and lead bromide
sodium perchlorate and lead nitrate
sodium perchlorate, lead bromide and ammonium nitrate
None of the above.
loodnitraat
loodnitraat en ammoniumperchloraat
ammoniumperchloraat
loodbromied
loodbromied en ammoniumbromied
natriumperchloraat
natriumperchloraat en loodbromied
natriumperchloraat en loodnitraat
natriumperchloraat, loodbromied en ammoniumnitraat
Nie een van bogenoemde nie.
or
A
B
C
D
E
F
G
H
I
J
Vraag 8
[2]
Die volgende drie oplossings word gemeng:
• lood(II)perchloraat
• ammoniumbromied
• natriumnitraat
Watter een van die volgende neerslae sal gevorm word?
Pr
et
Question 8
The following three solutions are mixed:
• lead(II) perchlorate
• ammonium bromide
• sodium nitrate
Which one of the following precipitates will be formed?
Vrae 9 tot 11 handel oor die volgende redoksreaksie wat in
waterige basiese medium plaasvind. Balanseer die reaksie
op die agterkant van die vorige bladsy en antwoord dan die
vrae.
from −1 to −2 and it is reduced
from −1 to −2 and it is oxidised
from +1 to 0 and it is reduced
from +1 to 0 and it is oxidised
from −1 to 0 and it is reduced
from −1 to 0 and it is oxidised
rs
A
B
C
D
E
F
Vraag 9
[2]
Die verandering in die oksidasietoestand van chloor is
ity
Question 9
The change in the oxidation state of chlorine is
of
Cr(OH)3(𝒔) + CℓO- (𝒂𝒒) → CrO42- (𝒂𝒒) + Cℓ2 (𝒈)
van −1 tot −2 en dit word gereduseer
van −1 tot −2 en dit word geoksideer
van +1 tot 0 en dit word gereduseer
van +1 tot 0 en dit word geoksideer
van −1 tot 0 en dit word gereduseer
van −1 tot 0 en dit word geoksideer
Vraag 10
[2]
Die aantal mol elektrone wat in die finale gebalanseerde
reaksie oorgedra word is
©
U
ni
ve
Question 10
The number of moles of electrons transferred in the final
balanced reaction equation is
A 1
B 2
C 3
D 6
E 10
F
None of the above /
Geen van bogenoemde nie
A
B
C
D
E
F
CHM 172
Finale Eksamen
11/15
© Universiteit van Pretoria
2014
STUDENT NUMBER
STUDENTENOMMER
Question 11
The amount of water in the final balanced equation is
ia
1 mol H2O on the reactant side
1 mol H2O on the product side
2 mol H2O on the reactant side
2 mol H2O on the product side
6 mol H2O on the reactant side
6 mol H2O on the product side
None of the above
or
A
B
C
D
E
F
G
Vraag 11
[2]
Die hoeveelheid water in die finale gebalanseerde vergelyking
is
A 1 mol H2O aan die reagens kant
B 1 mol H2O aan die produk kant
C 2 mol H2O aan die reagens kant
D 2 mol H2O aan die produk kant
E 6 mol H2O aan die reagens kant
F
6 mol H2O aan die produk kant
G Geen van bogenoemde nie
of
−35.2 kJ per mol H2O2
−19.4 kJ per mol H2O2
−1.20 kJ per mol H2O2
−28.2 kJ per mol H2O2
−14.1 kJ per mol H2O2
None of the above /
Geen van bogenoemde nie
2H2O2 (𝒂𝒒) → O2 (𝒈) + 2H2O (𝓵)
ity
A
B
C
D
E
F
Pr
et
Question 12
Vraag 12
[3]
30.0 𝑚𝐿 of a 0.882 𝑀 hydrogen peroxide solution is placed in
30.0 𝑚𝐿 van ‘n 0.882 𝑀 waterstofperoksied oplossing word in
a coffee-cup calorimeter. 7.5 𝑚𝐿 of a dilute iron(III) nitrate
‘n koffiekoppie-kaloriemeter geplaas. 7.5 𝑚𝐿 van ‘n verdunde
solution, which only acts as a catalyst for the decomposition of yster(III) nitraatoplossing, wat slegs as ‘n katalisator dien vir
the hydrogen peroxide, is added. The density and specific heat
die ontbinding van die waterstofperoksied, word bygevoeg.
capacity of the total solution can be regarded as being the same Die digtheid en spesifieke hittekapasiteit van die totale
as that of water.
oplossing kan beskou word as dieselfde as die van water.
How much heat is generated by the reaction if the temperature
Hoeveel hitte word gegenereer deur die reaksie indien die
of the mixture increased from 18.24℃ to 24.18℃?
temperatuure van 18.24℃ tot 24.18℃ styg?
rs
Questions 13 and 14 refer to the following process:
One mole of a compound decomposes at 25℃. This results
in a change in reaction enthalpy of +3.5 𝒌𝑱 and a change in
reaction entropy of +258 𝑱 ∙ 𝑲−𝟏 .
©
U
ni
ve
Question 13
The energy available to perform useful work is
A
B
C
D
E
F
Vraag 13
[2]
Die energie wat beskikbaar is om bruikbare werk te verrig is
7.69 × 104 kJ
3.24 kJ
255 kJ
1.04×103 kJ
73.4 kJ
None / Niks
Question 14
The total change in the entropy of the universe at 25℃ is
A
B
C
D
E
F
Vrae 13 en 14 verwys na die volgende proses:
Een mol van ‘n verbinding ontbind by 25℃. Dit het ‘n
verandering in reaksie-entalpie van +3.5 𝒌𝑱 en ‘n
verandering in reaksie-entropie van +258 𝑱 ∙ 𝑲−𝟏 tot gevolg.
Vraag 14
[2]
Die totale verandering in entropie van die heelal by 25℃ is
117 J/K
255 J/K
−0.865 J/K
246 J/K
118 J/K
270 J/K
CHM 172
Final Examination
12/15
© University of Pretoria
2014
STUDENT NUMBER
STUDENTENOMMER
Vrae 15 tot 17 verwys na die volgende:
‘n Voltaïese sel bestaan uit ‘n kadmiumelektrode in ‘n
1.0 𝑴 CdSO4(aq) oplossing, en ‘n nikkelelektrode in ‘n
1.0 𝑴 NiSO4(aq) oplossing by 25°C. Die soutbrug bevat ‘n
oplossing van NaNO3(aq).
Question 15
The anode and positive electrode of this cell are respectively:
E
F
G
Cd(s), Cd2+(aq)
Cd(s), Ni2+(aq)
None of the above /
Geen van bogenoemde nie
Question 16
The direction of flow of electrons, and the direction of
movement of nitrate anions in this cell are respectively:
From Cd(s) to NaNO3(aq);
From Ni(s) to Cd(s);
From Ni(s) to Cd(s);
From Ni(s) to NaNO3(aq);
From Cd(s) to Ni(s);
From Cd(aq) to Ni(aq);
None of the above.
Vraag 16
[2]
Die rigting van elektronvloei, en die rigting van beweging van
nitraatanione in hierdie sel is respektiewelik:
To the CdSO4(aq) half-cell.
To the CdSO4(aq) half-cell.
To the NiSO4(aq) half-cell.
To the NiSO4(aq) half-cell.
To the CdSO4(aq) half-cell.
To the NiSO4(aq) half-cell.
A
B
C
D
E
F
G
Van Cd(s) na NaNO3(aq); Na die CdSO4(aq) halfsel.
Van Ni(s) na Cd(s);
Na die CdSO4(aq) halfsel.
Van Ni(s) na Cd(s);
Na die NiSO4(aq) halfsel.
Van Ni(s) na NaNO3(aq); Na die NiSO4(aq) halfsel.
Van Cd(s) na Ni(s);
Na die CdSO4(aq) halfsel.
Van Cd(aq) na Ni(aq);
Na die NiSO4(aq) halfsel.
Nie een van bogenoemde nie.
of
A
B
C
D
E
F
G
or
Cd(s), Ni(s)
Ni(s), Cd(s)
Ni(s), Ni(s)
Cd(s), Cd(s)
Pr
et
A
B
C
D
Vraag 15
[2]
Die anode en positiewe electrode van hierdie sel is
respektiewelik:
ia
Questions 15 to 17 refer to the following:
A Voltaic electrochemical cell consists of a cadmium
electrode in a 1.0 𝑴 CdSO4(aq) solution, and a nickel
electrode in a 1.0 𝑴 NiSO4(aq) solution at 25°C. The salt
bridge consists of a solution of NaNO3(aq).
Vraag 17
[2]
Die spesie wat gereduseer word, en die halfsel-elektroliet
waarvan die konsentrasie van sy eie metaalkatioon toeneem,
is respektiewelik:
©
U
ni
ve
rs
ity
Question 17
The species that is reduced, and the half-cell electrolyte that
increases in concentration of its own metal cation, are
respectively
A Ni2+(aq);
NiSO4(aq)
B Ni(s);
Na2SO4(aq)
C Na(s);
Na2SO4(aq)
D Cd2+(aq);
CdSO4(aq)
E Cd(s);
NiSO4(aq)
2+
F Ni (aq);
CdSO4(aq)
G None of the above / Geen van bogenoemde
Question 18
Vraag 18
[3]
Silver tableware is tarnished by foods such as eggs that contain Silwer tafelware word aangeslaan deur kosse soos eiers wat
certain sulphur compounds. A proposed reaction is:
sekere swaelverbindings bevat. ‘n Voorgestelde reaksie is:
[Hint: Estimate the sign of ∆𝑆 by inspection]
[Wenk: Bepaal die teken vir ∆𝑆 deur inspeksie]
𝟒𝐀𝐠 (𝐬) + 𝟐𝐇𝟐 𝐒 (𝐠) + 𝐎𝟐 (𝐠) → 𝟐𝐀𝐠 𝟐 𝐒 (𝐬) + 𝟐𝐇𝟐 𝐎 (𝓵)
Which statement is true for this reaction?
A Spontaneous at all temperatures.
B Spontaneous at high temperatures only.
C Non-spontaneous at all temperatures.
D Non-spontaneous at low temperatures only.
E Non-spontaneous at high temperatures only.
F
None of the above.
∆𝐇 < 𝟎
Watter stelling is waar vir hierdie reaksie?
A Spontaan by alle temperature.
B Spontaan by hoë temperature alleenlik.
C Nie-spontaan by alle temperature.
D Nie-spontaan by lae temperature alleenlik.
E Nie-spontaan by hoë temperature alleenlik.
F
Geeneen van bogenoemde nie.
Check that you have coded answers 1 to 18 on Side 2 of the answer sheet
Maak seker dat jy antwoorde 1 tot 18 op Kant 2 van die antwoordblad gekodeer het
CHM 172
Finale Eksamen
13/15
© Universiteit van Pretoria
2014
THE PERIODIC TABLE OF THE ELEMENTS
Key / Sleutel
1
H
1.01
2. 1
3
4
Li
Be
6.94
1. 0
DIE PERIODIEKE TABEL VAN ELEMENTE
Atomic number / Atoomgetal
W Element symbol / Element simbool
183.84 Relative Atomic mass / Relatiewe atoommassa (u)
1 . 7 Electronegativity / Elektronegatiwiteit
74
2
He
4.00
5
6
7
8
9
10
B
C
N
O
F
Ne
9.01
10.81
12.01
14.01
16.00
19.00
20.18
1. 5
2. 0
2. 5
3. 0
3. 5
4. 0
12
13
14
15
16
17
18
Mg
Al
Si
P
S
Cl
Ar
22.99
24.31
26.98
28.09
30.97
32.07
35.45
39.95
0. 9
1. 2
1. 5
1. 8
2. 1
2. 5
3. 0
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
39.10
40.01
44.96
47.87
50.95
52.00
54.94
55.85
58.93
58.69
63.55
65.39
69.72
72.61
74.92
78.96
79.90
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
85.47
87.62
88.91
91.22
92.91
95.94
0. 8
1. 0
1. 2
1. 4
1. 6
1. 8
1. 9
2. 2
2. 2
2. 2
1. 9
1. 7
1. 7
1. 8
1. 9
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
1. 3
132.91 137.33 138.91
0. 7
0. 9
87
88
89
Fr
Ra
Ac
223.02 226.03 227.03
0. 7
1. 5
1. 6
1. 6
1. 5
1. 8
1. 9
1. 8
1. 9
1. 6
1. 6
1. 8
2. 0
2. 4
2. 8
83.80
3. 0
or
1. 0
36
Kr
52
53
54
Te
I
Xe
98.91 101.07 102.91 106.42 107.87 112.41 114.82 118.71 121.76 127.60 126.90 131.29
2. 1
2. 5
84
85
2. 6
86
Pr
et
0. 8
ia
11
Na
Hg
Tl
Pb
Bi
Po
At
Rn
178.49 180.95 183.84 186.21 190.23 192.22 195.08 196.97 200.59 204.38 207.20 208.98 208.98 209.99 222.01
1. 3
1. 5
1. 7
1. 9
2. 2
2. 2
2. 2
2. 4
104
105
106
107
108
109
110
111
Rf
Db
Sg
Bh
261.11 262.11 263.12 262.12
Hs
Mt
Ds
Rg
265
266
271
272
0. 9
58
59
60
61
62
63
64
65
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
1. 9
1. 8
1. 9
1. 9
2. 0
112
114
116
Cn
Fl
Lv
2. 2
285
66
67
68
69
70
71
Dy
Ho
Er
Tm
Yb
Lu
140.12 140.91 144.24 144.91 150.36 151.97 157.25 158.93 162.50 164.93 167.26 168.93 173.94 174.97
90
91
92
93
94
95
96
97
98
99
100
101
102
103
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
of
Th
Fm
Md
No
Lr
232.04 231.04 238.03 237.05 244.06 243.06 247.07 247.07 251.08 252.08 257.10 258.10 259.10 262.11
Vergelykings / Equations
𝑞 = 𝑚𝐶Δ𝑇
ity
∆𝑟 𝐻 𝑜 = ∑𝑛Δ𝑓 𝐻 𝑜 (𝑝𝑟𝑜𝑑) − ∑𝑚Δ𝑓 𝐻 𝑜 (𝑟𝑒𝑎𝑐𝑡)
∆𝑟 𝑆 𝑜 = ∑𝑛𝑆 𝑜 (𝑝𝑟𝑜𝑑) − ∑𝑚𝑆 𝑜 (𝑟𝑒𝑎𝑐𝑡)
∆𝑟 𝐺 𝑜 = ∑nΔ𝑓 𝐺 𝑜 (𝑝𝑟𝑜𝑑) − ∑𝑚Δ𝑓 𝐺 𝑜 (𝑟𝑒𝑎𝑐𝑡)
𝑜
𝑜
𝑜
= Δ𝑆𝑠𝑦𝑠
+ Δ𝑆𝑠𝑢𝑟𝑟
Δ𝑆𝑢𝑛𝑖𝑣
Konstante / Constants
ℎ = 6.626 × 10-34 J⋅s
𝑅 = 1.097 × 107 m-1
c = 2.998 × 108 m⋅s-1
NA = 6.022 × 1023
ℱ = 9.6485 x 104 C⋅mol−1 of/or J⋅V−1⋅mol−1
𝑒 = 1.602 x 10−19 C
Omsettingsfaktore / Conversion Factors
1 u = 1.661 × 10−24 g
1 L = 10−3 m3 = 1 dm3
1 bar = 1.000 x 105 Pa
1 C = 1 Amp⋅1 sec
©
𝑜
∆𝐻𝑠𝑦𝑠
𝑞𝑠𝑢𝑟𝑟
=−
𝑇
𝑇
rs
𝑜
∆𝑆𝑠𝑢𝑟𝑟
=
Δ𝑟 𝐺 𝑜 = Δ𝑟 𝐻 𝑜 − 𝑇Δ𝑟 𝑆 𝑜
Δ𝑟 𝐺 = Δ𝑟 𝐺 𝑜 + 𝑅𝑇𝑙𝑛𝑄
U
ni
ve
𝑉 = 𝜋𝑟 2 ℎ (cylinder / silinder)
𝐸 = ℎ𝑣
𝑍2
𝐸𝑛 = −𝑅ℎ𝑐 2
𝑛
1
1
∆𝐸 = −𝑅ℎ𝑐𝑍 2 � 2 − 2 �
𝑛𝑓𝑖𝑛 𝑛𝑖𝑛𝑖𝑡
ℎ
𝜆=
𝑚𝑣
𝑝𝐻 = −log[𝐻3 𝑂+ ]
𝑝𝐻 + 𝑝𝑂𝐻 = 14
Δ𝑈 = 𝑞 + 𝑤
0°C = 273.15 K
C (H2O(ℓ)) = 4.184 J⋅g−1⋅K−1
C (H2O(s)) = 2.06 J⋅g−1⋅K−1
C (H2O(g)) = 1.92 J⋅g−1⋅K−1
∆𝑓𝑢𝑠 𝐻 𝑜 (𝑤𝑎𝑡𝑒𝑟) = 6.01 𝑘𝐽 ∙ 𝑚𝑜𝑙−1
∆𝑣𝑎𝑝 𝐻 𝑜 (𝑤𝑎𝑡𝑒𝑟) = 40.65 𝑘𝐽 ∙ 𝑚𝑜𝑙−1
𝑤𝑚𝑎𝑥 = 𝑛ℱ𝐸
Δ𝑟 𝐺 𝑜 = −𝑛ℱ𝐸 𝑜
Δ𝑟 𝐺 𝑜 = −RT𝑙𝑛K
𝑅𝑇
𝑙𝑛Q
𝐸 = 𝐸𝑜 −
𝑛ℱ
0.0592
𝑙𝑜𝑔Q (25°C)
𝐸 = 𝐸𝑜 −
𝑛
𝑜
𝑛𝐸
𝑙𝑜𝑔𝐾𝑒𝑞 =
(25°𝐶)
0.0592
𝑙𝑛K = 2.303𝑙𝑜𝑔K
𝑙𝑛Q = 2.303𝑙𝑜𝑔Q
Gas constant / gaskonstante, R:
= 0.08206 L⋅atm⋅mol−1⋅K−1
= 8.3145 J⋅mol−1⋅K−1
= 8.3145 kPa⋅L⋅mol-1⋅K-1
= 62.36 mm Hg⋅L⋅mol-1⋅K-1
1 inch = 2.54 cm
1 ft = 12 inches
1 yard = 3 feet
1 duim = 2.54 cm
1 voet = 12 duim
1 tree = 3 voet
1 cal = 4.184 J
1 J = 0.2390 cal = 1 Pa⋅m3 = 1 m2⋅kg⋅s−2 = 1 V⋅C = 1 Watt⋅1 sec = 9.9 x 10-3 L.atm
1 atm = 1.013 x 105 N⋅m−2 = 1.013 x 105 Pa = 760 mm Hg = 760 torr
Ewewig / Equilibrium
By ewewig:
𝑎𝐴(𝑎𝑞) + 𝑏𝐵(𝑎𝑞) ⇌ 𝑐𝐶(𝑎𝑞) + 𝑑𝐷(𝑎𝑞) 𝑎𝐴(𝑎𝑞) + 𝑏𝐵(𝑎𝑞) ⇌ 𝑐𝐶(𝑎𝑞) + 𝑑𝐷(𝑎𝑞)
At equilibrium:
𝑃𝐶𝑐 𝑃𝐷𝑑
[𝐶]𝑐 [𝐷]𝑑
E = 0 and
𝐾𝑝 of/or 𝑄 = 𝑎 𝑏
𝐾𝑐 of/or 𝑄 =
𝑎
𝑏
[𝐴] [𝐵]
𝑃𝐴 𝑃𝐵
Q=K
Liquids and solids are given a value of 1 unit in the Q and K expressions
CHM 172
Final Examination
14/15
Information page / Inligtingsblad
For insoluble salts / 𝑉𝑖𝑟 𝑜𝑛𝑜𝑝𝑙𝑜𝑠𝑏𝑎𝑟𝑒 𝑠𝑜𝑢𝑡𝑒
𝑀𝑚 𝑋𝑥 (𝑠) ⇌ 𝑚𝑀 𝑥+ (𝑎𝑞) + 𝑥𝑋 𝑚− (𝑎𝑞)
𝐾𝑠𝑝 = [𝑀 𝑥+ ]𝑚 [𝑋 𝑚− ]𝑥
© University of Pretoria
2014
Standard Reduction Potentials in Aqueous, Acidic or Basic Solution at 25°C
Standaard Reduksie Potensiale in Waterige, Suur- of Basiese Oplossing by 25°C
..........................................................................................................................................
Reduction Half-Reaction / Reduksie Half-Reaksie
H2O2 (aq) + 2 H+ (aq) + 2 e
−
4
H+
MnO (aq) + 8
−
(aq) + 5 e
Au3+ (aq) + 3 e−
−
Cℓ2 (g) + 2 e
−
Cr2O7 2-(aq) + 14 H+ (aq) + 6 e
O2 (g) + 4
H+ (aq)
Br2 (ℓ) + 2 e
−
+4e
−
−
−
3
NO (aq) + 4 H+ (aq) + 3 e
−
Hg2+ (aq) + 2 e
Ag+
−
(aq) + e
−
Hg+ (aq) + 2 e
−
−
I2 (s) + 2 e
O2 (g) + 2 H2O (ℓ) + 4 e
−
−
2 H+ (aq) + 2 e
−
Sn2+ (aq) + 2 e
−
V3+ (aq) + e
−
Cd2+ (aq) + 2 e
©
U
ni
ve
Fe2+ (aq) + 2 e−
rs
Ni2+ (aq) + 2 e−
−
Cr3+(aq) + e
−
Zn2+ (aq) + 2 e
2 H2O (ℓ) + 2 e−
[Zn(CN)4]2- (aq) + 2 e
−
Aℓ3+ (aq) + 3 e
Mg2+ (aq) + 2 e−
−
K+ (aq) + e
⇌
2 H2O (ℓ)
+1.77
⇌
Mn2+ (aq) + 4 H2O (ℓ)
+1.52
⇌
Au (s)
+1.50
⇌
−
⇌
2 Cr3+ (aq) + 7H2O (ℓ)
⇌
2 H2O (ℓ)
⇌
+1.360
−
+1.33
+1.229
−
⇌
2 Br (aq)
+1.08
⇌
NO (g) + 2 H2O (ℓ)
+0.96
Hg (ℓ)
+0.855
⇌
Ag (s)
+0.80
⇌
2 Hg (ℓ)
+0.789
⇌
Fe2+ (aq)
+0.771
−
−
2 I (aq)
+0.535
⇌
4 OH (aq)
+0.40
⇌
Cu (s)
+0.34
⇌
H2 (g)
0.00
⇌
Sn (s)
−0.14
⇌
Ni (s)
−0.25
⇌
V2+ (aq)
−0.255
⇌
Cd (s)
−0.403
⇌
Fe (s)
−0.44
⇌
Cr2+(aq)
-0.50
⇌
Zn (s)
⇌
H2 (g) + 2 OH− aq)
⇌
−0.763
−
−0.8277
⇌
Zn (s) + 4 CN (aq)
−1.26
⇌
Aℓ (s)
−1.66
Mg (s)
−2.37
⇌
Na (s)
−2.714
⇌
K (s)
−2.925
Li (s)
−3.045
⇌
Li+ (aq) + e−
*Versus the standard hydrogen electrode /
Teen die standaard waterstofelektrode
CHM 172
Final Examination
2 Cℓ (aq)
ity
Cu2+ (aq) + 2 e
Na+ (aq) + e
+2.87
⇌
−
−
2F (aq)
of
Fe3+ (aq) + e
⇌
ia
−
or
F2 (g) + 2 e
E° (V)*
−
Pr
et
−
15/15
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2014