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Answer: B
14. “Year after year, men cruising timber or hunting deer in the Blue Mountains of
eastern Oregon had come back with the same story. Near the little hamlet of
Kamela, they had heard a faraway tinkling, a ghostly bell ringing. No one was
able to track down the strange sound. It would fade away in the sighs of the wind
through the big pines. Skeptics accused the men of hearing things. Last week,
while slashing a right -of-way for a po wer line from Bonneville Dam, lumberjacks
brought down a ponderosa pine. Tied by a shriveled leather thong, high in the
treetop, was the answer to the mystery of Kamela; a bronze cattle bell inscribed
with the date 1878. The people of Kamela guessed that a pioneer had tied it to a
sapling that had grown to a towering pine.” ( Time Magazine , 1937) Which of the
following is the best appraisal of the concluding sentence of the above report?
A. Logical because this tree could have attained great height since 1878 .
B. Logical because a tree trunk elongates from the base up.
C. Illogical because no one knows with certainty when the bell was tied to the sapling.
D. Illogical because elongation of a tree trunk occurs from the apical meristem up.
E. There is no basis for appraisin g the concluding sentence of the report.
Answer : D
15. Arrange the following five events in an order that explains the bulk flow of
substances in the phloem.
1. Sugar moves down the stem
2. Leaf cells produce sugar by photosynthesis
3. Sugar is transported from c ell to cell via the apoplast and/or symplast
4. Solutes are actively transported into sieve elements
5. Water diffuses into the sieve tube elements
A. 2,1,4,3,4
B. 1,2,3,4,5
C. 2,4,3,1,5
D. 4,2,1,3,5
E. 2,4,1,3,5
Answer : C
16. A field of a common grain crop, Zea mays (corn, maize), was being cultivated for
genetic plant -studies. Over the years it was noted that rabbits were eating the
fallen kernels (corn seed) that were left in the field. The botanists were quick to
notice that the rabbits did not eat the most starchy port ion of the seed (taking
mostly the protein) when environmental conditions were good. As conditions
became more stressful, the rabbits came back to eat the most starchy portions of
the seeds remaining. These behaviors lead the botanists to an interesting
collaborative study with a team of zoologists and nutritionists.
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7 What part of the corn seed was eaten last by the rabbits under more stressful
conditions?
A. Cotyledon
B. Hypocotyl
C. Scutellum
D. Endosperm
E. Coleoptile
Answer: D
17. Which of the following statement (s) about the regulation of the stomatal opening
is true?
A. The concentration of abscisic acid in the guard cells increases
B. Higher K+ concentrations decrease guard cell´s more negative water potential
C. The Level of carbon dioxide in the spaces inside the lea f increases
D. Lower K+ concentrations give guard cells a more negative water potential
E. Potassium ions diffuse passively out of the guard cells
Answer: B
18. When an oak seedling is one year old, a small marker is inserted into its primary
phloem tissue. Two years later, where would you expect to find this marker?
A. External to the cork cambium
B. Between the secondary phloem and the cork cambium
C. Between the vascular cambium and the secondary phloem
D. Between the vascular cambium and the primary xylem
E. Internal to t he primary xylem
Answer : B
19. Plant hormones react on target tissues to activate a receptor. Thus, for a response
to occur the following five events must take place:
1. the binding of the hormone/receptor should initiate a change in the receptor
(amplifica tion). Calcium is often involved and its interaction is mediated by
the protein calmodulin
2. the target tissue recognizes the hormone (i.e., there must be a receptor to
which the hormone can bind)
3. the target tissue must be sensitive (sensitized) to the horm one
4. the hormone must be present in an appropriate quantity
5. the activated receptor initiates a physiological response
What is the appropriate order of these events?
A. 1, 2, 3, 4, 5
B. 2, 4, 3, 1, 5
C. 4, 3, 2, 1, 5
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8 D. 3, 4, 2,1,5
E. 4, 3,1, 2, 5
Answer : C
20. Which of the following combinations have only primary walls in a mature plant?
A Collenchyma cells Fibers Sieve tube member
B Meristematic cells Tracheary elements Collenchy ma cells
C Sclereids Collenchyma cells Sieve cells
D Sieve elements Meristemati c cells Collenchyma cells
E Vessels members Meristematic cells Parenchyma cells
Answer : D
21. Blue light plays a variety of roles in guard cell action. Which of the following is
NOT a role of blue light in guard cell action?
A. Activates a calcium tr ansporter
B. Activates a membrane bound proton ATPase
C. Stimulates potassium uptake
D. Stimulates malate synthesis
E. Stimulates starch hydrolysis
Answer: E
ANIMAL ANATOMY AND PHYSIOLOGY
Questions 22 and 23 refer to the diagram shown below. The diagram illustrates
feedback loops. Increased or decreased stimulation is indicated by + or —.
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22. Which of the following would lead to a DECREASE in activity of the anterior
pituitary gland?
A. A lack of receptors for cortisol on the hypothalamus
B. A lack of receptors for co rtisol on the anterior pituitary
C. An intravenous injection of a large amount of ACTH
D. A tumor in the hypothalamus causing it to secrete excess corticotropin - releasing
hormone
E. An increased sensitivity of the anterior pituitary to corticotropin -releasing horm one
Answer: C
23. What would happen if the adrenal cortex was artificially stimulated to produce
large amounts of cortisol?
A. Less corticotropin -releasing hormone would be released
B. More ACTH would be released
C. The activity of the hypothalamus would increa se
D. The activity of the anterior pituitary would increase
E. The hypothalamus would become insensitive to cortisol
Answer: A
24. The graph below represents the contraction patterns for three different kinds of
muscles. Find the correct sequence of pattern s for smooth muscle, skeletal
muscle and cardiac muscle.
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A. ABC
B. ACB
C. BAC
D. BCA
E. CBA
Answer B
25. The numbers in the first column correspond to human, elephant, bat, mouse, and
carp. Which number below indicates each organism?
Number Body Temp
(oC) Heart Rate
(beats/min) Maximal speed of
locomotion (m/s)
1 1-30 30-40 1.5
2 38 450-550 3.5
3 31 500-660 14
4 36.2 22-28 11
5 36.6 60-90 10
1 2 3 4 5
A. Human Elephant Bat Mouse Carp
B. Mouse Bat Elephant Human Carp
C. Carp Mouse Bat Elephant Human
D. Carp Mouse Elephant Bat Human
E. Bat5 Mouse Carp Human Elephant
Answer C
26. Why does the stomach use both pepsin and HCl to break down proteins in food?
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11 A. HCl denatures the tertiary and quaternary structure of proteins, while pepsin breaks
the primary structure
B. HCl removes the tertiary structure of proteins, while pepsin breaks the primary and
quaternary structure
C. HCl removes the quaternary structure of proteins, while pepsin breaks the primary
and tertiary structure
D. pepsin removes the tertiary and quaternary str ucture of proteins, while HCl breaks
the primary structure
E. pepsin removes the tertiary structure of proteins, while HCl breaks the primary and
quaternary structure
Answer: A
27. A patient is diagnosed with a disorder in which the pituitary gland overprod uces
anti-diuretic hormone (ADH). Which of the following combinations of symptoms
would you expect to be associated with this disorder?
A. Reduced urine volume and low blood osmolarity
B. Reduced urine volume and high blood osmolarity
C. Increased urine volume and low blood osmolarity
D. Increased urine volume and high blood osmolarity
E. Increased urine volume with no change in blood osmolarity
Answer: A
28. In the diagram below, the cells indicated by “y” are most likely responsible for
which of the following ss?
A. Estrous cycle
B. Growth
C. Glucose metabolism
D. Increasing blood calcium concentration
E. Regulating the rate of oxygen use by cells
Answer: E
29. The pathology associated with Vibrio cholerae is due to cholera toxin (CT) leads
to which of the fol lowing?
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12 1. Active pumping of Ca2+ out of the cell
2. Increased concentration of intracellular cAMP
3. Increased osmolarity of the intestine and diarrhea
4. Ribosylation and activation of Gi protein
5. Ribosylation and activation of Gs protein
A. 1, 2, 4
B. 1, 3, 5
C. 2, 3, 4
D. 2, 3, 5
E. 3, 4, 5
Answer : D
30. Students conducted a renal function experiment. Before the experiment began,
students in one group consumed their usual volume of fluids, while students in
the other group limited their fluid intake. At time t = 0, both group s drank 750
mL of water.
Answer: D
31. What observations would you NOT make when the body responds to a rapid
increase of organic acids?
A. Decreased blood pH
A
B
B
A According to the data in the figures to the left ,
which of the following sta tements is TRUE?
A. Group A stude nts’ blood level of
aldosterone was pro bably highest at t = 60
minutes
B. Group B excreted approximately 140 mL
of urine from t = 20 to 40 minutes
C. Group B probably repre sents the fluid -
deprived groups
D. Group B students’ collectin g ducts were
less permeable to water than those of
Group A students at t = 60 minutes
E. None are correct
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13 B. Decreased respiration rate
C. Increased alveolar ventilation
D. Increased blood pressure
E. Increa sed heart rate
Answer: B
32. A preadolescent boy suffers an injury to the anterior pituitary. Although FSH is
no longer produced, ICSH (LH) levels are normal. At 25 years of age, one would
expect that he would
A. Be impotent
B. Be sterile
C. Have impaired inters titial cell function
D. Not develop secondary sex characteristics
E. Produce large amounts of inhibin
Answer B
33. If adenyl cyclase, the enzyme that converts ATP to cAMP is blocked, the :
A. Activity of progesterone at the endometrium would increase
B. Endometrial cells would not respond to estradiol
C. Corpus luteum would not respond to HCG
D. Level of progesterone produced b y the ovary during the luteal phase would increase
E. Mammary glands would not respond to progesterone
Answer: C
34. All of the following would re sult from severing the sensory fibers of the vagus
nerve from the lung, EXCEPT
A. Atelectasis
B. Disappearance of the Hering -Breuer reflexes
C. Less inhibition of the inspiratory center during forced breathing
D. Less stimulation of the expiratory center during forced breathing
E. Potential damage o the lungs due to overinflation
Answer: A
35. In the graph below, what factors would account most directly for the shape of the
plot?
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A. Number of amoebcytes per sponge
B. Number of choanocyte s per sponge
C. Number of spicules per sponge
D. Rate of cribrostatin synthesis (molecules/time unit)
E. Spongin concentration (g/volume unit)
Answer : B
36. ____________ are to termites as _______________ are to plants.
A. Cellulose molecules, water molecules
B. Gut mi crobes, mycorrhizal fungi
C. Gut microbes, water molecules
D. Hours of darkness, hours of light
E. Predators, prey
Answer: B
Ethology
37. Which one of the following statements is NOT a general precondition necessary
for the evolution of reciprocal altruism ?
A. Benefits of receiving aid must exceed the cost of donating
B. Donors must be able to recognize partners, remember their previous actions and
refuse to cooperate with individuals that did not reciprocate
C. Interacting organisms must have a high coefficient of rela tedness
D. Repeated pair -wise interactions are needed to permit role exchanges between donor
and beneficiary
E. All of the above are necessary pre -conditions
Answer C
38. What is the proximate cause of the cuteness response?
A. Individuals who didn’t find babie s cute were more likely to destroy their own genes
B. Infants who did not look cute were more likely to be killed
C. Infants have relatively large eyes located in the middle of the face H2O purity
exiting the
osculum
?
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15 D. A and B are both proximate causes
E. B and C are both proximate causes
Answer : C
39. Which of the following is a model of parental care paired with its correct
description?
A. Association model – strength of the pair -bond between parents is directly associated
with the amount of parental care
B. Conflict model – females who compete aga inst other females for resources provide
less care
C. Parental provision model – males are most likely to provide for offspring in species
with internal fertilization
D. Symbiosis model – Bidirectional exchange of resources between parent(s) and
offspring
E. Paren tal provision model – biological parents recruit alloparents to care for offspring
Answer : D
Genetics and Evolution
40. Genes A and B are linked 12 map units apart. A heterozygous individual whose
parents were Aabb and aaBB would be expected to produce gametes in the
following frequencies:
A. 44% AB 6% Ab 6% aB 44% ab
B. 6% AB 44% Ab 44% aB 6% ab
C. 12% AB 38% Ab 38% aB 12% ab
D. 6% AB 6% Ab 44% aB 44% ab
E. 38% AB 12% Ab 12% aB 38% ab
Answer: B
41. A particular plant is variegated, that is all parts of some branches are green and
other branches are white. An investigator wishes to find how this character is
inherited. She removes mature anthers from a white flower and pollinates a
pigmented flower. All the offspring produce pigmented flowers. However, b efore
jumping to conclusions about the pattern of inheritance she decides she needs to
make a reciprocal cross. Therefore she removes mature anthers from a pigmented
flower and pollinates a white flower. All the offspring produce white flowers.
What was he r conclusion about the pattern of inheritance?
A. a Mendelian trait with white being recessive to colored.
B. a Mendelian trait with colored being recessive to white
C. a sex -linked (X -linked) recessive gene was involved
D. cytoplasmic inheritance was involved
E. incompl ete dominance was involved
Answer: D
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16 42. For a dominant trait that is FAVORED by natural selection, the frequency of the
relevant allele in a population will change fastest when the allele
A. Frequency is initially low
B. Frequency is initially intermediate
C. Frequency is initially high
D. Frequency is fixed at zero or one
E. Frequency will not change; it will remain constant
Answer : A
43. Which of the following is true?
A. Gametes have a homologous pair of each chromosome
B. Gametes have two copies of each autosome
C. After anaphase II of meiosis each chromosome consists of two sister chromatids
D. After anaphase I of meiosis a chromosome consists of one sister chromatid
E. Immediately after S phase a chromosome consists of two sister chromatids
Answer: E
44. An individual has t he following genotype 49XXXXY.Which description below
best describes this individual?
A. Genetic female with two Barr bodies
B. Genetic male with two Barr bodies
C. Genetic female with three Barr bodies
D. Genetic male with three Barr bodies
E. More than one answer abov e is possible description of this individual
Answer : D
For Questions 45 and 46, use the pedigree below that illustrates a relatively rare trait
that is non -lethal.
45. Which of the following best describes the inheritance pattern ?
A. Autosomal dominant
B. Autosomal recessive
C. X-linked, dominant
D. X-linked, recessive
E. Mitochondrial I
II
III
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Answer: D
46. Using the sequenc e of I -4, II-2, II-4, and III -6 for the individuals in the diagram
above, which of the following best describes their genotypes ?
A. XXa, XaY, XXa, XY
B. XXa, XY, XXa, XaY
C. XY, XXa, XaY, XXa
D. XaY, XXa,XY, XXa
Answer: C
47. Which of the following statements is FALSE concerning Fox’s proteinoid
microspheres?
A. Movement within the microspheres resembles cytoplasmic streaming
B. They swell in hypertonic medium and shrink in hypotonic medium
C. Chemical reactions can occur within the microspheres
D. They possess a defined int ernal structure
E. They incorporate and adsorb various materials from the medium
Answer : B
48. Suppose that the traits below occurred just once in evolution. What traits were
most likely possessed by the most recent ancestor of Deuterostomia, Ecdysozoa,
and Lophotrochozoa?
1. Cephalization
2. Exoskeleton
3. Indeterminate development
4. Segmentation
5. Triploblastic
A. 1, 2, 5
B. 1, 4, 5
C. 2, 3
D. 2, 3, 4
E. 3, 4
Answer: B
49. The presence of a vertebral column is with respect to all vertebrates [first term],
but with respect to all chordates [second term].
A. homologous; analogous
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18 B. primitive; derived
C. adaptive; maladaptive
D. analogous; homologous
E. analogous; derived
Answer: B
50. Population genetics shows us that certain traits of a species will become more
abundant if they benefit the species. The diagram below illustrates the change
that occurred in the frequency of phenotypes in an insect population over 10
generations. A probable explanation for this change would be that over time there
was:
A. A decrease in the population of this insect
B. An increase in the population of this insect
C. A decrease in the adaptive value of gene a
D. An increase in the adaptive value of gene a
E. A decrease in the mutation rate of gene A
Answer : D
51. In foxes, there are 9 coat colors: red, standard silver, Alaskan silver, double -black,
smoky red, cross -red, blended -cross, substandard silver, and sub -Alaskan silver.
A red fox was crossed with a double -black fox and their offspring were then
crossed with each other. The F2 phenotypes were; 10 red : 18 smoky r ed : 20 cross -
red : 39 blended -cross : 9 standard silver : 19 substandard silver : 12 Alaskan silver
: 22 sub -Alaskan silver : 8 double -black. How many genes are involved in this
cross?
A. 18 genes
B. 4 pairs of genes
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19 C. 9 genes
D. 9 pairs of genes
E. 4 genes
Answer : B
Ecology Questions 52 to 57
52. For populations that exhibit __________ population growth, maximum
harvesting yield is achieved when _________________.
A. exponential, population growth rate is highest
B. logistic, the population reaches its carrying c apacity
C. logistic, population growth rate is highest
D. logistic, population growth rate is not changing with population size
E. exponential, the population reaches its carrying capacity
Answer : C
53. The best description of the relationships between f undamental niches (FN) and
realized niches (RN) of two competing species that coexist is:
A. FNA = RNA; FNB = RNB
B. FNA > RNA; FNB = RNB
C. FNA < RNA; FNB < RNB
D. FNA > RNA; FNB > RNB
E. FNA = RNA; FNS > RNS
Answer : D
54. Which of the following is a description of protostomes?
A. Radial and determinate cleavage, blastopore becomes mouth, schizocoelous
development
B. Radial and determinate cleavage, enterocoelous development, blastopore becomes
anus
C. Spiral and determinate cleavage, blastopore becomes mouth, schizocoelous
development
D. Spiral and determinate cleavage, enterocoelous development
E. Spiral and indeterminate cleavage, coelom forms as split in solid mass of mesoderm
Answer : C
55. All of the major body plans seen today appeared in the fossil record over 500
millio n years ago at the beginning of the:
A. Burgess period
B. Cambrian period
C. Carboniferous period
D. Cretaceous period
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20 E. Ediacaran period
Answer : B
56. You have been asked to begin to develop some genetic testing molecular markers
to determine diversity of two equally sized groups of humans. Group A lives in
remote regions of Borneo without outside contact and is highly inbred, while
group B lives in New York with wide genetic origins and divergent populations
that freely interbreed with others. What is the most accur ate statement
concerning a strategy for developing DNA markers for these species:
A. Each group will require a balance between conserved and polymorphic DNA
markers
B. Group B will require greater numbers of conserved DNA markers than group A
C. Group B will have g reater numbers of polymorphic DNA markers than group A
D. Group B will have fewer numbers of poly morphic DNA markers than group
E. The same DNA markers for each group regardless of their polymorphic nature
Answer: C
57. A man with the blood type A and Rh+ marr ies a woman with A and Rh - blood -
type. They have two boys who both have O - blood. Assuming simple Mendelian
inheritance of blood type and Rh, what is the probability that their next two
children will be male and will have O - blood, too?
A. 1/8
B. 1/16
C. 1/32
D. 1/64
E. 1/256
Answer : C
Biosystematics Questions 58 to 60
58. For which of the following fungi does one mitotic division follow the meiotic
division that occurs after karyogamy?
I. Ascomycota
II. Basidiomycota
III. Zygomycota
A. I only
B. II only
C. III only
D. I and II
E. I, II, and III
Answer : B
59. Which of the following characteristics is unique to the protist phylum Ciliophora?
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A. Autotrophic nutrition
B. Concentration of organelles at one end of the cell
C. Differentiation of macro - and micronuclei
D. Paramylon storage molecule
E. Use of r hizopoda for movement
Answer: C
60. Heterospory first arose in the:
A. Bryophytes
B. Ferns
C. Lycophytes
D. Psilophytes
E. Sphenophytes
Answer: C
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22 PART B
Cell and Molecular Biology (20%) 12 questions 61-73
Plant Anatomy and Physiology (15%) 9 questions 74-82
Animal Anatomy and Physiology (25%) 15 questions 83-97
Ethology (5%) 3 questions 98-100
Genetics and Evolution (20%) 12 questions 101-112
Ecology (10%) 6 questions 113-118
Biosystematics (5%) 3 questions 119-120
Cell and Molecular Questions 61 to 73
Use the labels in the diagram below to answer the questions 61 to 65. (8 points)
Using the labels of the picture to answer the questions below:
You are a scientist hunting for enzymes involved in various aspects of cellular
metabolisms. Which organelle would you want to isolate and analyze if you are
looking for (fill in the relevant organelle label in the sentences below):
61. Enzyme that is involved in DNA replication _____ ___
62. Enzyme involved in producing large amounts of ATP _______
63. Enzyme attaching sugar molecules to proteins _________
64. Enzyme involved in protein synthesis _________
65. DNA ________
Answer: 61, A & B, 62, b, 63, C, 64, blank, 65 A& B
66. How do polypeptides find their way from the site of syn thesis on the cytoplasmic
ribosome to the place of their destination in the peroxisome?
A. Without signals
B. By specific transport along the cytoskeleton
C. By specific carboxy -terminal targeting signals
D. By specific vesicular transport
E. By transport within the ER
Answer C
67. In the lab, you are running an SDS -PAGE, but you realize that you forgot to add
DTT, a reducing agent, when you were loading the gel. Which amino acid
residue will this affect?
A
B
C D
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A. Cysteine
B. Serine
C. Glutamate
D. Lysine
E. Methionine
Answer: A
68. Sequ encing reactions involve many of the same reagents as PCR, with one major
exception: a modified version of dNTPs is used. Which of the following should
be included in a sequencing reaction but not a PCR?
A.
B.
C.
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24 D.
E.
Answer : C
69. A human liver cell and a plant cell from the cortex of a root are placed in a beaker
of distilled water. What will most likely occur ?
A. Both cells will be in osmotic balance with the water and neither will swell or shrink
B. The liver cell will shrivel, while the plant cell will becom e flaccid and the plasma
membrane will pull away from the cell wall
C. The liver cell will shrivel, while the plant cell will become very turgid
D. The plant cell will become flaccid and the plasma membrane will pull away from the
cell wall, while the liver cell will swell and may even burst
E. The plant cell will become very turgid, while the liver cell will swell and may even
burst
Answer : E
70. You are given an unknown organism to study. Upon examination, you find that it
does not have a nuclear membrane or mito chondria, which structure (s) below
would it possess?
A. Chloroplast
B. Endoplasmic reticulum
C. Ribosomes
D. Lysosome
E. 9+2 Cilia
Answer : C
71. The velocity of carrier -mediated diffusion across cellular membranes:
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25 A. Can increase up to a maximum value
B. Does not depend o n saturation of the carrier
C. Is always proportional to substrate concentration
D. Is greater in uniporters than symporters
E. Varies with substrate concentration in the same way as that observed for simple
diffusion
Answer : A
72. Why can glycolysis proceed under anaerobic or aerobic conditions while the citric
acid cycle is strictly aerobic?
A. Enzyme catalysis during the citric acid cycle is regulated by allosteric effectors which
include oxygen
B. Enzymes of glycolysis do not use oxygen as cofactors, while enzymes o f the citric
acid cycle require oxygen for proper folding
C. NAD+ for glycolysis can be regenerated through the conversion of pyruvate into
lactate under anaerobic conditions
D. NAD+ is produced during three steps of glycolysis and only one step of the citric
acid cycle
E. Oxygen is a byproduct of the conversion of the glycolytic intermediate glucose -6-
phosphate to fructose -6-phosphate
Answer : C
73. Which of the following are likely to cause cellular changes when injected directly
into the appropriate cells cytopl asm?
I. Triiodothyronine (T3)
II. Cortisol
III. Insulin
IV. Epidermal growth factor
A. I, II
B. I, IV
C. II, III
D. II, IV
E. III, IV
Answer : A
Plant and Anatomy Questions 74 to 82
74. The tissues in bryophytes which are analogous to the xylem and phloem,
respectively, are made of :
A. Tracheids, leptoids
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26 B. Hydroids, leptoids
C. Tracheids, albuminous cells
D. Sieve cells
E. None of the above
Answer : A & B
75. The water potential ( ) of cells in a plant’s root are typically around -0.2 MPa. If
a plant is placed in solution with a water potenti al () o f -.05 MPa, the plant
roots will
A. Become turgid from water moving into the roots
B. Become flaccid from water moving out of the roots into the water
C. Be largely unaffected due to the minor difference in water potential
D. Be unaffected as water potential has nothing to do with water flow
Answer: A
76. The table below shows data from an experiment with two groups of plants. For two
weeks, Group I was not fertilized and Groups II was fertilized. The growth in millimeters of
each plant was measured for the each group. Based on the results in the chart below, which
statement is true?
Growth (mm)
A. Experimental populations contained different sample sizes
B. After two weeks the mean size of fertilized and unfertilized plants was similar
C. The unfertilized group has a larger standard deviation than the fertilized
D. In order to observe any differences the experiment should have been run for longer
than two weeks
E. The fertilized group has a large r standard error than the unfertilized
Answer: E
77. The diagram below shows an agar block inserted beneath the tip of a growing
grass seedling that is exposed to light from the left. After exposure to light,
the block is removed and cut into two halves as s hown by the dotted line.
These halves are then placed onto a pair of new, straight grass seedlings
whose tips have been removed. These seedlings are then grown in the dark. If Group I Group II
3.2 6.7
3.5 10.4
4.0 8.7
2.9 9.8
4.1 11.5
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27 the agar blocks are placed on these two plants as indicated, which of the
diagra ms blow shows the expected result?
Answer: A
78. All but which of the following stat ements are true about photosynthesis except:
A. A plant that could carry out cyclic photophosphorylation but could not perform any
other aspects of photosynthesis would release O2 as a by -product of its metabolism
B. ATP is synthesized as high energy electrons that are passed along a chain of electron
acceptor molecules
C. Light energy to drive photophosphorylation may first be trapped by the chlorop hyll
pigments in P hotosystem I and then passed to a molecule of P700, which traps it
D. There are two light events in n oncyclic photophosphorylation
Answer: A
79. There is a large difference in pH across the thylakoid membrane between the
thylakoid compartment and the stroma. From the list given below, choose those
that are appropriate in explaining the difference. (Choo se A, B, C, or D)
i. The transport of protons into the thylakoid compartment by the electron
transfer system
ii. The transport of protons out of the thylakoid compartment into the
stroma by the electron transfer system
iii. Protons splitting from water remaining in the thylakoid compartment
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28 iv. Protons splitting from water exiting the thylakoid compartment
v. The removal of hydrogen from the stroma during the reduction of
NADP to NADPH
vi. The retention of hydrogen in the stroma during the reduction of NADP
to NADPH
A. i, iv, and vi only
B. ii, iv and vi only
C. I, ii, iii only
D. i, iv and v only
E. i, iii and v only
Answer : E
80. Chloroplasts and mitochondria share similarities . From the following
characteristics, use A, B, C, D, or E to select those to both chloroplasts and
mitochondria .
i. both contain ribosomes
ii. both are single -membrane bounded structures
iii. both are double -membrane bounded structures
iv. both undergo reactions that are primarily oxidative in nature
v. both undergo chemiosmosis
vi. both have similar, although not identical, electron trans fer systems
vii. both undergo reactions that are primarily reductive in nature
viii. both contain DNA and RNA
A. i, iii, v i, vii, and viii only
B. i iv, vi, vii and viii only
C. ii iv, v, vii and viii only
D. ii, iii, vi , vii, and viii only
E. ii, iv, vi, vi, and viii only
Answe r: A
81. Plant cells have large vacuoles that comprise as much as 90% of mature cell
volume, in contrast to animal cells that typically have vesicles comprising less
than 10% cell volume. Which of the following explains this characteristic
difference in plant vs animal cells?
A. Most plants do not have well -developed excretory systems and use vacuoles as
waste dumps
B. Vacuoles are an energetically inexpensive way to increase cell size and therefore
surface area to intercept light for photosynthesis
C. Vacuoles a ccumulate ions and water, contributing to turgor pressure which keeps
the plants rigid, despite their lack of skeletons
D. All of the above
E. None of the above
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29 Answer: D
82. The group of plants that is sister to the euphyllophytes is
A. Bryophytes
B. Lycophytes
C. Ferns
D. Charales
E. Green algae
Answer: B
Animal Anatomy and Physiology 83 -97
83. A patient previously diagnosed with Graves disease (an antibody -mediated
autoimmune disorder) comes to your clinic for a routine follow -up exam. Your
patient complains of ner vousness, fatigue, and diarrhea all of which are worse
when they forget to take their medications for Graves. Upon finding a goiter of
the thyroid gland, you draw a complete blood screen and assay iodine uptake by
the thyroid gland. You expect to find the following:
A. Decreased iodine uptake by the thyroid gland, decreased serum cholesterol
B. Decreased serum TSH, decreased T3, increased T4
C. Decreased T3 and T4, increased serum cholesterol
D. Increased serum TSH, decreased triiodothyronine (T3) and thyroxine ( T4) levels,
increased iodine uptake by the thyroid gland
E. Increased calcium, increased serum TSH
Answer : C
Questions 84 to 88 are based on a virtual experiment . A confluent layer of
macropha ge cells are grown in two (2), 25 cm2 tissue culture flasks w ith appropriate
medium containing 10% plasma serum. For the experiment, all medium is removed
from the flask and the cells are washed with appropriate buffers to remove all traces
of serum and media. E. coli bacteria, suspended in serum free media, are th en added
or “fed” to the macrophages. The process of E. coli death from phagocytosis was
then studied.
84. It is necessary in a phagocytosis study to remove the serum because:
A. The complement proteins would have independently killed the bacteria.
B. The B cells would have independently killed the bacteria.
C. The natural killer cells would have independently phagocitized the bacteria.
D. Serum proteins would have inhibited the membrane attack complex (MAC).
E. Serum proteins would have inhibited the MHC complex.
Answer: A
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30 After 30 minutes the cells in flasks in tissue culture flask one (1) and two (2) are
washed with appropriate buffers to remove all E. coli. Serum free media is added
and the flasks incubated.
85. The macrophage cell notifies other cells of an immunological invader by:
i. Up-regulating expression of MHC I molecules upon activation.
ii. Up-regulating expression of MHC II molecules upon activation.
iii. Interacting with the complement system.
iv. Acting as an antigen presenting cell.
v. Inactivating viruses once they enter the macrophage.
vi. Decreasing enzymatic production
vii. Increasing enzymatic production
A. i, iii, iv, vi
B. iv, v, vii
C. ii, iii, iv, vii
D. ii, iii, iv, v
E. i, ii, iv, vi
Answer: C
86. During incubation the macrophage and microbe are at war. Which statement is
true.
A. The macrophage digests the E. coli through the oxidative burst.
B. The E. coli protects itself through the oxidative burst.
C. The oxidative burst is characterized by an increased production of hydrogen
peroxide.
D. A and B
E. A and C
Answer E
The macroph ages are then lysed by adding ice cold water. The lysed macrophages
and water are then centrifuged and the bacteria isolated from the macrophage debris.
The bacteria were suspended, diluted and plated on agar. After 90 minutes , the same
procedure was fo llowed f or tissue culture F lask 2 and the bacteria were again plated.
Bacterial P lates 1 and 2 were incubated for 24 hours.
87. Why were the macrophages lysed with a brief exposure to ice cold water and not
the E. coli ?
A. Rapid cold -shock results in osmo tic lysis.
B. Gram positive bacteria are impervious to osmotic insult.
C. The membrane lipid bilayer is impermeable to water.
D. Lysis occurs only with eukaryotic cells.
E. Cell walls limit osmotic lysis.
Answer : E
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31 88. When you compare bacteria plate 1, (45 minutes ) and bacteria plate 2 (90
minutes) you would predict (3 points) :
A. Equal colonies on P lates 1 and 2
B. No colonies on P late 1
C. No colonies on P late 2
D. More colonies on Plate 1 than P late 2
E. More colonies on Plate 2 than P late 1
Answer : D
89. Why do adults usu ally die when their body temperature exceeds 105°F, but
young children may survive that temperature, at least for a longer period of time?
A. Children have a larger relative surface area to volume ratio than adults
B. Children perspire more profusely than adu lts
C. At a higher temperature children’s hemoglobin carries more oxygen than adult
hemoglobin
D. Children’s proteins are denatured at higher temperatures than adults
E. Adults lose their ability to sense warmth as they age
Answer : A
90. When someone already has t he flu, they can be given the drug Tamiflu to prevent
further spread of the virus. Tamiflu inhibits neuraminidase, an enzyme on the
surface of the influenza virus. Below is a diagram of the activity of
neuraminidase. How would inhibiting neuraminidase prev enting further
spreading of influenza?
A. Influenza particles will no longer be able to attach to host cells
B. Receptors containing sialic acid will not be cleaved. These receptors are
needed to make influenza virulent.
C. Receptors containing sialic acid wi ll not be cleaved. Thus, new virions will
not be released and no new infections can occur.
D. If neuraminidase is not active, hemagglutinin can not function. This would
render the virus non -virulent.
Answer : C
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32
91. The following graph shows the pressure -volume curve of a mammalian lung
when it is removed from the body and filled under two conditions. One condition
is when the lung is inflated with air and the second is when it is inflated with
saline solution (the entire lung is filled with saline). In both instances the excised
lungs retained the thin film of liquid that lines the respiratory alveoli.
Based on the d ata presented in the graph, which of the following statements is
true?
A. The lung cannot increase its volume if it is filled with saline instead of air.
B. Surface tension created by the air -liquid interface on the alveolar surface increases
the work needed t o inflate the lung.
C. Resistance to flow of fluids through the respiratory passageways is greater than is the
resistance to flow of air, so more work is needed to inflate the lung with saline than
with air.
D. Lung compliance is independent of infusion media.
E. The work required to inflate the lungs with air at small volumes is less than the work
required to inflate the lung at large volumes.
Answer : B
92. Polarity in the developing Drosophila embryo is determined by a(n):
A. Expression of the gap protein hunchb ack throughout the embryo
B. Protein gradient of the segmentation protein engrailed
C. Protein gradient of the bicoid protein expressed from maternal mRNA
D. Protein gradient of the gap protein hunchback
E. Expression of the segmentation protein engrailed throughout t he embryo
Answer : C
93. An extremely high dose of HMG -CoA reductase inhibitor would severely disrupt
which of the following processes?
I. Synthesis of bile salts
II. Formation of bilirubin
III. Production of surfactant
IV. Aldosterone synthesis
V. Vitamin D synthesis
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33 A. I, II, and IV
B. I, III, and IV
C. I, IV, and V
D. II and IV
E. III, IV, and V
Answer : C
94. A child is born with erythroblastosis fetalis. Which of the following is an acceptable
explanation for the observed pathology ?
A. The child must be the first conceptus of the mot her
B. The mother was treated with the carcinogen Rhogam during pregnancy
C. Anti-D antibodies were generated into the mother by a previous pregnancy
D. The father must be Rh -negative
E. Erythroblastosis fetalis leads to decreased red blood cell synthesis
Answer : C
95. Substances that excite β2 adrenergic receptors (β -agonists) are used to treat
asthma and COPD (Chronic Obstructive Pulmonary Disease). These drugs can
have significant side effects including:
A. Decreased sensitivity to inflammation, inhibition of u terine contraction, hypotension
B. Hypotension, increased blood pressure, decreased plasma potassium
C. Hypertensi on, dizziness, vasoconstriction
D. Bronchial -dilation, increased uterine contracti on, increased serum angiotensin
E. Dizziness, bronchial -constriction, decrea sed sensitivity to inflammation
Answer : A
96. The 4 layers of the digestive tract from the inside out are:
A. Serosa, muscular layer, mucose, submucosa
B. Serosa, muscular layer, submucosa, mucosa
C. Mucosa, submucosa, muscular layer, serosa
D. Submucosa, mucosa, se rosa, muscular layer
E. Muscular layer, serosa, mucosa, submucosa
Answer: C
97. Trace the flow of filtrate through the nephron as it enters the kidney through the renal
artery.
1. Proximal convoluted tubule
2. Collecting duct
3. Descending limb of loop of H enle
4. Bowman's capsule
5. Distal convoluted tube
6. Ascending limb of loop of H enle
A. 4,1,3,6,5,2
B. 4,1,6,3,5
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34 C. 4,5,6,3,1
D. 4,5,3,6,1,2
E. 4,6,3,1,5,2
Answer: A
Ethology 98 -100
98. Which of the following may represent advantage(s) of a dominance hierarchy?
A. Suppression of a ggression among conspecifics
B. Ensures food access for high ranking individuals
C. Reduction of the effects of lethal combat
D. A and B only
E. A, B, and C
Answer : E
99. Select the mating system that best describes the following exemplary
species/characteristics:
Very large testes relative to body size
No male parental care
Females advertise oestrus
A. Dispersed
B. Promiscuity
C. Monogamy
D. Polyandry
E. Polygyny
Answer : B
100. For certain fishes that maintain feeding territorie s that they must patrol against
intruders, the most successful individuals will be those maintaining:
A. Very large territories compared to their size
B. Very small territories compared to their size
C. Intermediate sized territories compared to their size
D. Cooperation with individuals of the same species
E. Grou ps of females to capture prey
Answer : C
Genetics and Evolution 101 -112
101. Two cultivars of the 4 -o' clock plant ( Mirabilis jalapa ) are reciprocally crossed.
When pollen from cultivar A pollinates cultivar B, all the offspring are nonvariegated
(all g reen). When pollen from cultivar B pollinates cultivar A, some of the offspring
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35 are variegated (having green and white patches), some are green, and some are white
but die as seedlings.
Which of the following statements are true?
1. Cultivar A is variegated
2. Cultivar B is variegated
3. A portion of chloroplasts in variety A carry a mutation for chlorophyll
synthesis
4. The pollen's genotype determines variegation in this cross
5. Variegation is a lethal recessive phenotype
A. 2, 4 only
B. 2, 4, 5 only
C. 1, 3 only
D. 1, 2, 5 onl y
E. 1, 2, 3, 4, 5
Answer: C
102. In bacteria, the elongation factor EF -Tu binds aminoacyl -tRNAs and transports
them to the A site of the ribosomal large subunit after translation has been
initiated. Which amino acid's aminoacyl -tRNA does not need EF -Tu to enter
the ribosome ?
A. Phenylalanine
B. Proline
C. Lysine
D. Glycine
E. Formylmethionine
Answer: E
103. The Telome Theory describes the evolution of
A. Euphyllophyte leaves
B. Lycophyte leaves
C. Plant meristems
D. Roots from shoots
E. Vascular tissues
Answer : A
104. Plesiomo rphy refers to a character that is
A. Ancestral
B. Derived
C. Distributed among taxa
D. No longer found in extant taxa
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36 E. Unique to a given taxon
Answer : C
105. A recessive phenotype was discovered in two unrelated families. A man from
one family marries a woman from the other family. They have five children [see
the pedigree below]. Assume the recessive phenotype is completely penetrant.
If so, what is the genotype of the starred individual?
A. D1D2
B. D1d2
C. d1D2
D. d1d2
Answer : D
106. Which statement is true accord ing to Darwin’s theory of evolution through
natural selection?
A. Species adapt until reaching an optimal form
B. Individuals evolve as a function of their environment
C. Species producing many offspring are more adapted than those producing few
D. The number of offs pring produced by an individual is related to both the traits of
the parents and the environment
E. In a population undergoing natural selection most mutations lead to new species
Answer: D
For Questions 107 to 110, use the following information: You are stu dying a
population of penguins found on an island near Antarctica, and discover that some
have blue feet, while others have orange.
107. You count 16 blue -footed penguins and 84 orange -footed penguins. Assuming
that foot color is determined by two differ ent alleles of one gene, where the
orange foot allele is dominant to the blue foot allele, and that the population is
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37 in Hardy -Weinberg equilibrium, what is the allele frequency of the orange foot
allele?
A. 0.40
B. 0.50
C. 0.60
D. 0.75
E. 0.84
Answer: C
108. Under t he same assumptions as above, what is the frequency of heterozygotes
for the foot color gene in this population of penguins?
A. 0.0
B. 0.24
C. 0.4
D. 0.48
E. 0.6
Answer : D
109. Upon further investigation, it is discovered that 20 of the orange -footed
penguins have a s mall black spot on the bottom of their feet, and that these
individuals are the heterozygotes for the foot color gene. Assuming random
mating over many generations, one would expect the frequency of blue -footed
penguins to approach
A. 0
B. 0.04
C. 0.07
D. 0.1
E. Cannot be determined
Answer: C
110. Assume the black -spotted orange feet are caused by another gene epistatic to
the foot color gene . With random mating over many generations, one would
expect the frequency of blue -footed penguins to approach
A. 0
B. 0.04
C. 0.06
D. 0.07
E. Cannot be determined
Answer : E
111. In Lycopersicon esculentum (tomato) , plants that were homozygous tall with
smooth fruit were crossed with plants that were homozygous recessive dwarf
with hairy fruit. The F1 plants were test -crossed with the following results:
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38 Tall, smooth fruits 36
Dwarf, hairy fruits 39
Tall, hairy fruits 12
Dwarf, smooth fruits 13
These data indicate that the genes for:
A. Height and fruit surface are on different chromosomes
B. Height and fruit surface are completely linked
C. Height and fruit surface are linked and 12.5 units apart
D. Height and fruit surface are linked and 25 units apart
E. Height and fruit surface are linked and 33 units apart
Answer : C
112. A woman with blood type A, RH+ gives birth to a child with type O, Rh- blood.
There are five men who may be the father. The blood types of all five are
determined in order to identify the father. Regrettably, only one of the men can
be removed from consideration. Of the blood types below, which man can
definitely be removed?
A. A, Rh+
B. O, Rh+
C. B, Rh+
D. B, Rh-
E. AB, Rh-
Answer : E
Ecology 113 -117
113. Which of the following factors are believed to influence the global latitudinal
gradient of species richness? Choose all that apply.
1. Stability of tropical biomes over time
2. Actual evapotranspirat ion
3. Potential evapotranspiration
A. 1
B. 2
C. 1, 2
D. 1, 3
E. 1, 2, 3
Answer:E
114. The region of the world with the greatest biodiversity of marine taxa (in total
and per area) is the:
A. Abyssal plain of the Atlantic Ocean
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39 B. Southern Ocean (the ocean surrounding Antar ctica)
C. Seas between Southeast Asia and northern Australia
D. Caribbean Sea
E. Pelagic zone of the North Pacific
Answer: C
115. Which of the following threaten coral reefs worldwide?
1. Pollution
2. Destruction of mangrove and seagrass communities, which trap and f ilter
pollution from runoff
3. Destructive forms of fishing, such as fishing with dynamite
4. Ocean acidification due to increasing atmospheric CO2 concentration
5. Diseases of corals
A. 1, 2, 5
B. 1, 3, 4
C. 3, 4, 5
D. 1, 2, 3, 5
E. 1, 2, 3, 4, 5
Answer: E
Questions 116 to 1 17 refer to the following paragraph. In Argentina at 32˚ S latitude
there are deserts east of the Andes Mountains. In Chile at 18˚ S latitude there are
deserts west of the mountains in. Select from the following list of explanations those
that bes t explain these two situations (6 Points /3 ea ch).
i. Easterly winds
ii. Westerly winds
iii. Descending dry air masses
iv. Ascending dry air masses
v. Rain shadow on east side of mountain
vi. Rain shadow on west side of mountain
116. The Argentina deserts mentioned above are most likely the result of:
A. i, ii, and ii only
B. i, iv and v only
C. i, iii, and vi only
D. iii, and v only
E. ii, iv and vi only
Answer: C
117. Chile deserts mentioned above are most likely the result of:
A. i and ii only
B. i and v only
C. i and iv only
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40 D. ii and v only
E. ii and iv only
Answer : A
Biosystemati cs 118 -120
118. The Kingdom of Tonga is an archipelago consisting of 176 islands in the South
Pacific. Only 51 of these islands are inhabited by man. A plant population that
reproduces annually was found on one of the remote uninhabited islands of
Tonga. Two flowe r-color variants were present, yellow and orange. Flower color
is known to be a monogenic trait. The frequencies of the two flower colors were
observed annually over a ten -year period. It was noted the proportion of plants
with oran ge flowers steadily decl ined each year. From these data, it may
reasonably be concluded that:
A. The increased frequency of yellow -flowered plants was a result of genetic drift.
B. Migration of yellow -flowered plants into the population was the most likely cause of
the observed change .
C. Mutation occurred more frequently in yellow -flowered plants than in orange -
flowered plants.
D. Orange -flowered plants had a lower genetic fitness than the yellow -flowered plants.
E. Yellow -flowered plants were capable of crossing with other yellow -flowered pla nts
or with orange -flowered plants but orange -flowered plants could only cross with
other orange -flowered plants.
Answer : D
119. The region of the world with the greatest biodiversity of marine taxa (in total
and per area) is:
A. Abyssal plain of the Atlan tic Ocean
B. Southern Ocean (the ocean surrounding Antarctica)
C. Seas between Southeast Asia and northern Australia
D. Caribbean Sea
E. Pelagic zone of the North Pacific
Answer: C
120. Some populations are characterized by the presence of balanced polymorphism.
This condition may be maintained by all of the following EXCEPT:
A. Balancing selection
B. Directional selection
C. Disruptive selection
D. Frequency -dependent selection
E. Natural selection
Answer : B
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41
PART C
1. Match the animals and their characteristics. Mark the app ropriate box with an
“X” in the space provided (Points Possible 20. Incorrect answers will be
subtracted from correct answers for Total Points)
Fresh water
fish Bird Marine
fish
Lizard Marine
mammal Terrestrial
mammal
Drinks
water
regularly
Does not
drink water
Wastes are
discarded
as ammonia
Wastes are
discarded
as urea
Wastes are
discarded
as uric acid
Actively
secretes salt
Actively
absorbs salt
Excretes
hypotonic
urine
relative to
the body
fluids
Excretes
isotonic
urine
relative to
the body
fluids
Excretes
hypertonic
urine
relative to
the body
fluids
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42
Answer
Fresh
water
fish Bird Marine
fish
Lizard Marine
mammal Terrestrial
mammal
Drinks
water
regularly X X X
Does not
drink water X X X
Wastes are
discarded as
ammonia X X
Wastes are
discarded as
urea X X
Wastes are
discarded as
uric acid X X
Actively
secretes salt X
Actively
absorbs salt X
Excretes
hypotonic
urine
relative to
the bo dy
fluids X
Excretes
isotonic
urine
relative to
the body
fluids X
Excretes
hypertonic
urine
relative to
the body
fluids X X X X
2. Movement from an aquatic environment to the land presented a number of
challenges for plant reproductions. D escribe these challenges and explain how
the angiosperms have adapted to survive and reproduce on land. Include a
diagram of the angiosperm life cycle and describe with some detail the process of
sexual reproduction in plants (Total 20 points) .
Answer
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| 43
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43 Primary problems :
1. Lack of water for flagellated sperm (2)
2. Zygote and embryo in danger of desiccation (2)
(extreme fluctuation in temperature, humidity, wind, light… can affect embryo)
Life cycle altered to avoid necessity of flagellated sperm swimmi ng to egg and to
roved protection fro embryo.
3. Insert Life Cycle sketch (10 points)
Haploid (2)
Gametes, spores, multicellular haploid (Total 3)
Diploid (2)
Zygote, multicellular diploid (2)
Meiosis (1)
Fertilization (1)
4. Details (6)
Any six of th e following:
Gametophytes are much reduced in angiosperms
Major advance is evolution of pollen
Pollen grain falls on stigma and pollen tube grows down style and delvers 2
sperm directly to embryo sac —no need for flagellated sperm
Double fertilization gi ves rise to zygote and triploid endosperm (provides
food source to maintain embryo and early growth)
Seed coat protects zygote and embryo
Ovary wall will ripen into fruit that often provides means of dispersal.
Congratulations on being a S emi-finalist!
We hope to see you as a F inalist!!
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1
Inorganic Chemistry 411/511 Final Exam
115 minutes; 200 points total
Show your work for partial credit.
1. Draw the molecular geometry and i ndicate any deviations from ideal VSEPR coordination
angles. Give the point group for each molecule.
(a) [10 pts] CH
2Cl2 (b) [10 pts] XeO3
Td coordination, C 2v trig pyramidal, C 3v
OXeO < 109.5°
(c) [7 pts] Is either of the above molecules chiral ? Explain using symmetry rules.
No, they both have mirror plane operations.
2. (a) [12 pts] Write down a Born -Haber analysis (give all the reaction steps) needed to
estimate ∆H
f for CaO(s) from the elements in standard states. Label each reaction
step with an appropriate energy term (such as I, E a, ΔH L, etc).
Ca(s) → Ca (g) ΔH at(Ca)
Ca(g) → Ca2+(g) + 2 e- I1(Ca) + I 2(Ca)
½ O 2 (g) → O(g) ½ D 0 (O-O)
O(g) + 2 e- → O2-(g) - (Ea1(O) + E a2(O))
Ca2+(g) + O2-(g) → CaO (s) - ΔH L (CaO)
_________________________________________________
Ca(s) + ½ O 2 (g) → CaO (s) ΔH f (CaO(s))
Correct reactions steps [ 5]
Correct label for energetics (I, Ea, ΔHL, etc) [ 4]
Correct summation and signs [ 3]
(b) [7 pts] Is CaO(s) more or less soluble in polar solvents than KF (s) ? Explain
briefly .
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2
Much less soluble. CaO has a much larger lattice enthalpy due to the higher
ion charges, this opposes ion s olvation.
3. (a) [1 4 pts] Construct a n MO diagram for the cyanide anion , CN¯(g), including valence
atomic and molecular orbitals , symmetry labels, and the electron filling of the MO's.
CN¯ is isoelectronic and isolobal with CO. See the MO for CO
in text Fig. 2.22.
(b) [7 pts] Draw a figure showing the geometry of the LUMO orbital on C N¯.
(c) [5 pts] Show with a simple picture how the LUMO in part (b) can act as a π –
acceptor for transition metal cations .
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3
4. A Latimer diagram for Cl at pH=0 is:
1.20 V 1.18 1.67 1.63 1.36
ClO4¯ ClO3¯ HClO 2 HClO Cl2 Cl ¯
(a) [10 pts] What is the standard potential for reduction of ClO
3¯ to Cl ¯ at pH=0 ?
[2(1.18) + 2(1.67) + 1(1.63) + 1(1.36)] / 6 = +1.45 V
(b) [10 pts] Does C l
2 disproportion ate spontaneously in aqueous acid? Explain .
No. The reaction Cl 2 → Cl- + HClO has a negative potential (1.36 V – 1.63
V). Reactions to form chloride and HClO 2 or Cl(V) or Cl(VII) also have a
negative potential.
(c) [10 pts] Write a balanced half -reaction for the reduction of H ClO to Cl
2 in acidic
solution.
2 e
- + 2H+ + 2 HClO → Cl2 + 2 H 2O
(d) [10 pts] Calculate the potential for the half -reaction in part (c) at pH = 3.
The Nernst equation is E = E
0 – (0.059 V/ n) log Q
E = +1.63 – (3)(0.059) = +1.45 V
5. (a) [6 pts] Write out the full name of the octahedral complex [ Mn(NH
3)6]2+
hexammineman ganese (II) or hexaammineman ganese(2+)
(b) [10 pts] Write a d orbital energy level diagram for BOTH the strong -field (low
spin) and weak -field (high spin) cases for the complex in part (a ). Label the d -
orbitals with (z
2, x2-y2, xy, xz, and yz) and also give the orbitals symmetry labels .
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4
(c) [10 pts] Calculate the ligand field stabilization energies (LFSE) in terms of ∆O for
BOTH of the two electron configurations in part (a).
High spin LFSE = zero; low spin is LFSE = 2. 0 ΔO
(d) [8 pts] How would you experimentally determine the actual configuration (high or
low-spin) for an O h complex of Mn (II)?
Magnetic moment should be proportional to sqrt [N(N+2)]. [8 pts] Trying to measure LFSE by calorimetry is not a good method. [4 pts if this]
6. (a) [10 pts] Write a balanced reaction showing how H
2 is generated on an
industrial scale.
CH 4 (g) + H 2O (g) → CO (g) + 3 H 2 (g)
Or
C (s) + 2 H
2O (g) → CO 2 (g) + 2 H 2 (g)
Or, electrolysis is OK
(b) [8 pts] Describe briefly 2 important ways that hydrogen bonding affects our daily
lives.
1. Lower density of ice than water
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5
2. DNA replication
3. High bp of liquid water.
7. [4 pts each] Circle the ONE best choice.
(a) Wurtzite structure has the stacking sequence of (A aBb)n, where uppercase letters
are anions and lowercase are cations . What is the coordination geometry around
anions in the lattice ?
(1) tetrahedral (2) square planar (3) octahedral
(4) trigonal prismatic (5) linear (6) trigonal bipyramidal
(b) The following complexes are all high spin. Which has a LFSE equal to zero ?
(1) [Fe(OH 2)6]2+ (2) [Fe(OH 2)6]3+ (3) [Mn(OH 2)6]2+
(4) [Co(OH 2)6]3+ (5) all these complexes have the same LFSE
(either choice was OK)
(c) Which of the following octahedral complexes has two geometric isomers?
(1) [FeCl(OH 2)5]2+ (2) [IrCl 3F3]2- (3) [RuCl 4(bipy)]2-
(4) [CoBr 2Cl2(NH 3)2]+ (5) [W(CO) 6] (6) [CrCl 5(NH 3)]2-
(d) Which ligand will form the strongest complex with the hard acid Al3+ ?
(1) O2- (2) S2- (3) Se2-
(4) Te2- (5) all complexes will have the same K f
(e) Which ligand is a π -acceptor with a strong ligand field effect?
(1) N(CH 3)3 (2) H¯ (3) CN¯ (4) Cl¯ (5) NH 3 (6) P(CH 3)3
(f) Based on d electron configurations, which of the following is NOT likely to be square planar?
(1) [AgCl
4]¯ (2) [AuF 4]¯ (3) [Pd(CN) 4]2- (4) [RhF 4]2- (5) [PtCl 4]2-
(g) The anion XeF
5─ shows only a single 19F NMR peak over a wide temperature
range (ignoring coupling to 129Xe). Which geometry is consistent with this data?
(1) trigonal bipyramidal (2) square pyramidal
(3) pentagonal planar (4) linear
(5) rhomboidal (6) cubic
(h) Which of the following will rapidly react with water to form H 2 ?
(1) O 2 (2) F 2 (3) HF 2¯ (4) CH 4 (5) SF 6 (6) CaH 2
(i) Using Pauling’s rules, what is the expected pK a for nitrous acid, HNO 2?
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6
(1) 0 (2) 1 (3) 3 (4) 5 (5) 7 (6) 8
Oh E 8C3 6C2 6C4 3C2 i 6S4 8S6 3
h 6
d
A1g 1 1 1 1 1 1 1 1 1 1 x2+y2+z2
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7
A2g 1 1 -1 -1 1 1 -1 1 1 -1
Eg 2 -1 0 0 2 2 0 -1 2 0 (z2, x2-y2)
T1g 3 0 -1 1 -1 3 1 0 -1 -1
T2g 3 0 1 -1 -1 3 -1 0 -1 1 (xz, yz, xy)
A1u 1 1 1 1 1 -1 -1 -1 -1 -1
A2u 1 1 -1 -1 1 -1 1 -1 -1 1
Eu 2 -1 0 0 2 -2 0 1 -2 0
T1u 3 0 -1 1 -1 -3 -1 0 1 1 (x, y, z)
T2u 3 0 1 -1 -1 -3 1 0 1 -1
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T h e o r e t i c a l P r o b l e m s
S o l u t i o n s
M O C K I C H O 2 0 2 1
Chemistry Online Discord Server
Pensive Insular and Scoob
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
T a b l e o f C o n t e n t s
Instructions
Problem 1.
Cool Chemistry
13% of the Total
Problem 2.
Interesting Oxidation States of Copper
15% of the Total
Problem 3.
Wolfram,
钨
,
вольфрама,
اﻟﺗﻧﻐﺳﺗن
8% of the Total
Problem 4.
Quantum Kinetics
16% of the Total
Problem 5.
Three in One
16% of the Total
Problem 6.
Adamantane, Symmetry , and NMR
9% of the Total
Problem 7.
Build Your Own Frogger
15% of the Total
Problem 8.
Non-Classical Chemistry
8% of the Total
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
1
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
I n s t r u c t i o n s
●
This theoretical exam booklet contains 22 pages.
●
You may begin writing as soon as the
START
command
is given.
●
You have
5 hours
to complete the exam.
●
All results and answers must be clearly written
in
PEN in their respective
designated areas
on the exam papers. Answers written
outside the answer boxes
will NOT be graded.
●
You may use as much scratch paper as you want (due
to the online format of this
competition). Remember that nothing outside the designated
areas will be graded.
●
The periodic table is not part of this booklet; it
is provided separately .
●
Use only the pen and calculator provided.
●
This is the official English version
of the exam booklet
●
The supervisor will announce a 30-minute warning before
the
STOP
command.
●
You must stop your work immediately when the
STOP
command is announced.
Failure to stop writing within 30 seconds will lead
to nullification of your theoretical
exam.
●
After the
STOP
command has been given, place your
exam booklet back in your
exam envelope and wait at your seat. The exam supervisor
will come to collect the
envelope.
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
2
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
T h e o r e t i c a l # 1
1 .
2 .
3 .
4 .
5 .
6 .
7 .
8 .
9 .
T o t a l
1 3 % o f t h e t o t a l
1
2
3
1
3
3
4
3
3
2 3
P r o b l e m 1 . C o o l C h e m i s t r y
1.
Calculate the standard change in internal ener gy
and change in enthalpy for the vaporization of
one mole of NKT -19.
Δ
U =
Δ
A +
Δ(
TS) =
Δ
A + T
Δ
S = -5100 + 298(140-20) =
30.66 kJ/mol
Δ
H =
Δ
U +
Δ(
PV), since the volume of the liquid is
negligible:
Δ
H =
Δ
U + R T
Δ
n = 30660 + (8.314)(298)(1) = 33.1376
≈
33.14 kJ/mol
If you were to account for the volume of liquid, V
l
= 60·10
-3
/ρ = 0.01796
L
∆H = ∆U + P(V
gas
- V
liquid
) = ∆U + P(nR T/P - 0.044776)·0.1
≈ 33.14 kJ/mol
Note: L·bar = 100 J = 0.1 kJ
∆Uº =
30.66 kJ/mol
∆Hº =
33.14 kJ/mol
2.
Calculate the boiling point of NKT -19 at P
ext
=
1 bar .
Δ
G =
Δ
H-T
Δ
S = 0 at P = 1 bar .
T =
Δ
H/
Δ
S =
276.1 K
T
boil
=
276.1 K
3.
What is the vapor pressure of the liquid?
Use the Clausius-Clapeyron equation. ln(P/1 bar) =
-Δ
H/R (1/270 - 1/276.1)
P =
0.7217 bar
P
vap
=
0.7217 bar
4.
Which of the following should be true of a reversible
process?
(Choose one)
1.
▢
P < P
ext
▢
P = P
ext
← This is the answer
▢
P > P
ext
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
5.
The piston then expands isothermally and reversibly until all the liquid is vaporized, removing
heat from the refrigerator . How much work is done by the expanding gas, and how much heat is
absorbed?
w = R T
Δ
n =
2.245 kJ
(Δ
n = 0.9), q ~ n·
Δ
H = 0.9· ∆H
=
29.83 kJ
(all or nothing)
w
done
=
2.020 kJ1
q
ads
=
29.83 kJ
6.
What is the vapor pressure of the coolant at T
H
?
Use the Clausius-Clapeyron equation. ln(P/1 bar) =
-Δ
H/R (1/330 - 1/276.1)
P = 10.57 bar
Alternatively ∆G = ∆H – T∆S = –6.426 kJ/mol, ∆G =
–RT lnK = –R(330) ln K
K =
P
vap
= 10.57 bar
P
vap
=
10.57 bar
7.
Calculate the work done during the compression
step, and the temperature after compression.
The initial pressure is .7217 bar , and the initial
volume is 31.81 L from our answer to part (3).
(.7217)(31.81)
1.3
= (10.57)V
1.3
V = 4.2275 L
T = PV/nR =
537.5 K
At such a high temperature, P
vap
>>> 10.57 bar .
We could integrate the adiabat, but this is hard,
and Alec hates calculus, so it seems impossible
(it’s not actually that hard though but takes a long
time so don't do it lol)
BUT WAIT!
Δ
U = q + w , but q = 0, so we only need to
find
Δ
U!
Δ
U = C
v
Δ
T, (C
v
+ R)/C
v
=
𝛾
= 1.33
C
v
= 24.94 J/K, so
Δ
U = w =
6.672 kJ
w
done
=
6.672 kJ
T
f
=
537.5 K
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
4
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
8.
Finally ,
the
gas
must
be
cooled
isobarically
from
X
to
Y
before
it
begins
condensing.
Calc-
ulate the heat removed from the gas and work done
on the gas during this step.
At P = 10.57 and T = T
H
= 330 K, V = 2.596 L.
w = -P
Δ
V = -(10.57)(2.596-4.2275) =
1.724 kJ/mol
q =
Δ
U - w = C
v
(330-537.5) - w =
-6.900 kJ/mol
q
rem
=
-6.900 kJ
w
done
=
1.724 kJ
9.
Compute
the
refrigerator
efficiencies
of
the
NKT -19
refrigerator
and
that
of
a
maximally
efficient heat pump refrigerator operating between
270 K and 330 K.
The heat transfer is simply equal to 0.90
Δ
H = 29.83
kJ
The work was found earlier to be 6.672 kJ.
Thus, ef ficiency = 29.83/6.672 =
447%
!!!! This is
why refrigerators (and heaters) are such
powerful devices: they can move more heat than they
use work.
Ideal ef ficiency = 1/(1-T
C
/T
H
) =
550%
.
η
NKT-19
=
447%
η
Ideal
=
550%
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
T h e o r e t i c a l # 2
1 .
2 .
3 .
4 .
5 .
6 .
7 .
8 .
9 .
1 0 .
1 1 .
1 2 .
1 3 .
T o t a l
1 5 % o f t h e t o t a l
6
2
2
2
1
1
2
1
1
2
3
6
9
3 8
P r o b l e m 2 . I n t e r e s t i n g O x i d a t i o n S t a t e s o f C o p p e r
1.
Determine the formula of
A
, and write the overall
reaction for the formation of the copper (III)
periodate complex
A
according to the procedure.
Ratio of Cu:I:O:H = 1:2:24:26
Leftover char ge: 1(3)+2(7)+24(–2)+26(1) = –5
Excess mass: 131 g/mol
Possible counterions are K, Na → 4Na, 1K
A
=
KNa
4
Cu(HIO
6
)
2
·12H
2
O or KNa
4
Cu(IO
4
)
2
O
3
·13H
2
O etc.
-
3 points for work
-
1 point for finding correct ratio of Cu,I,O,H
-
1 point for finding total mass of counterions
-
1 point for finding corresponding char ge
-
1 point for final formula
Reaction
K
+
+ 4Na
+
+ Cu
2+
+ 2HIO
6
4-
+ ½ S
2
O
8
2-
+ 11 H
2
O→
A
+ SO
4
2-
or
8NaIO
4
+ 2CuSO
4
+ K
2
S
2
O
8
+ 12 KOH + 20H
2
O → 2
A
+ 4K
2
SO
4
+ 4 KIO
4
-
2 points for a correct reaction
-
All or nothing
No penalty for:
-
Multiplying by 2,
-
Using IO
4
-
and OH
-
instead of HIO
4
-
-
Using any form of orthoperiodic acid
except
H
5
IO
6
.
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
6
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
2.
Draw the structure of the copper containing ions
in the complex and draw a d-orbital splitting
diagram for Cu (III) in its coordination environment.
Fill in the electrons and label the orbitals.
Structur e
-
1 point
Splitting Diagram
-
1 point
3.
Draw and fill the crystal field of
A
if (a) H
2
O
is a stronger field ligand than periodate and if
(b)
H
2
O is weaker . Be sure to label each orbital.
(a)
stronger
(either accepted)
-
1 point
(b) weaker
(either accepted)
-
1 point
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
4.
Draw
approximately
the
shape
of
the
cyclic
voltammetry
curve
(i
vs
E)
and
the
potential
waveform
(E
vs
time)
for
the
pH
=
13.3
trial.
Label
the
axes
of
the
potential
waveform
as
you
see fit.
Cyclic V oltammetry Graph
-
1 point
Potential W aveform
-
1 point
5.
Based on the experiment above which of the following
can the researchers conclude?
(Check all that apply)
▢
The reaction is reversible
▢
The reaction is quasi-reversible ← This answer
▢
The reaction is irreversible
▢
The reaction becomes more reversible at high pH
← This answer
▢
The reaction becomes less reversible at high pH
-
0.5 points for each correct answer
-
0.5 point penalty for each incorrect answer
6.
Which of the following best accounts for your answer
to question 5?
(Check only one)
▢
The reaction produces a Cu
2+
periodate complex which
is more stable than other possible products
▢
The reaction produces Cu(OH)
2
precipitate which
is more stable than other possible products
▢
The reaction produces a Cu
2+
periodate complex which
is more stable in acidic conditions and competes
with the precipitation of Cu(OH)
2
which is favorable
at highly basic pH
▢
The reaction produces a Cu
2+
periodate complex which
is more stable in basic conditions and competes
with the precipitation of Cu(OH)
2
which is favorable
at intermediate high pH. ← This answer
-
1 points for correct answer
-
1 point penalty for each incorrect answer
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
7.
Construct a mathematical equation for the calibration
curve.
Perform a linear regression.
C
0
= 0.1032
i
pa
- 0.2593, r
2
= 1.000
or
i
pa
= 9.690C
0
+ 2.513
(This plot is linear as we were told in the preceding
paragraph. For more details see
Randles-Sevcik equation)
-
2 points for a correct linear equation (either acceptable)
8.
Determine the concentration of copper (III) periodate
in the sample.
Plug into formula.
C
Cu
= 4.729 mM
-
1 points for correct answer
9.
Write the balanced equation for the reaction of
Cu
2+
and I
–
.
Either of
2Cu
2+
+ 4I
–
→ 2CuI + I
2
2Cu
2+
+ 5I
-
→ 2CuI + I
3
–
-
1 points for correct answer
-
Fractional coef ficients acceptable
-
I
3
–
acceptable for this equation
10.
Calculate standard reduction potentials of Cu
2+
and I
2
with r espect to the SCE
.
Simply add the potential of the reverse SCE equation,
i.e. subtract 0.25.
Eº(Cu) = -0.1 V, Eº(I2) = 0.29 V
E(Cu
2+
)/SCE =
-0.1 V
E(I
2
)/SCE =
0.29 V
-
1 points for correct answer
-
Each is all or nothing
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
11.
Calculate the voltage read by the indicator electrode in the beaker with 30 mmol added KI.
Overall reaction: Cu
2+
+ 2I
-
→ CuI + ½ I
2
We use the simplified reaction Cu
2+
+ I
-
→ Cu
+
+ ½
I
2
has Eº = -.39V , so K = 2.529*10
-7
, K/Ksp is a huge
number!
Thus, it is not hard to see that the reaction goes
to completion.
Thus,
past
the
endpoint
the
dominant
species
are
I
2
and
I
-
,
with
concentrations
of
5
mM
and
10
mM. These can be plugged directly into the Nernst
equation.
E
=
Eº(I
2
)
-
RT/nF
ln([I
-
]/[I
2
]
0.5
)
=
0.5902
→
0.340
V
vs.
SCE
(Note
that
if
the
student
forgets
to give with respect to SCE, only one point will be
deducted from problems 1 1, 12, and 13.)
E
30
=
0.340 V
-
2 points for setup
-
1 point for determining [I
-
] = C
I
- 2C
Cu
-
1 point for determining [I
2
] = 0.5 C
Cu
-
1 point for final answer
-
Each is all or nothing
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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|
M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
12.
Calculate the voltage read by the indicator electrode
in the beaker with 10 mmol added KI.
This
is
before
the
equivalence
point.
Since
the
reaction
goes
to
completion,
[I
2
]
can
be
considered to be ≈2.5 mM and [Cu
2+
] ≈ 5 mM.
[I
2
]
0.5
[Cu
+
]/([I
-
][Cu
2+
]) = 2.529E-7, hence [Cu
+
]/[I
-
]
= 2.529E-8
Since [Cu
+
][I
-
] = Ksp = 1.1E-16, [I
-
]
2
= 4.34955E-9
[I
-
] = 6.5951 1E-5 M
E = Eº(I
2
) - RT/nF ln([I
-
]/[I
2
]
0.5
) =
0.460 V
vs.
SCE
E
10
=
0.460 V
-
5 points for setup
-
1 point for assuming [I
2
] = 2.5 mM
-
1 point for assuming [Cu
2+
] = 5 mM
-
1 point for setting up a first equation with Cu
+
and
I
–
(i.e. K
sp
)
-
1 point for setting up a second equation with Cu
+
and I
–
-
1 point for setting up nernst
-
1 point for final answer
-
If [I
-
] = [Cu
+
] is assumed
-
If computation is done correctly (E = 0.69 V), 2 points
-
If computation is incorrect max score is 1 point
-
Each is all or nothing
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
13.
Calculate the voltage read by the indicator electrode
in the beaker with 20 mmol added KI.
Again, since the reaction goes to completion, assume
[I
2
] = 5 mM.
There
is
a
bunch
of
CuI,
and
we
don’t
know
exactly
how
much
there
is,
so
mass
balance
is
useless.
However ,
we
can
still
use
charge
balance.
The
NO
3
-
and
K
+
exactly
cancel
out,
so
we
have the following char ge balance:
[I
-
] = [Cu
+
] + 2[Cu
2+
]
[Cu
+
] = Ksp/[I
-
]
[Cu
2+
] = [I
2
]
0.5
[Cu
+
]/[I
-
]/2.529E-7 = 3.0759E-1 1/[I
-
]
2
Solve:
x = Ksp/x + 2*3.0759E-1 1/x
2
, assume the Ksp/x term
is negligible.
x = (2*3.0759E-1 1)
1/3
= 3.947E-4M << [I
2
] so our assumption worked.
E = Eº(I
2
) - RT/nF ln([I
-
]/[I
2
]
0.5
) =
0.423 V
E
20
=
0.423 V
-
8 points for setup
-
1 point for assuming
[I
2
] = 5 mM
-
2 points for char ge balance
[I
-
] = [Cu
+
] + 2[Cu
2+
]
-
1 point for using
[Cu
2+
] = [I
2
]
0.5
[Cu
+
]/[I
-
]/2.529E-7
-
1 point for
[Cu
+
] = Ksp/[I
-
]
-
2 point for setting up equation and computing [I
–
]
with method of choice
-
1 point for Nernst Equation
-
1 point for final answer
-
0 points overall if [I
-
] = [Cu
+
] is assumed
-
Each is all or nothing
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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| 14
|
M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
T h e o r e t i c a l # 3
1 .
2 .
3 .
4 .
5 .
6 .
7 .
8 .
T o t a l
8 % o f t h e t o t a l
1
2
3
5
1
7
8
2
2 9
P r o b l e m
3 .
W o l f r a m ,
钨
,
в о л ь ф р а м а ,
اﻟﺗﻧﻐﺳﺗن
1.
Draw the lewis structure of tungstate (WO
4
2-
).
-0.5 for no lone pairs
2.
Draw
the
structure
of
a
tungstate
polymer ,
and
write
a
generic
formula
in
terms
of
X
n
,
the
number of monomer units in the polymer (henceforth
referred to as the polymer length).
Anything reasonable will be accepted.
Formula = (
WO
3
)
Xn
O
2-
3.
Many
cyclic
oxyanions
are
known.
Draw
3
examples,
each
based
on
a
different
group
of
the
periodic table.
1
2
3
Similar structures are known
for vanadates.
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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|
M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
4.
Calculate the value of the equilibrium constant
K
eq
.
General formula of polymer is [(WO
3
)
Xn
O]
2-
X
n
= (2200-16)/(183.81+16*3) = 9.4219.
1pt
The number of polymer molecules = 100 mmol/X
n
= 10.614
mmol.
1pt
This
corresponds
to
21.228
mmol
of
terminal
groups
(since
each
polymer
molecule
has
two
terminal groups). Note that we start of f with 200
mmol of terminal groups.
This
means
that
200-21.228
=
178.772
mmol
of
terminal
groups
have
reacted
to
form
linkages,
so
there
are
178.772/2
=
89.386
mmol
of
linkages
(each
linkage
requires
two
terminal
groups).
1pt
And through this line of thinking, we get the mass
balance:
n(terminal) + 2n(linkage) = n(terminal)
i
= 200 mmol
K
eq
= [Linkages]/([T erminal]
2
[H
+
]
2
) = 1.984*10
10
2pt
We can also apply Carothers equation to get the same
answer once we have X
n
:
X
n
= 1/(1-p) where p is the degree of conversion which
is defined in this case as the number of
ends that have reacted to form linkages divided by
the initial total number of linkages.
X
n
= 9.4219 = 1/(1–p) → p = 0.8918
p = (n
i, terminal
- n
terminal
)/(n
i, terminal
) = (200
- n(terminal))/(200) = 0.8918
n(terminal) = 21.227 → [terminal] = 0.021227 M
from mass balance, [linkages] = 0.0893 M
K
eq
= [Linkages]/([T erminal]
2
[H
+
]
2
) = 1.984*10
10
K
eq
=
1.984·10
10
5.
Pick the statements that are true regarding reaction
(1).
(Check all that apply)
▢
Δ
H
298
> 0
▢
Δ
H
298
< 0
▢
Δ
G
298
> 0
▢
Δ
G
298
< 0
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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|
M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
6.
Compute the statistical average molecular mass
of tungstate species in this new system.
Note that the overall initial concentration of linkages
for 100 mmol added to 1 L of water is
0.1 M. We can thus set up the mass balance:
[terminal] + 2[linkage] = 2[terminal]
i
= 0.2 M
From K
eq
[linkage] = K
eq
[H
+
]
2
[terminal]
2
= 12 [terminal]
2
Plugging this into mass balance, we get a quadratic
which we can solve
24x
2
+ x – 0.2 = 0, where [terminal] = x
[terminal] = 0.07280 M → [linkage] = 0.0636 M
X
n
= C
i
(WO
4
2-
)/[polymers], [polymers] = [end]/2, X
n
= 2.74
Thus the final molecular mass is X
n
(183.8 + 3·16)
+ 16 =
652.81 g/mol
I’ll demonstrate the solution using Car others equation
as well just for completeness
Once we have [terminal] from the above methods, we
can apply Carothers:
X
n
= 1/(1-p) where p is the degree of conversion which
is defined in this case as the number of
ends that have reacted to form linkages divided by
the initial total number of linkages.
Thus p = (n
i, terminal
- n
terminal
)/(n
i, terminal
)
= (200 - 72.80)/(200) = 0.636
And so, X
n
= 1/(1-0.636) = 2.74 → MW
avg
=
652.81 g/mol
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
7.
Determine
the
weight
%
of
Tungsten
in
the
alloy
sample
and
write
balanced
chemical
equations for steps (1)-(5).
The white precipitate is WO
3
. 8.3 mg corresponds to
≈36 μ
mol of W in the 20 ml aliquot.
Scaling up from 15 ml to 20 ml aliquot, n(Cr
VI
) =
2* 20/15 *0.000765*0.01723 = 35.15 μ
mol.
The 5 ml aliquot contains the same precipitate as
before. Thus the ion exchange column
removes nickel, but preserves W
VI
.
Since there is a 1:1 redox ratio of W to Cr , lead
must reduce W
VI
to W
III
.
(Using the approximate error bounds for WO
3
precipitate
given confirms that the reduction
step produces W
III
and not W
IV
or W
II
etc.)
We obtain 35.15
μ
mol of W in the 20 ml aliquot, so
there is 70.30
μ
mol in the sample.
This corresponds to 12.92 mg of W, i.e.
4.74%.
1)
6NO
3
-
+ W + 6H
+
→ 6NO
2
+ WO
3
+ 3H
2
O (½ credit for forming
NO)
2)
WO
3
+ 2OH
-
→ WO
4
2-
+ H
2
O (HWO
4
-
is also acceptable)
3)
WO
4
2-
+ 2H
+
→ H
2
O + WO
3
4)
3Pb + 2WO
4
2-
+ 6Cl
-
+ 16H
+
→ 2W
3+
+ 3PbCl
2
+ 8H
2
O
(NH
4
+
is also fine, ½ if Pb
2+
)
5)
2W
3+
+ Cr
2
O
7
2-
→ W
2
O
7
2-
+ 2Cr
3+
(either tungstate
or ditungstate is fine)
8.
Propose a reason why the titration procedure is
less accurate without ammonium chloride.
Ammonium chloride is both an acid and a source of
chloride. The ammonium is a source of
acid, which makes dichromate a stronger reducing agent,
as well as producing H
2
. The chloride
precipitates the Pb
2+
formed, preventing it from interfering
with the titration with tungsten (if
for example it was oxidized).
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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| 18
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
T h e o r e t i c a l # 4
1 .
2 .
3 .
4 .
5 .
6 .
7 .
8 .
9 .
1 0 .
1 1 .
1 2 .
T o t a l
1 6 % o f t h e t o t a l
1
2
2
7
2
2
2
3
2
4
2
6
3 5
P r o b l e m 4 . Q u a n t u m K i n e t i c s
1.
Write
down
the
acid-base
reaction
between
HOAc
and
H
2
O.
Draw
structures
indicating
the
relevant bonds being broken and formed.
H-O
Ac + H
2
O → OAc
-
+
H-O
H
2
+
-
1 point
-
All or nothing
2.
Calculate,
in
amu,
the
reduced
mass
of
the
O-H
bond
and
the
O-D
bond
respectively .
You
should use integer molar masses: m(H) = 1 amu, m(O)
=16 amu.
μ(
O-H) = (1*16)/(1+16) =
.94118 amu
, μ(
O-D) = 2*16/(2+16)
=
1.7778 amu
µ(O-H) =
.94118 amu
µ(O-D) =
1.7778 amu
-
1 point each
-
Each is all or nothing
3.
Calculate the stretching frequency of the O-D bond
in DOAc and D
3
O
+
, respectively .
ṽ
is
proportional
to
sqrt(1/mu),
and
we
may
assume
force
constants
don’t
change
with
isotopic
substitution.
ṽ(O-D, DOAc) = ṽ(O-H, HOAc)*sqrt(.941 18/1.7778) =
2037.3 cm
-1
ṽ(O-D, D
3
O
+
) = ṽ(O-H, H
3
O
+
)*sqrt(.941 18/1.7778) =
2182.8 cm
-1
ṽO-D(D
3
O
+
) =
2182.8 cm
-1
ṽO-H (DOAc)=
2037.3 cm
-1
-
1 point each
-
Each is all or nothing
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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| 19
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
4.
Compute the approximate pK
a
of DOAc in D
2
O using
the harmonic oscillator model.
Δ
G
H
= -RTlnK
eq
= -RT/log(e)*logK
eq
= RT/log(e)*pK
a
=
27.0408 kJ/mol
The
difference
in
Gibbs
free
change
is
equal
to
the
change
in
ZPE
(this
can
be
proved
using
the general approach of description for the other
solution presented below this one)
ΔΔ
ṽ = [ṽ(O-D, D
3
O
+
) - ṽ(O-D, DOAc)] - [ṽ(O-H, H
3
O
+
)
- ṽ(O-H, HOAc)] = -54.5 cm
-1
ΔΔ
ZPE = ½ hcN
a
*ΔΔ
ṽ = -326.197 J/mol =
ΔΔ
G
Δ
G
D
= Δ
G
H
+ ΔΔ
G = 26.7146 kJ/mol
pK
a
= Δ
G*log(e)/R T =
4.6828
Another
acceptable
(and
more
intuitive/less
handwavey!)
solution
is
to
just
apply
K
=
k
f
/k
r
,
and
K
H
/K
D
=
KIE
f
/KIE
r
.
You
can’t
necessarily
assume
that
the
transition
states
of
the
protium
and
deuterium
reactions
are
the
same,
but
this
ends
up
not
being
an
issue
since
the
TS
terms
in
Ea
=
TS
–
ZPE
parts
of
KIE
f
cancel
with
the
TS
terms
in
the
Ea
=
TS
–ZPE
parts
of
KIE
r
!
(all
praise exponentials for having nice rules)
pK
a
(DOAc) =
4.6828
-
7 points total
-
2 points if ∆G = ∆ZPE is used (incorrect)
-
5 points for correct setup but incorrect computation
5.
Determine the rate law and rate constant for this
reaction.
First order in [H]
+
, first order in [
К],
zeroth order
in [Br
2
].
Rate = k[H
+
][К],
k = 1.1527*10
2
M
-1
h
-1
R =
k[H
+
][К]
k
=
1.1527*10
2
M
-1
h
-1
-
1 point each
-
Each is all or nothing
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
6.
Using
the
steady
state
approximation,
derive
the
rate
law
for
the
proposed
mechanism.
Under
what
conditions
does
it
match
the
experimental
rate
law?
What
step
of
the
mechanism
do
you
think is rate-limiting? Justify your answers.
Rate = k
3
[Br
2
][enol] = k
2
[H
К
+
]
k
1
[К][
H
+
] = k
2
[H
К
+
] + k
-1
[H
К
+
]
[H
К
+
] = k
1
[H
+
][К]/(
k
2
+k
-1
)
Rate =
k
2
k
1
[H
+
][К]/(
k
2
+k
-1
)
. This is actually valid
for all values of k
2
, k
1
, and k
-1
.
BUT WAIT!
A
proton
transfer
should
be
very
rapid
(most
proton
transfers
are
diffusion
limited),
whereas
the
second
step
of
the
mechanism
involves
breaking
a
C-H
bond.
Thus
it
is
likely
that
k
2
<<<
k
1
or k
-1
.
R =
k
2
k
1
[H
+
][К]/(
k
2
+k
-1
)
The rate law matches under
all conditions
Rate Limiting Step:
2nd step
Justification: A proton transfer should be very rapid
(most proton transfers are dif fusion
limited), whereas the second step of the mechanism
involves breaking a C-H bond. Thus it is
likely that
k
2
<<< k
1
or k
-1
.
-
1 point for rate law
-
1 point for justification
-
All or nothing
*Points should be awar ded for leaving the k
2
on the
denominator , however the logic given
above is needed to solve the r emaining parts.
7.
К
has
a
pK
b
of
approximately
21.
Estimate
the
value
of
the
rate
constant
of
the
rate
limiting
step.
k
eff
= k
2
k
1
/(k
2
+k
-1
). Since k
2
<<< k
-1
, this becomes
k
2
(k
1
/k
-1
) = k
2
K
eq
pK
eq
= pK
b
- pK
w
= 7. Thus, k
2
= k
eff
/10
-7
=
1.1527*10
9
M
-1
h
-1
k
rds
=
1.1527*10
9
M
-1
h
-1
-
1 point for setup
-
1 point for correct computation
-
2 points total
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
-
1 point only if pKb is used instead of pKa, producing
k = 1.1527*10
23
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
8.
The
reaction
К
+
H
+
→
H
К
+
has
Δ
H
~
30
kJ/mol.
If
the
rate
of
bromination
doubles
at
308
K,
maintaining
the
same
buffer
pH,
estimate
the
activation
energy
and
pre-exponential
factor
of
the
rate limiting step.
k
eff
(308 K)/k
eff
(298 K) = k
2
(308 K)*K
eq
(308 K)/[k
2
(298
K)*K
eq
(298 K)]
ln(K
eq
(308
K)/K
eq
(298
K))
=
-Δ
H/R*(1/308-1/298)
=
.393137,
K
eq
(308
K)/K
eq
(298
K)
=
1.48162
Hence k
2
(308 K)/k
2
(298 K) = 1.34987
ln(k
2
(308 K)/k
2
(298 K)) = -E
a
/R*(1/308-1/298)
E
a
= 22.894 kJ/mol
, k
2
= Ae^(-E
a
/RT) = 1.1527*10
9
M
-1
h
-1
at T = 298
A = 1.188*10
13
.
E
a
=
22.894 kJ/mol
A
=
1.188*10
9
-
1 point for setup for Ea
-
1 point for Ea value
-
1 point for A
-
ECF if k = 1.1527*10
23
is used, producing A = 1.188*10
27
-
Each is all or nothing
9.
Calculate
initial
rate
of
bromination
in
H
2
O
solution
containing
1M
HOAc,
0.1M
NaOAc,
0.001M
К,
and 0.001M Br
2
.
pH = pK
a
+ log(.1/10) = 3.74, [H
+
] = 1.8197*10
-4
Rate = k
eff
[H
+
][К] = 2.098*10
-5
M/h
-
1 point each for setup and final answer
-
Each is all or nothing
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
10.
Calculate
initial
rate
of
bromination
in
a
D
2
O
solution
containing
1M
DOAc,
0.1M
NaOAc
0.001M
К-
D, and 0.001M Br
2
. ṽ(C-H) = 2950 cm
-1
.
pH = pK
a
+ log(.1/10) = 3.6828, [D
+
] = 2.07587*10
-4
ṽ(C-D) = ṽ(C-H)*sqrt(.941 18/1.7778) = 2146.4
Assume -
Δ
E
a
is equal to change in ZPE
Δ
E
a
= -½ hc*N
a
Δ
ṽ = 4.810 kJ/mol
k
D
= k
H
* k
D
/k
H
= 1.1527*10
9
*e^(-4810/R T) = 1.654*10
8
Rate = k
D
*K
eq
[D
+
][К-
D] = 3.434*10
-6
M/h
R =
3.434*10
-6
M/h
-
3 points for setup (numerical results can be wrong
as long as set up correct)
-
1 point for computation of [D
+
]
-
1 point for computation of ṽ(C-D)
-
1 point for computation of E
a
-
1 point for final answer
-
Each is all or nothing
11.
Write the mechanism for this reaction. Note that
the step involving Cl
2
is rate limiting.
К-
H + OH
–
⇌
К
–
(enolate)+ H
2
O
К
–
+ Cl
2
→ К-
Cl + Cl
–
-
2 points total
-
All or nothing
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
12.
Calculate
the
concentrations
of
all
isotopic
variants
of
К
and
К-
Cl
35
/К-
Cl
37
after
3
hours,
ignoring species of negligible concentration.
The formation of the enolate is not the rate determining
step, and both are treated as having the
same pKa, so
К
and
К-
D behave kinetically the same
in this scenario!
We can express the relative amounts of
К-
Cl
35
and
К-
Cl
37
that form according to the relative
rates:
[К-
Cl
35
]/[К-
Cl
37
] = (1 + 0.5·1/1.07)/(0.5·1/1.07 +
1/1.14) = 1.0913
t
1/2
= 3400·5 = 17000 seconds in these conditions.
Note this half life accounts for the presence
of isotopically mixed Cl
2
. Thus after 3 hours, 3·3600/17000
= 0.6352 half lives have passed.
From half life
[К-
Cl
x
]
tot
= (1– 0.5
0.6352
) · 0.002 M = 7.12·10
-4
M
[К-
Cl
35
] = 1.0913/(1.0913+1) ·[К-
Cl
x
]
tot
=
3.71·10
–4
= [К-
Cl
35
]
[К-
Cl
37
] = 1/(1.0913+1) ·[К-
Cl
x
]
tot
=
3.405·10
–4
=
[К-
Cl
37
]
Returning to the [
К-
D] and [
К-
H] species:
Since they are both treated as having the same pKa,
and being much faster than the rate
determining attack on Cl
2
(i.e. a fast equilibrium),
we can assume that they equilibrate with
water quickly and adopt the general isotopic ratios
of natural water (2000:1).
[К-
H] + [
К-
D] = 0.5
0.6352
· 0.002 M = 1.2810
-3
M
[К-
D] = 6.43 · 10
-7
M
[К-
H] = 1.28 · 10
-3
M
[К-
Cl
35
] =
3.71·10
–4
[К-
H] =
1.28 · 10
-3
M
[К-
Cl
37
] =
3.41·10
–4
[К-
D] =
6.43 · 10
-7
M
(grade leniently here)
-
4 points for correct setup
-
1 point for
[К-
Cl
35
]/[К-
Cl
37
]
-
1 point for
[К-
Cl
x
]
tot
from half life
-
1 point for set up of computation of
[К-
Cl
35
]
-
1 point for observation that
[К-
H], [
К-
D]
adopt isotopic
ratios of natural water
-
(can be taken implicitly)
-
0.5 points for each final answer
-
Each is all or nothing
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
Theoretical #4
1 .
2 .
3 .
4 .
5 .
6 .
7 .
8 .
9 .
Total
16% of the total
1 4
1
2
3
2 2
2
4
2
1 0
60
P r o b l e m 5 . T h r e e i n O n e
1.
Give the structures of compounds
A-G
.
A
B
C
D
E
F
G
Use this area if you made a mistake, indicate the
letter
clearly
.
2.
What is the primary purpose of HNO
3
in step
A
→
B
:
▢
Oxidant
▢
Nitration Reagent
▢
Acid Catalyst
▢
Reductant
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
3.
Reagent
Z
1
is made by reaction of BH
3
in THF with
cycloocta-1,5-diene. Draw its structure.
4.
Draw ligand
L
H
2
, and provide the identity of reagent(s)
Z
2
.
LH
2
Reagent(s)
Z
2
NaH, then Allyl Bromide (any base/allylating agent
will do)
5.
Draw
structures
of
the
compounds
in
the
scheme,
and
identify
metal
M
.
Note:
Some
of
the
compounds ar e ionic, and you ar e not r equir ed to draw
any counterions that do not contain
M
.
M
Ruthenium
N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
O
[NH
4
]
3
[RuCl
6
], RuCl
6
3-
is octahedral
P
RuCl
3
Q
trig. Bipyr . also accepted
R
must be trig. bipyr!
S
T
[Ru(Bipy)
3
]Cl
2
U
V
W
Resonance structures are fine
Use this area if you made a mistake, indicate the
letter
clearly
.
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
6.
What
is
the
phosphorus-containing
byproduct
of
the
reaction
P
→
Q
?
Note
that
this
reaction
is
usually done in the presence of water .
Triphenylphosphine oxide
7.
Reaction
of
Q
with
ligand
L
H
2
gives
cationic
octahedral
complex
I
(contains
1
chlorine
ligand),
which
when
treated
with
a
suitable
reagent
gives
neutral
complex
II
.
Draw
the
structures
of
I
and
II
(feel
free
to
abbreviate
the
structure
of
L
as
something
similar
to
RN-N
～
N
～
N-NR,
but draw any hydrogens bonded to nitrogen!) and propose
a suitable reagent.
I and II
8.
Both
I
and
II
react
with
MeOH
in
the
presence
of
KN(SiMe
3
)
2
to
give
III
.
Draw
the
structure
of
III
; again feel free to abbreviate the structure
of
L
.
III
Use this ar ea if you made a mistake, indicate
the letter clearly .
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
9.
Draw
structures
of
complexes
IV-VII
;
again
feel
free
to
abbreviate
the
structure
of
X
;
you
do
not need to draw out the -R groups.
IV
V
VI
VII
Use this area if you made a mistake, indicate the
letter
clearly
.
Use this area if you made a mistake, indicate the
letter
clearly
.
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
T h e o r e t i c a l # 6
1 .
2 .
3 .
4 .
5 .
6 .
7 .
8 .
9 .
T o t a l
9 % o f t h e t o t a l
3
4
1 6
1
7
7
3
2
3
4 6
P r o b l e m 6 . A d a m a n t a n e , S y m m e t r y , a n d N M R
1.
How many hydrogen NMR signals are
theoretically
expected for adamantane according to its
molecular symmetry? Draw in and label at least 3 of
each chemically distinct hydrogen atom by
environment H
a
, H
b
, H
c
etc. (Hydrogen atoms in the
same environment should have the same
label). State the integration values expected for
each environment.
Number of Signals:
2 ( 1 point)
(3 of each environment must be shown) (1 point)
All CH
2
groups are the same (this can be seen by drawing
some symmetry axes)
All CH groups are the same (same reasoning)
H
a
H
b
H
c
H
d
H
e
H
f
Integration Values
12
4
(Ther e are not necessarily 6 distinct envir onments)
As long as integration for H
a
is 3 times lar ger than
that of H
b
, full points will be awar ded.
(1 point)
2.
Draw the structures of
A
and
B
.
A
B
-
2 points
-
2 points
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
3.
Draw
the
structures
of
intermediates
C
-
I
,
and
state
whether
each
step
Z
1
-Z
4
you
have
proposed
is
a
carbocation
rearrangement
or
intramolecular
hydride
shift.
There
may
be
more
than
one
unique
answer ,
so
all
reasonable
solutions
that
are
consistent
with
the
chosen
Z
1
-Z
4
will
be
accepted.
A
chemically
reasonable
answer
will
receive
full
points,
while
a
suspect
one
will
only
obtain partial.
Note:
you
may
abbreviate
carbocation
rearrangement
as
“CR”
and
intramolecular
hydride
shift
as “intraHT” for Z
1
- Z
4
.
C-I: 2 points max, 1 point for reasonable……. Z1-Z4:
1 point each
C
D
E
F
G
H
I
Z
1
CR
Z
2
intraHT
Z
3
CR
Z
4
intraHT
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
4.
Which solvent would 21st camper Harrison have used to carry out the reaction?
▢
Benzene
▢
Toluene
▢
THF
▢
Nitrobenzene
← This Answer
▢
Phenol
▢
Pyridine
▢
Pentane
(1 point)
5.
Based on the data given above, draw all products
formed in this reaction and their molar ratios
in the NMR sample.
Work
Possible Products
We can figure out the integration values from the
NMR data for each one to see the
dibrominated and tribrominated products are present
Answer:
Their ratio is
(dibr omo/tribr omo)
= 5.15:1
-
2 points for evaluating the possible products
-
2 points for using the NMR integration values to get
the correct ratio
-
1.5 points for each correct final structure
-
1.5 point penalty for each incorrect final structure
drawn
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
6.
Assign
every
NMR peak to a compound and the correct
proton on that compound. You may
use the letter labels from the table (i.e. label protons
H
a
, H
b
, etc.)
Protons i and k may be swapped for no penalty , same
with e and g.
Score = (# of correct protons)(7/13), round down to
nearest integer .
7.
How much Br
2
actually reacted with Adamantane?
Explain this result, with regard to the
amount (≈2 ml) of Br
2
added.
There are 14.68 mmol of adamantane. It takes 3 eq.
of Br
2
to make the tribromo adamantane,
and 2 eq. to make the dibromo version.
14.68[2(5.15/6.15) + 3(1/6.15)] = 31.45 mmol of Br
2
consumed.
However , about 39 mmol of Br
2
were added, which is
much more! The extra Br
2
was
consumed by reaction with Fe to form the necessary
FeBr
3
catalyst
in situ
.
-
2 points for determining the amount of Br
2
reacted
with adamantane
-
1 point for explanation involving reaction with Fe
to from FeBr
3
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
8.
Draw the structure of
I
with the correct orientation
of atoms/stereochemistry .
(2 points, all or nothing)
The computationally derived enthalpy of activation
∆H
†
for the conversion between the epimers
is 81 kJ/mol at 298 K.
9.
Estimate the entropy of activation ∆S
†
for the
conversion. Assume that conversion of the
transition state to the product occurs with 100% probability .
k = ln(2)/t
1/2
= 23.1 s
-1
= k
b
T/h e
-∆G‡/RT
∆G
‡
= 65.2 kJ/mol.
∆S
‡
= (∆H
‡
-∆G
‡
)/T =
53 J/mol*K
-
2 points for use of Eyring Equation
-
1 point for correct final answer
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
T h e o r e t i c a l # 7
1 .
2 .
3 .
T o t a l
1 5 % o f t h e t o t a l
2 7
6
3 0
6 3
P r o b l e m 7 . B u i l d y o u r o w n f r o g g e r
3 points each. 2 points if bad ster eochem (penalized
once), 1 point for the first r easonable
but incorr ect structur e, 0 points for further . 0 points
for “for cing” a solution.
1.
Draw the structures of intermediates
A
-
I
with stereochemistry
A
B
C
D
E
F
G
H
I
(no stereo needed on OH)
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
2/3.
Draw the structures of
J
and intermediates
K
-
T
with stereochemistry
(The answer boxes for these parts wer e combined to
save space)
J
4 pts for Z isomer
K
L
M
N
O
P
Q
Note: ambig ster eochem
R
Note: ambig ster eochem
S
T
(No stereo needed on OH)
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
T h e o r e t i c a l # 8
1 .
2 .
3 .
4 .
T o t a l
8 % o f t h e t o t a l
6
1 0
5
9
3 0
P r o b l e m 8 . N o n - C l a s s i c a l C h e m i s t r y
1.
Draw
the
structure
of
intermediates
of
X1
and
X2
,
which
are
intermediates
in
the
mechanism
for the synthesis of
A
.
X1
X2
2.
Draw structures of compounds
A
through
D1/2
.
A
B
C1
C2
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
D1
D2
3.
Propose reagent(s)
Y
(note that reagent(s)
Y
may
involve many steps).
Some examples:
1) Br
2
/
hv
2) Na 3) Pd, heat, -H
2
1) SeO
2
2) Pinacol Coupling 3) Corey-W inter Olefination
1) SeO
2
2) CH
2
PPh
3
3) Grubbs Cat.
Use your imagination!
4.
Draw
the
structures
of
reactive
intermediate
E
,
F
,
and
the
“classical”
and
“non-classical”
resonance structures
G1
and
G2
. Draw the dimer of
G
as well.
E
F
G1
G2
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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M o c k I n t e r n a t i o n a l C h e m i s t r y O l y m p i a d / 2 0 2 0
U S A - 2 1
G Dimer
T H E O R E T I C A L P R O B L E M S , O F F I C I A L E N G L I S H V E R S I O N
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CChO Local 2012 Solutions
Chinese Chemistry Olympiad
2012
3 Hours
Question 1 2 3 4 5 6 7 8 9 10 11 Total
Points 7 7 10 5 8 7 10 16 12 6 12 100
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CChO Local 2012 Solutions
Problem 1 [7%]
1-1 Aqueous ammonia of appropriate concentration is added dropwise to the aqueous solution of
zinc sulfate to an excess, and two main reactions occur . Briefly describe the experimental
phenomena and write the ion equation for the two-step main reaction.
The solution will first exhibit a white gelatinous precipitate of Zn(OH) 2 followed by
redissolution due to complexation to form Zn(NH 3 ) 4 2+
1)Zn 2+ + 2NH 3 + 2H 2 O → Zn(OH) 2 (s) + 2NH 4 +
2)Zn(OH) 2 + 2NH 4 + + 2NH 3 → [Zn(NH 3 ) 4 2+ ](aq) + 2H 2 O
1-2 The compound [Cu(pydc)(amp)]·3H 2 O has formula C 11 H 14 CuN 4 O 7 (pydc and amp are
organic ligands containing aromatic rings). Thermogravimetric analysis shows that the
compound thermally decomposes in two steps. The first weight loss peak accounts for a weight
loss of approximately 15%. The second decomposition occurs between 400 and 500 ºC leaving
behind a solid residue with a mass of 21% of the original compound mass. Determine the
following:
Relevant Paper:
https://www .researchgate.net/publication/221765752_Supramolecular_assembled_of_hexameric
_water_clusters_into_a_1D_chain_containing_H2O6_and_H2O4O2_stabilized_by_hydrogen_b
onding_in_a_copper_complex
(1) What causes the weight loss occurring during the first step?
The answer is likely loss of H 2 O. MW(C 11 H 14 CuN 4 O 7 ) = 377.55 g/mol.
0.15·377.55 = 56.63 which is in the realm of experimental error for 3·18 = 54 g/mol
Indeed, 3·18/377.55 = 14.3% ≈ 15%
(2) What is the solid residue remaining after further heating to 500ºC? Justify your answer .
377.55·0.21 = 79.29 g/mol which is close to MW(CuO) = 79.55 g/mol (within experimental
error)
Thus the solid residue is CuO. Note that the original question listed a mass % of 20.0 which
was the actual figure reported in the paper . This leads to a molar mass of 75.51 g/mol. As a
result the official answer says something like a mixture of CuO and Cu 2 O, but at high
temperature equilibrium I find this answer unlikely , so I modified the numbers.
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CChO Local 2012 Solutions
Problem 2 [7%]
A and X are two common non-metallic elements. The sum of their atomic numbers is 22, and the
sum of their valence layer electron counts is 10. Under certain conditions, AX , AX 3 (a common
Lewis acid), A 2 X 4 and A 4 X 4 can be generated.
A (s) + 3/2 X 2 (g) → AX 3 (g)
After being passed through a mercury electrical dischar ge, the following reactions can follow
AX 3 (g) → AX (g) + 2 X (g)
2Hg + 2 X (g) → Hg 2 X 2 (g)
AX (g) + AX 3 (g) → A 2 X 4 (g)
4 AX (g) → A 4 X 4 (s)
2-1 Determine the identities of A and X
A: B X: Cl
2-2 A 4 X 4 has 4 three-fold rotation axes, and there are 4 atoms around each A atom. Draw the
structural formula of A 4 X 4
2-3 Write the equation for the reaction of AX 3 with CH 3 MgBr in a 1:3 molar ratio.
All acceptable:
BCl 3 + 3CH 3 MgBr → B(CH 3 ) 3 + 3 MgBrCl
AX 3 + 3CH 3 MgBr → A(CH 3 ) 3 + 3 MgBrX
BCl 3 + 3CH 3 MgBr → B(CH 3 ) 3 + 3/2 MgBr 2 + 3/2 MgCl 2
2-4 Write the equation for the alcoholysis of A 2 X 4 by ethanol.
All acceptable:
B 2 Cl 4 + 4 C 2 H 5 OH → B 2 (OC 2 H 5 ) 4 + 4HCl
A 2 X 4 + 4 C 2 H 5 OH → A 2 (OC 2 H 5 ) 4 + 4HX
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CChO Local 2012 Solutions
Problem 3 [10%]
Aqueous solutions of CuSO 4 can react with K 2 C 2 O 4 to form a blue crystal of unknown
composition. The formula of the crystal was determined by the following experiments
(a) Weigh 0.2073 g of the sample, put it into an Erlenmeyer flask, add 40 mL of 2 mol L -1
H 2 SO 4 , and dissolve the sample with slight heat. Add 30 mL of water , heat to near boiling, titrate
with 0.02054 mol L -1 KMnO 4 solution to the end. This step requires 24.18 mL of titrant.
(b) Next, the solution is sufficiently heated until the color changes from lavender to blue. After
cooling, 2g of KI solid and an appropriate amount of Na 2 CO 3 were added. The solution turned
brown and a precipitate was formed. The solution was titrated with 0.04826 mol L -1 Na 2 S 2 O 3
solution, with starch indicator added near the end point, to the end point, consuming 12.69 mL.
3-1 Write the equation for the titration reaction occurring in step (a).
2MnO 4 - + 4 H 2 C 2 O 4 + 6H + → 2Mn 2+ + 10 CO 2 (g) + 8H 2 O *
3-2 Write the equation for the reaction causing the color change from lavender to blue in step (b).
4MnO 4 - + 12H + → Mn 2+ + 5O 2 (g) + 6H 2 O
3-3 Write the reaction occurring upon addition of KI in step (b). Also write the equation for the
Na 2 S 2 O 3 titration reaction.
KI reacts with Cu 2+ to form a white precipitate of CuI and a brown solution of I 3 -
2Cu 2+ + 5 I - → 2CuI (s) + I 3 -
3-4 Determine the chemical formula of the blue crystal by calculation (all coefficients in the
formula are integers)
n(C 2 O 4 2- ) = 0.02054 M·24.18 ml·5/2 = 1.241 mmol
n(Cu 2+ ) = 0.04826 M·12.69 ml = 0.6124 mmol
n(C 2 O 4 2- ) ≈ 2 n(Cu 2+ ) → Solid contains [Cu(C 2 O 4 ) 2 2- ]
Between Cu 2+ and K + for the counterion, K + is more likely (although you can test both).
So the crystal has the formula K 2 Cu(C 2 O 4 )· x H 2 O
If we try using the formula:
n(H 2 O) · 18 g/mol = 0.2073 - n(Cu 2+ )·MW(K 2 Cu(C 2 O 4 ) 2 )
n(H 2 O) · 18 g/mol = 0.2073 - 0.5·n(C 2 O 4 2- )·MW(K 2 Cu(C 2 O 4 ) 2 )
We get x = n(H 2 O)/n(Cu 2+ ) = 1.15 and 0.989 respectively , which tells us that the complex is a
monohydrate.
K 2 Cu(C 2 O 4 ) 2 ·H 2 O
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CChO Local 2012 Solutions
Problem 4 [5%]
Neutron diffraction experiments carried out in 1967 determined that there are only three ions in
the crystal structure of trans-[Co(en) 2 Cl 2 ]Cl·HCl·2H 2 O: X + , Cobalt-containing A + and Cl - . All
atoms in X + are coplanar . X + also has a center of symmetry and three mirror planes that are
perpendicular to each other . Note: en is an abbreviation for ethylenediamine.
Relevant Paper:
https://www .sciencedirect.com/science/article/abs/pii/0020165067800862?via%3Dihub
4-1 Draw the structural formula of the stereoisomers of A + and its stereoisomers
A + :
Stereoisomers
4-2 Draw the structural formula of X + .
X + = H 5 O 2 + (very technically this is not planar , but oh well)
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CChO Local 2012 Solutions
Problem 5 [8%]
A supramolecular crystal is obtained by mixing aqueous solutions of urea and oxalic acid. X-ray
diffraction experiments showed that the crystal belongs to the monoclinic system, with unit cell
parameters a = 505.8 pm, b = 1240 pm, c = 696.4 pm, β = 98.13°. The crystal’ s supramolecular
structure is caused by hydrogen bonding between adjacent molecules leading to a
two-dimensional lattice. The crystal has density D = 1.614 g·cm -3 .
5-1 Determine the ratio of oxalic acid molecules to urea molecules present in the crystal.
V = abc sin (β) = 4.32·10 8 pm
D = MW(formula unit)/V → MW(formula unit) = 420.25 g.mol
MW(H 2 C 2 O 4 ) = 90 g/mol, MW(OC(NH 2 ) 2 ) = 60 g/mol
From guess and check we get 4·60 + 2·90 = 420 g/mol
Thus the ratio of oxalic acid molecules to urea molecules is 1: 2
5-2 Using structural formulas draw the hydrogen bonding in a formula unit of the crystal
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CChO Local 2012 Solutions
Problem 6 [7%]
2,3-pyridinedicarboxylic acid, commonly known as quinolinic acid, is a central neurotoxin
related to Alzheimers and Parkinsons. At room temperature, quinolinic acid exists as a solid, and
upon heating at 185-190 ºC, CO 2 is released forming niacin.
6-1 In the solid state, quinolinic acid adopts its lowest energy configuration. Draw this
configuration (you do not need to draw lone pairs or irrelevant hydrogen atoms).
6-2 the pK a1 of quinolinic acid in aqueous solution is 2.41, write the equation for its first
ionization (draw organic molecules as structural formulas).
6-3 Draw the structure of niacin.
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CChO Local 2012 Solutions
Problem 7 [10%]
In aqueous solutions of boric acid with a total boron concentration of ≥ 0.4 mol L -1 , ions like
tetraborate (B 4 O 5 (OH) 4 2- ), pentaborate (which has charge –1), and two distinct kinds of triborate
ions with charges -1 and -2 can form. These polyborate ions are formed by condensation of
B(OH) 3 and B(OH) 4 - . The boron atoms in the structure are connected to form a ring by oxygens
in a B-O-B fashion.
Source Paper:
https://onlinelibrary .wiley .com/doi/10.1002/0471238961.0215181519130920.a01.pub2
7-1 In the above pentaborate ions, the chemical environment of all three coordinated boron
atoms is exactly the same. Draw the structural formula for pentaborate (you do not need to draw
lone pairs)
7-2 The figure on the right shows the
relationship between the existence form and
pH of the boric acid-borate system when the
total concentration of boron is 0.4 mol L -1 .
The numbered areas (labeled 1,2,3,4) between
the curves represent the fractional
composition of the 4 kinds of polyborate ions
at each pH. Determine the chemical formulas
of the polyborate ions 1-4.
1: B 5 O 6 (OH) 4 -
2: B 3 O 3 (OH) 4 -
3: B 4 O 5 (OH) 4 2-
4: B 3 O 3 (OH) 5 2-
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CChO Local 2012 Solutions
Problem 8 [16%]
The pourbaix diagram on the right shows the
relationship between the most stable species of
uranium over a range of cell potentials and pH in a
the carbonic acid-carbonate system (total carbonate
concentration 1.0 × 10 -2 mol L -1 ) using the standard
hydrogen electrode as the reference value.
For comparison, the dashed lines show the E-pH
relationship for H + /H 2 and O 2 /H 2 O pairs.
8-1 Calculate the concentrations of the main species in a carbonic acid-carbonate system at pH of
4.0 and 6.0 respectively
H 2 CO 3 : K a1 = 4.5·10 -7 , K a2 = 4.7·10 -11
Note that this first part has nothing to do with Uranium (it's just about the bicarbonate system)
C tot = 1.0 ·10 -2 M = [H 2 CO 3 ] + [HCO 3 - ] + [CO 3 2- ]
[HCO 3 - ] = K a1 ·[H 2 CO 3 ]/[H + ], [CO 3 2- ] = K a1 K a2 [H 2 CO 3 ]/[H + ] 2
At both pH values H 2 CO 3 dominates.
At pH = 4 , it is safe to assume [H 2 CO 3 ] ≈ 1.0 ·10 -2 M
It follows that [HCO 3 - ] = 4.5·10 -5 M , [CO 3 2- ] = 2.115·10 -9 M
At pH = 6 , [HCO 3 - ] and [H 2 CO 3 ] are the primary species, [CO 3 2- ] is comparatively negligible.
[HCO 3 - ]/[H 2 CO 3 ] = K a1 /[H + ] = 0.45 , so C tot = 1.45 [H 2 CO 3 ], so [H 2 CO 3 ] = 6.90·10 -3 M
It follows that [HCO 3 - ] =3.10·10 -3 M , [CO 3 2- ] = 1.46·10 -7 M
8-2 In the figure, a and b are two straight lines with pH = 4.4 and 6.1, respectively . Write the
equations for the transformation of uranium species corresponding to a and b, respectively .
Recall that H 2 CO 3 is the primary species at these pH values
a: UO 2 2+ + H 2 CO 3 → UO 2 CO 3 + 2H +
b: UO 2 CO 3 + H 2 CO 3 → UO 2 (CO 3 ) 2 2- + 2H +
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CChO Local 2012 Solutions
8-3 Write the half cell potential equations for the reactions corresponding to the straight lines c
and d, respectively , and explain the reason for the positive or negative slope.
The reaction for line c is
4UO 2 + + H 2 O + 2e - → U 4 O 9 + 2H +
The reaction for line d is
U 4 O 9 + 2H + + 2e - → 4UO 2 + H 2 O
The equations for the lines comes from the nernst equation for each half reaction. Above this
potential, the oxidation is spontaneous, while below it, the reverse reaction (reduction) is
spontaneous.
If all non H + species are at standard concentrations then the half
E = Eº – 0.0592/n log([H + ] m ) = Eº + m/n · 0.0592pH
c: E = Eº – 0.0592/2 log ([H + ] 2 ) = Eº + 0.0592 pH
The slope is positive because [H + ] is a product.
d: E = Eº – 0.0592/2 log([H + ] –2 ) = Eº – 0.0592 pH
The slope is negative because [H + ] is a reactant.
8-4 Write the reaction that occurs upon adding UCl 3 to a solution buffered at pH = 4.0
2UCl 3 + 4H 2 O → 2UO 2 + 6H + + 6Cl – + H 2
8-5 Can UO 2 (CO 3 ) 3 4- and U 4 O 9 (s) coexist between pH = 8-12? Can UO 2 (CO 3 ) 3 4- and UO 2 (s)
coexist? Justify your answers.
UO 2 (CO 3 ) 3 4- and U 4 O 9 (s) can coexist between these pH values because their stability regions
share a boundary (along which the two species can coexist).
UO 2 (CO 3 ) 3 4- and UO 2 (s) cannot coexist for the same reason.
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CChO Local 2012 Solutions
Problem 9 [12%]
The Knoevenagel reaction is a useful type of condensation reaction. As shown in the figure
below , diethyl malonate and benzaldehyde react in the presence of piperidine to form diethyl
2-benzylidenemalonate.
9-1 Draw the nucleophile in the reaction.
9-2 Briefly describe the role of piperidine
Piperidine is a base catalyst for the
aldol/Knoevenagel condensation. It can also
form an iminium ion with benzaldehyde
rendering it more electrophilic.
9-3 Compound A is a precursor to the anticonvulsant drug gabapentin D . Propose a synthesis of
A using 2 organic reagents.
Cyclohexanone + Malonic Acid
Gabapentin ( D ) is synthesized as follows:
9-4 Draw the structural formulas of B , C and D in the above scheme.
B
C
D
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CChO Local 2012 Solutions
Problem 10 [6%]
The taste of pepper is mainly derived from capsaicinoids. The synthetic route of capsaicin F is
shown below:
Draw the structural formulas of compounds A - F .
A
B
C
D
E
F
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CChO Local 2012 Solutions
Problem 11 [12%]
11-1 The oxymercuration-demercuration reaction of alkenes is similar to alkene bromination.
The main product formed when 4-penten-1-ol is reacted under the same conditions is B . Draw
the structural formulas of A and B .
A
B
11-2 Which of the following organic compounds are aromatic?
2 and 5
Compound A reacts in the following two steps to give compound D . Answer the following
questions:
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CChO Local 2012 Solutions
11-3 Write the formula for compound B
O 3
11-4 Circle the oxygen atom in C that came from A
11-5 Draw the structural formula of compound D .
D forms from an intermediate which under goes an aldol condensation
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THE CANADIAN CHEMISTRY CONTEST 202 2
PART A – MULTIPLE CHOICE QUESTIONS (60 minutes)
All contestants should attempt this part of the conte st before proceeding to Part B and/or Part C .
The only reference material allowed is the CIC/CCO Periodic Table provided . You must complete your answers online, directly in the TestInvite
program . Students may use a scientific calculator. No phones or communication device s are allowed.
1) Which of the following substances does not require the
following WHMIS symbol ( oxidizer )?
A) Cl2 B) O 2 C) NaClO D) Pb(NO 3)2 E) H 2
2) Which combination of atoms will form stable compounds with the
following molecular geometries: bent, see -saw, octahedral ?
A) sulfur and fluorine D) two of the combinations of atoms
B) xenon and fluorine E) all three combinations of atoms
C) bromine and fluorine
3) Consider the two species He and Li+. Which statement best describes the
comparison of their ionization energies and radii?
A) Li+ has a smaller radius and greater ionization energy than He
B) Li+ has a larger radius and lower ionization energy than He
C) He has a n equal radius and ionization energy compared to Li+
D) He has a smaller radius and lower ionization energy than Li+
E) Li+ has a smaller radius and lower ionization energy than He
4) A chemist combines 10.0 mL of 0.25 mol L-1 antimony (III) nitrate,
Sb(NO 3)3, with 10.0 mL of 0.35 mol L-1 sodium sulfide, Na 2S. What is the
theoretical yield of the precipitate ?
A) 0.40 g B) 0.42 g C) 0.60 g D) 0.85 g E) 3.6 g
5) 1. 0 L of 0.010 M NaOH is added to 0.50 L of water. What is the pH of the
solution?
A) 2.00 B) 2.18 C) 9.18 D) 11.82 E) 12.18
6) Chemistry students learn that , under many environmental conditions , the
behaviour of gases can be approximated using t he Ideal Gas La w.
However, in practice, many gases deviate somewhat from ideal gas
behaviour . Using your knowledge of int ermolecular forces, determine
which of the following gases should have the lowest pressure when all
other variable s remain constant .
A) Cl2 B) Xe C) CH 4 D) NH 3 E) CO
7) The graph represents the titration of 25.00 mL of 0.100 mol L-1 aqueous
HA with 0.100 mol L-1 aqueous sodium hydroxide .
Which of the following statements is false?
A) HA is not 100% ioniz ed in water .
B) At point A, other than H 2O and Na+, the major species are HA and A-.
C) Adding silver nitrate at point B will not affect the pH of the solution .
D) At point A, the pK a of HA is equal to the pH of the solution .
E) The acid dissociation constant of HA is less than one .
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CCC 202 2 page 2 of 4
8) If 2.38 g of a gas are released into an evacuated 2.00 L vessel at 22.5oC and
the pressure inside the container rises to 104.4 kPa, what is a possible
identity of the gas? The ideal gas constant is 8.314 kPa L mol-1 K-1; assume
ideal behaviour of the gas.
A) CO 2 B) N 2 C) F 2 D) Xe E) CH 4
9) If you prepared 0.25 mol L-1 aqueous solutions of each of three
electrolytes , which list of three would all have a pH > 7?
A) NH 4NO 3, NaOH, Na2CO 3 D) Na3PO 4, NaF, NaClO
B) Al(NO 3)3, NaCH 3COO, NaHCO 3 E) Na2S, Ca(NO 3)2, Fe(NO 3)3
C) NaNO 3, NaClO 4, Na2SO 4
10) Biodiesels are an alternative to fossil fuels. Unlike fossil fuels, biodiesels
are derived from recently grown plants that consumed greenhouse gases
while they were growing. A 0.435 g sample of C14H28O2 biodiesel fuel
combusts in a bomb calorimeter contain ing 500.0 g of water initially at
20.00oC. After the combustion, the final temperature of the water is
49.70oC. The calorimeter has a heat capacity of 390 J oC-1 and the specific
heat capacity of water is 4.184 J g-1 oC-1. What is the heat of combustion of
the biodiesel?
A) - 7.37 x 101 kJ mol-1 B) - 3.87 x 104 kJ mol-1 C) - 6.21 x 101 kJ mol-1
D) - 1.15 x 103 kJ mol-1 E) - 5.05 x 103 kJ mol-1
11) Many cars expose fuel to 1.5 0 L of atmospheric air every 2
revolutions. Assume atmospheric air contains 21.0 % oxygen gas by
volume. If the combustion reaction is standardized to SATP conditions and
uses an excess amount of pure octane (C 8H18), calculate the volume of
carbon dioxide produced in 30 seconds when the engine idles at 1500
revolutions per minute . Use the following balanced chemical reaction:
2 C 8H18 + 25 O 2 → 16 CO 2 + 18 H 2O
A) 360 L B) 263 L C) 184 L D) 118 L E) 75.6 L
12) How many chloride ions are present in 35 mL of a 0.70 mol L-1 solution of
CaCl 2 (aq)?
A) 0.025 B) 49 C) 1.5 x 1022 D) 3.0 x 1022 E) 6.0 x 1023
13) The ongoing COVID -19 pandemic has led to the development of an
antiviral drug compound ( A) which was approved for medical use in late
2021. The structure of compound A is shown below. How many carbon
atoms and hydrogen atoms are there in this compound?
A) 12 carbon a nd 18 hydrogen atoms
B) 12 carbon a nd 19 hydrogen atoms
C) 13 carbon and 18 hydrogen atoms
D) 13 carbon and 19 hydrogen atoms
E) 13 carbon a nd 20 hydrogen atoms
14) Which list of functional groups are all present in compound A?
A) alcohol, carboxylic acid , amide
B) amine , ether, carboxylic acid
C) alcohol, ketone, amine
D) ether , ketone , amide
E) ether, alcohol, ester
15) Two compounds structurally related to A were synthesized and subjected to
medical testing. These two new compounds were labelled B and C. In the
structure of c ompound B, the nitrogen atoms in compound A are replaced
by phosphorus atoms . In the of compound C, the oxygen atoms in
compound A are replaced by sulfur atoms. What is the difference in the
molecular weight of compounds B and C?
A) 29.4
B) 45.5
C) 61.5
D) 78.6
E) 94.8
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