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Chemistry & Materials Science: Second-round sample problems
Part 1. Choose one correct answer. (Each question is worth 1
point.) Total: 26 points
1.1. What is the molecular geometry of H2O and BF3? А) angular
and trigonal planar B) linear and trigonal planar C) angular and
trigonal pyramidal D) linear and trigonal pyramidal
Answer: A
1.2. The reaction between dilute nitric acid and copper
produces: А) copper(II) nitrate, nitrogen and water B) copper(II)
nitrate and hydrogen C) copper(II) nitrate, nitric oxide and water
D) copper(II) nitrate, nitrogen dioxide and water
Answer: C
1.3. How many geometrical isomers does [Rh(NH3)3(Br)2Cl] have?
A) 2 B) 3 C) 4 D) 6
Answer: B
1.4. Which factor decreases the rate of hydrolysis of copper
(II) nitrate in a water solution? А) Heating B) Adding nitric acid
C) Diluting D) Increasing pressure
Answer: B
1.5. The standard electrode potential value Е0(Cu2+/Cu0) is
0.337V. What is the potential of a copper electrode (T = 298 K)
immersed in a 0.01m solution of copper (II) sulfate?
А) 0.337V B) 0.169V C) 0.278V D) 0V
Answer: C
1.6. The isotope 63Ni undergoes β-decay with a half-life of
about 100 years. From which substances and in which quantities will
the sample consist in 300 years if it initially comprised 8 μg of
metal 63Ni?
A) 0 μg 63Ni and 8 μg 63Cu B) 1 μg 63Ni and 7 μg 63Cu
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C) 2 μg 63Ni and 6 μg 63Cu D) 4 μg 63Ni and 4 μg 63Cu
Answer: B
1.7. How much does the internal energy of a closed system
change, in which the gas-phase reaction H2 + 0.5O2 = H2O takes
place at a constant volume and T = 298 K and 18g of water vapor are
formed? ΔrH0 (298 K) = - 241.8 kJ/mol
A) -241.8 kJ B) +241.8 kJ C) -240.6 kJ D) +240.6 kJ
Answer: C
1.8. At the same temperature and atmospheric pressure, the
entropy of benzene cooled to -5 oC will behave as follows during
solidification:
А) decrease B) increase C) stay the same D) have a peak
Answer: A
1.9. Which of the following symmetry elements can be found in
the crystal structures with three-dimensional translational
symmetry?
A) axes of symmetry of order 6 and lower B) 1-, 2-, 3-, 4- and
6-fold axes of symmetry C) axes of symmetry of any order D) 1-, 2-,
3-, 4- and 6-fold axes of proper rotation and axes of improper
rotation of any
order Answer: B
1.10. Consider a non-stoichiometric solid material – iron(II)
oxide (Fe1-xO, x > 0). How will increasing x affect the
concentrations of Fe2+ and Fe3+ ions in the crystal structure of
the material?
A) Fe2+ concentration will increase and the Fe3+ concentration
will decrease B) Fe2+ concentration will decrease and Fe3+
concentration will increase C) Fe2+ and Fe3+ concentrations will
change non-monotonically D) Fe2+ and Fe3+ concentrations will not
change
Answer: B
1.11. The ruby and corundum differ in: A) the crystal structure;
their chemical composition is the same. B) that the ruby has a
small Cr3+ impurity; both materials have the same crystal
structure. C) that the ruby has a small amount of Fe3+ and Fe2+
impurities; both materials have the same crystal structure. D) both
chemical composition and crystal structure.
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Answer: B
1.12. Why will not an aluminum pan rust? А) Aluminum cannot be
oxidized by oxygen; B) A thin protective layer of aluminum oxide
quickly forms on the surface of aluminum: the layer stops the
transport of molecular oxygen and further oxidation of the
material. C) A thin protective layer of aluminum oxide quickly
forms over the surface of aluminum. The layer has neither ionic nor
electronic conductivity; it thus stops further oxidation. D) A thin
protective layer of aluminum oxide forms quickly on the surface of
aluminum. This layer has no electronic conductivity; ion
conductivity remains high nevertheless.
Answer: C
1.13. A sample is a mechanical mixture of two crystalline
substances if: A) the components neither are mutually soluble in
the solid state nor react with each other to produce a new
compound; B) the components are mutually soluble and form a
homogeneous solid phase with one of the initial components
preserving its crystal lattice; C) there is no short-range or
long-range order in the structure; D) the diffraction pattern does
not contain any sharp diffraction maxima.
Answer: A
1.14. With environmental influence on which properties do the
Rehbinder and Joffe effects h? A) thermal properties B) chemical
properties C) strength properties D) electrical transport
properties
Answer: C
1.15. A 28 mm-long surface crack appeared in a titanium alloy
plate, leading to fracture. To which tensile strength had the plate
been exposed? Titanium alloy has a plane strain fracture toughness
of 55MPa·m1/2. Take the dimensionless correction function Y (l/w)
dependent on the relationship between the length l and width w of
the crack as well as the method of strain application to be 1.
A) 18.55MPa B) 185.5MPa C) 1.85MPa D) 55MPa
Answer: B
1.16. Which technique measures the greatest number of
reflections for the same single crystal placed the same way in the
same high-pressure diamond anvil cell? It is assumed that the
dynamic range of the detector makes it possible to record
reflections of any, even very weak, intensity. Planck’s constant h
= 6.63·10-34J·s; the speed of light in vacuum c = 3·108 m/s.
A) measurement using a laboratory source of Mo Kα radiation (λ =
0.71Å)
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B) measurement using a laboratory source of Cu Kα radiation (λ =
1.54Å) C) measurement using a synchrotron source of a radiation
energy of 10 keV D) measurement using a synchrotron source of a
radiation energy of 20 keV
Answer: C
1.17. A scientist studied a new electrolyte by impedance
spectroscopy. The sample was a pressed tablet of a diameter of 5 mm
with silver electrodes applied. The impedance spectra showing the
relationship between the imaginary (Im Z) and real (Re Z) parts of
the complex resistance (electrical impedance) in Ohm divided by 105
were measured at several temperatures (Т11, Т22, Т33, Т44, Т55,
T66, T7, see the figure below). The sample was kept at each
temperature until equilibrium was reached. At the temperature T0,
the electrical conductivity σ0 = 4.33 μS/cm corresponds to the
resistance R0 = 120 kOhm. Which conductivity value corresponds to
the resistance R5 at the temperature T5?
А) 4.3 μS/cm B) 1.8 μS/cm C) 10.5 μS/cm D) 4.3 mS/cm
Answer: B
1.18. How many chiral carbon atoms are there in
1,2-dimethylcyclopentane? A) 7 B) 5 C) 2 D) 8
Answer: C
1.19. In the sequence acetic acid - chloroacetic acid -
fluoroacetic acid - trifluoroacetic acid acidity:
A) increases. B) decreases. C) does not change.
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D) has no clear pattern. Answer: А
1.20. What can be obtained from bromoacetic acid in one step? A)
acetic anhydride B) acetonitrile C) glycine D) glycerol
Answer: C
1.21. An HCl solution was prepared by diluting 9.0ml of
concentrated HCl solution of the density r = 1.198 g/mL (C (HCl) =
13.14 M) to 1.0l with water. What is the molar concentration of the
solution?
A) 0.1183M B) 0.1M C) 0.12M D) 0.11M
Answer: C
22. Which of the sulfides has the greatest solubility in water
(mol/L)? A) HgS (KL = 1.4 × 10-45) B) Ag2S (KL = 7.2 × 10-50) C)
Sb2S3 (KL = 2.2 × 10-90) D) Bi2S3 (KL = 8.9 × 10-105)
Answer: B
1.23. Order the solutions by increasing pH: 1) 0.10m H3PO4 (Ka1
= 7.1·10-3, Ka2 = 6.2·10-8, Ka3 = 5.0·10-13). 2) 0.10m Na2S
(Ka1(H2S) = 1.0·10-7, Ka2(H2S) = 2.5·10-13). 3) 300 mL 0.1m
CH3COONa and 250mL 0.1m CH3COOH (Ka = 1.74·10-5). 4) 0.050m C2H5NH2
(Kb = 6.5·10-4) + 0.050m (CH3)3N (Kb = 6.5·10-5) in equal volumes.
5) 0.10m tartaric acid (Ka1 = 9.1·10-4, Ka2 = 4.3·10-5) + 0.10М
NaOH in equal volumes. A) 4 - 1 - 5 - 2 - 3 B) 1 - 5 - 3 - 4 - 2 C)
5 - 1 - 3 - 2 - 4 D) 3 - 1 - 2 - 5 - 4
Answer: B
1.24. Which is the correct order of glucose oxidation in
carbohydrate catabolism? A) glucose – pyruvate – Acetyl-CoA – CO2
B) glucose – Acetyl-CoA – pyruvate – CO2 C) glucose –
α-ketoglutarate – pyruvate – CO2 D) glucose – α-ketoglutarate
–Acetyl-CoA – CO2
Answer: A
1.25. Which organic reactants are used in DNA
polymerization?
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A) dNMPs B) NMPs C) dNTPs D) NTPs
Answer: C
1.26. Which is the isoelectronic point (pI) of the Lys-Asn-Glu
tripeptide? pKa1 (α-carboxyl group) pKa2 (α-amino group) pKa3 (side
chain group) Lys 2.18 8.95 10.53 Asn 2.02 8.80 Glu 2.19 9.67
4.25
A) 6.8 B) 3.9 C) 9.9 D) 6.6
Answer: D
Part 2. There is more than one correct answer. Two points are
awarded for three correct choices; one point, for two; zero points,
for none. For each incorrect answer, one point is deducted. The
total score cannot be below zero.
Total: 14 points
2.1. Which of the statements are true for a reversible
exothermic elementary reaction? А) The activation energies of the
forward and reverse reactions are equal since they pass
through the same activated complex. B) The activation energy of
the forward reaction is greater than that of the reverse
reaction. C) The activation energy of the forward reaction is
lower than that of the reverse
reaction. D) The enthalpy of the reaction can be calculated
based on the difference between the
activation energies. E) Adding a catalyst will accelerate both
the forward and reverse reactions and reduce
their activation barriers. F) After reaching a state of
equilibrium between the forward and reverse reactions, their
activation energies become equal. Enter the correct answers
separated by spaces (A B C).
Answer: C D E
2.2. In which oxidation states does chromium form salts? A) +1.
B) +2. C) +3.
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D) +4. E) +5. F) +6.
Enter the correct answers separated by spaces (A B C). Answer: B
C F
2.3. Which of the substances discolor bromine water? A) styrene
B) trans-hexene-2 C) cyclohexane D) benzene E) xylene F)
isoprene
Enter the correct answers separated by spaces (A B C). Answer: A
B F
2.4. Select all the factors that increase the solubility of
poorly soluble compounds. A) the absence of competitive chemical
reactions B) the introduction of an extraneous electrolyte to the
solution (salt effect) C) adding the same ion D) adding precipitant
excess with complexing properties E) adding solvent excess F)
decreasing the pH of the solution
Enter the correct answers separated by spaces (A B C). Answer: B
D F
2.5. Select all the crystals that have the same chemical
composition but different crystal structures.
А) graphene B) graphite C) sphalerite D) lonsdaleite E)
stishovite F) fianite
Enter the correct answers separated by spaces (A B C). Answer: A
B D
2.6. Which cofactors are present in the citric acid cycle (the
Krebs cycle)? A) NAD+ B) flavin adenine dinucleotide C) pyridoxal
phosphate D) coenzyme A E) thiamine pyrophosphate F) lipoate
Enter the correct answers separated by spaces (A B C).
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Answer: A B D 2.7. Aluminum lowers the melting point of
germanium. Which of the following characteristics
should be known to estimate the crystallization temperature of a
germanium melt with aluminum additions?
A) the melting temperature and heat of fusion for aluminum B)
the melting temperature and heat of fusion for germanium C) the
solubility of germanium in solid aluminum D) the solubility of
aluminum in solid germanium E) the molar ratio of aluminum or
germanium in the melt F) the densities and melting points of
aluminum and germanium
Answer: B D E
Part 3. Open-ended questions. Each question is worth from 5 to 8
points. Total: 60 points
3.1. A blast furnace is a metallurgical furnace used for
smelting to produce industrial metals, most commonly pig iron. The
process involves many reactions at different temperatures. One of
them is the reduction of Fe2O3 by CO:
Fe2O3 solid + 3CO gas = 2Fe solid. + 3CO2 gas. The
thermodynamical parameters are known for all the compounds in the
reaction:
Compound Fe2O3 solid CO gas Fe solid CO2 gas
DfH0298, kJ/mol -823 -111 ? -394
C0p, 298, J/mol × K 104 29 25 37 Calculate the thermal effect of
this reaction (kJ) at 500 K (∆rH0500) assuming that ∆rC0p does not
depend on the temperature. Write out a detailed solution. Enter
your answer as an integer (15 kJ/mol). Solution. Since metallic
iron is a simple substance, its standard enthalpy of formation at
298 K is zero. The standard enthalpy of the reaction at 298 K
is:
The change in heat capacity in this reaction is:
Thus, the standard enthalpy of the reaction at 500 K is:
Answer: -32 kJ/mol
Points awarded: A correct calculation of the standard enthalpy
of formation of metallic iron is worth 1 point. A correct
calculation of the standard enthalpy of reaction at 298 K is worth
2 points. A correct calculation of the change in heat capacity is
worth 2 points. A correct calculation of the standard enthalpy of
the reaction at 500 K is worth 3 points Total: 8 points.
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3.2. The percentage of chromium(III) in the sample was
determined by back titration. 50.00 ml of a standardized EDTA
solution (C(EDTA) = 0.1000 mol/L) was added to the sample weights;
an excess of EDTA was titrated with a standard solution of Cu2+
(C(Cu2+) = 0.1000 mol/L). The table shows the weights of the sample
and the volumes of the Cu2+ solution used in the titration of the
excess of EDTA.
1 2 3 4
m, g 1.3248 1.3303 1.3569 1.5837 V(Cu2+), mL 23.63 23.52 22.73
18.78
Calculate the percentage of chromium(III) in the sample and
conduct analysis for t(3, 0.95)=3.18. Write out a detailed
solution. Enter the answer as follows: the number (rounded to the
nearest hundredth) ± the confidence interval. Solution: The formula
for calculating the chromium(III) content in a sample by back
titration is as follows:
Finding the chromium(III) content obtained in each
experiment:
Number of the experiment 1 2 3 4
msample, g 1.3248 1.3303 1.3569 1.5837 V(Cu2+), mL 23.63 23.52
22.73 18.78 w(Cr3+), % 10.35 10.35 10.45 10.25
Mathematical statistics methods can be used to find the average
value and the standard error:
= 10.35 = 0.082
t(3, 0.95)=3.18 = (10.35 ± 0.13)%
Answer: (10.35 ± 0.13) %
Points awarded: The formula for calculating the chromium(III)
content is worth 1 point. A correct calculation of the
chromium(III) content in each of the four experiments is worth 1
point. Finding the average value of the chromium(III) content is
worth 1 point. Statistical analysis is worth 2 points. A correct
presentation of the results of the analysis is worth 1 point.
Total: 6 points.
3.3. A 100g mixture of benzene, pyridine, and pyrrole was
dissolved in a non-polar solvent and treated with metallic
potassium. At the same time, 5.6 liters of gas were released
(T=273.15K, P=100kPa). When burning the same amount of the initial
mixture in an excess of
w1
)( 2
-
-= å
nS
i ww
ntS
±w
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oxygen, 11.2 liters (T=273.15 K, P=100 kPa) of nitrogen formed.
Determine the content of pyridine in the mixture (in grams rounded
to the nearest tenth). Write out a detailed solution.
Solution. Potassium releases gas only when reacting with
pyrrole:
The amount of the hydrogen substance is 5.6/22.4 = 0.25 moles.
Therefore, the amount of pyrrole is 0.5 moles. When the mixture is
burned, water, carbon dioxide and molecular nitrogen are formed.
The amount of the nitrogen substance is 11.2/22.4 = 0.5 moles. 0.5
moles of pyrrole give 0.25 moles of N2. Thus, during the combustion
of pyridine 0.25 moles of nitrogen were formed The amount of
pyridine substance is 0.5 moles. The pyridine mass is
0.5*79=39.5g.
Answer: 39.5g
Points awarded: The chemical equation is worth 1 point. The
amount of pyrrole is worth 1 point. The total amount of the
nitrogen substance is worth 1 point. The amount of nitrogen formed
during the combustion of pyridine is worth 1 point. The pyridine
mass is worth 1 point. Total: 5 points.
3.4. A tripeptide consisting of the residues of Val, Ala and Tyr
was treated with Sangers’ reagent (2,4-dinitro-1-fluorobenzene),
followed by complete hydrolysis. Mass spectrometry of the obtained
mixture showed the presence of a molecular ion peak at m/z = 255.
Determine the N-terminal amino acid in the tripeptide. Provide a
detailed solution. Write down the reaction scheme and identify the
reaction mechanism. Solution. Sangers’ reagent is used in the
analysis of peptides to determine the N-terminal amino acid. The
peptide is treated with the reagent, and a nucleophilic aromatic
substitution (the reaction mechanism) of the fluorine atom by an
amino group of the N-terminal amino acid occurs. This process is
enhanced by the presence of two electron-withdrawing nitro groups
in the benzene ring of the reagent. The hydrolysis of the modified
peptide leads to a mixture of amino acids, one of which contains
the 2,4-dinitrobenzene fragment. Mass spectrometry of the obtained
mixture makes it possible to identify the modified amino acid.
Completing the task requires writing the schemes of reactions
between the amino acids and 2,4-dinitrofluorobenzene, calculating
the molecular weights of the products and comparing them to the
mass obtained by mass spectrometry. Calculations: it is possible to
calculate the molecular weight of the modified amino acid by the
structural formula given in the reaction scheme. There are,
however, two more ways: m =m (amino acid) + М (Sangers’ reagent) –m
(HF) or m (amino acid) =m –m (Sangers’ reagent) + M(HF). М (Val) =
117 g/mol, М = 117 + 186 – 20 = g/mol М (Tyr) = 181 g/mol, М = 181
+ 186 – 20 = 347 g/mol
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M (Ala) = 89 g/mol, М = 89 + 186 – 20 = 255 g/mol The derivative
of alanine has the mass m =255 g/mol, which matches the mass
spectrometry data. Thus, alanine (the amino acid) is the N-terminal
amino acid of the peptide. Reaction scheme:
Points awarded: The name of the reaction mechanism is worth 1
point. The calculations are worth 2 points. The name of the amino
acid is worth 1 point. The reaction scheme is worth 2 points.
Total: 6 points
3.5. To prepare 19.2g of pure Na2S2O3·5H2O a student was given
solid Na2S2O3·5H2O, containing 15 wt % water-insoluble impurities.
The procedure for purifying this compound by recrystallization
requires that a saturated (40oC) aqueous solution of sodium
thiosulfate is first prepared. The suspension is heated to 50°С and
filtered hot. The filtrate is cooled to 0 °C for crystallization.
The solubility of Na2S2O3 at 40°С is 102.6g per 100g of water; at
0°С, 52.5g per 100 g. What mass of contaminated Na2S2O3·5H2O and
water (g) should the student use for recrystallization? Write out a
detailed solution. Enter the answer rounded to the nearest
integer.
Solution. The molar masses of Na2S2O3 and Na2S2O3·5H2O are 158.1
and 248.1g/mol respectively. To obtain 19.2g of crystalline hydrate
at 0°C, the required saturated solution should contain
9.2/248.1·158.1 = 12.2g of anhydrous salt in the precipitate and xg
in the solution. At 50°C, all the salt is contained in the
solution. Assuming that the mass of the solution is constant, we
set the total mass of water at 50°C equal to that at 0°C:
(12.2+x)·100/102.6 = x·100/52.5. Solving the equation gives x =
12.8g. Therefore, (12.2+12.8)/158.1·248.1·1.15 = 45g of crystalline
hydrate are needed for recrystallization. The amount of water
required for recrystallization is 12.8·100/52.5 –
(12.2+12.8)/158.1·18·5 = 10 g.
Answer: 45g Na2S2O3·5H2O and 10g H2O
Points awarded: The recrystallization process: the mass of the
salt in the solution at an increased temperature; the mass of the
salt in the solution and precipitate at low temperature: 1 point.
An equation for the mass of water is worth 2 points. Solving the
equation is worth 1 point. Calculating the mass of the crystalline
hydrate required for recrystallization is worth 1 point.
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Calculating the mass of the water required for recrystallization
is worth 1 point Total: 6 points.
3.6. A metal has a body-centered cubic crystal structure with a
unit-cell parameter of a Å(Angstrom). Calculate this unit-cell
parameter in Å (Angstrom) and round it to the next hundredth (2.89
Å). The atomic radius is 1.45 Å. Write out a detailed solution.
Solution. In a body-centered cubic unit cell, atoms are located
at the corners and in the center of the cube. The center and corner
atoms touch each other along the cube diagonals, so the length of
the cube diagonal equals four atomic radii (4r). The length of the
face diagonal of the cube, d, can be found by applying the
Pythagorean Theorem to the right-angled triangle formed by the two
edges, a, and the face diagonal, d, of the cube:
a2 + a2 = d2,
then d = a. Let us consider the right triangle formed by a cube
edge, the face diagonal and the cube diagonal:
By applying the Pythagorean Theorem to this triangle, we
obtain:
a2 + ( a)2 = (4r)2 3a2 = 16r2 a = r
Using the numerical values, we obtain: a = · 1.45 ≈ 3.35
(Å).
Answer: 3.35 Å Points awarded: Finding the direction in which
the atoms touch each other in the structure is worth 1 point.
Choosing a correct right triangle for the calculation of the
parameter a is worth 1 point. Applying the Pythagorean Theorem
correctly to the chosen right triangle is worth 1 point. The
formula for calculating the parameter a is worth 1 point. A correct
calculation of the parameter a is worth 1 point. Total: 5
points.
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3.7. By how many times will the intensity of Mo Kα1 X-ray
radiation (λ = 0.70930Å) weaken after penetrating an Al2O3 sample
to a depth of 100 microns? The density of Al2O3 is 3.99g/cm3. The
mass absorption coefficients for Al and O are 5.16cm2/g and
1.31cm2/g respectively. The atomic masses of Al and O are 27g/mol
and 16g/mol. Round the answer to the nearest tenth. Write out a
detailed solution.
Solution. When passing through a medium X-ray radiation is
weakened according to the law: ,
where I is the intensity of the radiation after passing through
the medium; I0 is the intensity of the initial radiation; µl is the
linear absorption coefficient of the medium; x is the depth of the
radiation penetration into the medium. The relationship between the
linear absorption coefficient of the medium and its mass absorption
coefficient is:
, where µl is the linear absorption coefficient of the medium;
µm is the mass absorption coefficient of the medium; is the density
of the medium. If the medium through which the radiation passes is
an individual substance, the mass absorption coefficient is defined
as the sum of the mass absorption coefficients of the elements
composing the substance. The weights in this sum are the mass
fractions of corresponding elements.
, where µm is the mass absorption coefficient of the substance;
(µm)i is the mass absorption coefficient of the ith element; is the
mass fraction of the ith element in the substance; mi is the sum of
masses of the atoms of the ith element in one mole of the
substance; m is the molar mass of the substance. These formulas
make it possible to calculate how much the intensity of X-ray
radiation will be weakened if it penetrates an Al2O3 sample to a
depth of 100 microns. First, the mass fractions of the elements in
Al2O3 should be calculated: ωAl = = = 0.529.
ωO = 1 – ωAl = 1 – 0.529 = 0.471. Next, the mass absorption
coefficient for Al2O3 is computed: µm (Al2O3) = µm,Al · ωAl + µm,O
· ωO = 5.16 · 0.529 + 1.31 · 0.471 = 3.347cm2/g. Thus, the linear
absorption coefficient for Al2O3 is:
(Al2O3) = ρ (Al2O3) · µm (Al2O3) = 3.99 · 3.347 = 13.35cm-1.
Finally, the ratio of the intensity of initial radiation and the
intensity of radiation after penetrating Al2O3 to a depth of 100
micron (10-2cm, since 1 micron = 10-6m = 10-4cm) is calculated:
= exp ( (Al2O3) · x) = exp (13.35 · 10-2) = 1.1. Thus, the
intensity of X-ray radiation will be weakened 1.1-fold after
penetrating Al2O3 to a depth of 100 microns.
Answer: 1.1
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Points awarded: Calculating the mass fractions of the elements
in the substance is worth 1 point. Calculating the mass absorption
coefficient correctly is worth 1 point. Calculating the linear
absorption coefficient is correctly is worth 1 point. Obtaining a
correct expression for the weakening of X-ray intensity after
passing through the medium is worth 1 point. The ratio of the
intensities is calculated correctly and the right conclusion is
drawn: 2 points. Total: 6 points.
3.8. Complete the chemical equations with reaction products and
stoichiometric coefficients. 1. KMnO4 + H2O2 + H2SO4 (diluted) =
... 2. CrCl3 (aq.) + Na2CO3 + H2O = ... 3. Cu + H2SO4
(concentrated) = … (heating) 4. SO3 + SCl2 = ... 5. NO2 + Ca(OH)2 =
... 6. H3PO2 + Ba(OH)2 (excess) = … Solution. Chemical equations:
1. 2KMnO4 + 5H2O2 + 3H2SO4 (diluted) = K2SO4 + 2MnSO4 + 5O2 + 8H2O.
2. 2CrCl3 + 3Na2CO3 + 3H2O = 2Cr(OH)3 + 3CO2 +6NaCl. 3. Cu + 2H2SO4
(concentrated) = CuSO4 + SO2 + 2H2O (heating). 4. SO3 + SCl2 = SO2
+ SOCl2. 5. 4NO2 + 2Ca(OH)2 = Ca(NO3)2 + Ca(NO2)2 + 2H2O. 6. 2H3PO2
+ Ba(OH)2 (excess) = Ba(H2PO2)2 + 2H2O.
Points awarded: Correct chemical equations with all the reaction
products are worth 0.5 points; correct stoichiometric coefficients
in the reactions are worth 0.5 points. The sum of the points
awarded should be rounded to an integer.
3.9. A cylindrical rod made of an alloy was exposed to reversed
tension-compression cyclic load along its axis. The maximum tensile
and compressive loads were 6000 N and -6000 N respectively. The
stress amplitude was 150 MPa. Calculate the diameter of the rod in
mm, rounded to the nearest tenth. Write out a detailed
solution.
Solution. The relationship between the amplitude σa, the maximum
tensile stress σtens and the compressive stress σcomp is as
follows:
σa = 0.5 · (σtens - σcomp) The relationships between the maximum
tensile and compressive loads (Ftens and Fcomp respectively) and
the maximum tensile and compressive stresses are as follows:
σtens =
σcomp = , where r is the radius of the rod. Combining the
equations gives:
-
DEMO VERSION OF THE SECOND ROUND TASKS
15
σa = 0.5 · ( - ) = · (Ftens – Fcomp)
r =
Ftens = 6000 N, Fcomp = -6000 N, σa = 150 MPa = 150·106 Pa,
thus:
r = = 0.0036 (m).
The diameter of the rod is therefore d = 2r = 2 · 0.0036m =
0.0072m, or 7.2mm. Answer: 7.2mm
Points awarded: A correct expression linking the stress
amplitude with the maximum tensile and compressive stresses is
worth 1 point. A correct expression linking the tensile and
compressive stresses, loads and the radius of the rod is worth 1
point. Correct formulae for calculating radius are worth 2 points.
Calculating the radius of the rod correctly is worth 1 point.
Calculating the diameter of the rod is worth 1 point. Total: 6
points
3.10. Using Vegard’s law, calculate the value of the unit cell
parameter (in Å) for the four-component solid solution AxByCD1-x-y.
x = 0.2, y = 0.4 and the unit cell parameters of binary compounds
composing the solution are aCA = 6.1 Å, aCB = 7.4 Å, aCD = 4.2 Å.
Enter the answer rounded to the nearest tenth. Write out a detailed
solution.
Solution. The solid solution AxByCD1-x-y consists of the
following binary compounds: CA, CB and CD. According to the
Vegard’s law, the unit cell parameter for the four-component solid
solution a (x, y) can be expressed as the sum of unit cell
parameters of constituent binary compounds weighted with the
corresponding stoichiometric coefficients of the separate
components aCA, aCB, and aCD:
a (x, y) = xaCA + yaCB + (1-x-y)aCD Using the numerical values
of the parameters, we obtain: a = 0.2 · 6.1 + 0.4 · 7.4 + (1 - 0.2
- 0.4) · 4.2 = 5.9(Å).
Answer: 5.9Å
Points awarded: A correct expression corresponding to the
Vegard’s law is worth 4 points. A correct calculation of the unit
cell parameter is worth 2 points Total: 6 points