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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:

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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