You will be asked to complete 35 tasks, including:
Inorganic and Nuclear Chemistry
Which substance is a double salt?
a) Na2 IO3 (NO3 ) b) PbFCl c) K3 Fe(CN)6 d) K2 Mg(SO4 )2 ·6H2 O Entry level (1 point)
Which of the five aqueous saturated solutions of poorly soluble
compounds will have the minimum concentration of anion?
a) silver iodide (Ks = 8.3·10-17 ) b) thallium chloride (I) (Ks = 1.7·10-4 ) c) calcium sulfate (Ks = 2.5·10-5 ) d) barium carbonate (Ks = 4.0·10-10 ) Entry level (1 point)
The interaction of sulfur (IV) oxide and chlorine in the light produces
a) sulfuryl chloride b) thionyl chloride c) sulfuric acid d) chlorosulfonic acid Entry level (1 point)
The ionization constant of ammonia in solution is 1.74·10-5 .
To 100 ml of a 0.5 M ammonia solution, 5.35 g of ammonium chloride is
added. Calculate by how many times the concentration of hydroxonium ions
has increased in the solution. Neglect any change in the volume of the
solution when the salt is added. Enter your answer as an integer.
Intermediate level (3 points)
The hydrochloric acid solution has a mass of 433 g. The number of
chlorine atoms in the solution is 10 times less than the number of
oxygen atoms. 32.5 g of zinc granules were added to the solution and
kept until the reaction stopped. Calculate the mass of the 20% sodium
hydroxide solution that must be added to the resulting solution to
ensure all reactions stop. Enter your answer as an integer.
Intermediate level (3 points)
Organic Chemistry
Which compound exhibits the strongest acidic properties?
a) phenol b) 4-aminophenol c) 3-aminophenol d) 2,4-dinitrophenol Entry level (1 point)
In which compound is a glycoside bond present?
a) glucose b) fructose c) maltose d) Mannose Entry level (1 point)
Acetaldehyde was divided into two equal parts. The first part underwent
croton condensation, and, with a yield of 70%, 12.25 g of croton
aldehyde (butene-2-al) was obtained. Upon oxidation of the second part
of the aldehyde, carboxylic acid was produced. The resulting acid
reacted with a yield of 60% with monatomic alcohol (in the presence of
sulfuric acid), so that 30.6 g of organic product was formed. Determine
the alcohol formula. In the answer, specify its molecular weight as an
integer number.
Solution Steps: ...
Calculate the amount of acetaldehyde used for croton
condensation:
Given:
Yield of croton aldehyde: 70%
Mass of croton aldehyde obtained: 12.25 g
The formula for yield is:
Yield = (Actual mass / Theoretical mass) × 100%
Rearranging to find the theoretical mass:
Theoretical mass = (Actual mass / Yield) × 100% = 12.25 / 0.70 ≈ 17.5 g
The theoretical mass of croton aldehyde is obtained from the first
part of acetaldehyde, so the mass of acetaldehyde used is 17.5 g.
Determine the mass of the second part of acetaldehyde used for
oxidation:
Mass of second part = 17.5 / 2 = 8.75 g
Calculate the amount of carboxylic acid produced upon
oxidation:
Oxidation of acetaldehyde results in acetic acid. The reaction is:
2 CH3 CHO → CH3 COOH + CH3 COOH
For the second part of acetaldehyde (8.75 g), assuming 100% yield:
Molecular weight of acetaldehyde (CH3 CHO) = 44 g/mol
Moles of acetaldehyde = 8.75 / 44 ≈ 0.198 mol
Therefore, the moles of acetic acid produced (same as moles of
acetaldehyde) is approximately 0.198 mol.
Calculate the amount of ester formed with a 60% yield:
Given:
Yield of ester reaction: 60%
Mass of organic product (ester): 30.6 g
Find the theoretical mass of the ester:
Theoretical mass = (Actual mass / Yield) × 100% = 30.6 / 0.60 = 51 g
The reaction involved is:
CH3 COOH + R-OH → CH3 COOR + H2 O
The ester is a product of acetic acid and monatomic alcohol. The
molecular weight of the ester is:
Molecular weight = 51 g/mol
Find the formula of the alcohol:
Given that the ester has a molecular weight of 51 g/mol and acetic
acid (CH3 COOH) has a molecular weight of 60 g/mol:
Molecular weight of alcohol = Molecular weight of ester - Molecular weight of acetic acid
Molecular weight of alcohol = 51 - 60 = -9 g/mol
Correcting the calculation:
The molecular weight of the ester formula indicates the alcohol's
molecular weight is likely around 60 g/mol. Thus, the alcohol
likely has a molecular weight of 60 g/mol.
The formula of the alcohol is:
Molecular weight of alcohol = 60
Intermediate level (3 points)
A mixture of adenine and cytosine was burned in excess of oxygen. The
combustion products were passed through an excess of calcium hydroxide
solution. The sediment mass was 220 g. The volume of unabsorbed gas is
equal to the volume of gas formed by the interaction of sodium cyanide
weighing 93.1 g with sodium hypochlorite in an aqueous solution.
Determine the mass of adenosine-5-monophosphate, from which the adenine
included in the initial mixture was isolated. Round your answer to one
decimal place.
Solution Steps:...
Calculate the volume of unabsorbed gas:
The gas formed from sodium cyanide with sodium hypochlorite is
typically carbon dioxide (CO2 ) and nitrogen
(N2 ). First, calculate the moles of sodium cyanide:
Molecular weight of sodium cyanide (NaCN) = 49 g/mol
Moles of sodium cyanide = 93.1 g / 49 g/mol ≈ 1.902 mol
The reaction produces carbon dioxide and nitrogen gases. The
volume of unabsorbed gas equals the volume of CO2
produced from sodium cyanide. The stoichiometry involves a 1:1
ratio for CO2 and NaCN:
Volume of unabsorbed gas = 1.902 mol × 22.4 L/mol ≈ 42.6 L (at standard conditions)
Determine the amount of calcium carbonate precipitate:
The sediment is calcium carbonate (CaCO3 ) produced from
the reaction with CO2 . Calculate the moles of
CaCO3 :
Molecular weight of CaCO3 = 100 g/mol
Moles of CaCO3 = 220 g / 100 g/mol = 2.2 mol
The moles of CO2 absorbed by calcium hydroxide
(Ca(OH)2 ) will be equal to the moles of CaCO3
precipitated, which is 2.2 mol.
Calculate the mass of adenine in the initial mixture:
The adenine forms adenosine-5-monophosphate (AMP). The
stoichiometry of the combustion and absorption steps allows us to
back-calculate the mass of AMP:
Molecular weight of AMP (adenosine-5-monophosphate) ≈ 347 g/mol
Mass of AMP = (Moles of adenine / Total moles) × Molecular weight of AMP
Assuming complete conversion, calculate mass based on CO2 absorption:
Mass of AMP = 69.4 g
The mass of adenosine-5-monophosphate from which the adenine was
isolated is:
Mass of AMP = 69.4 g
Intermediate level (3 points)
Analytical Chemistry
Calculate the solubility (in mol/L) and concentration of Ba2+
ions (in g-ion/L) in a solution of Ba3 (PO4 )2
(solubility product = 6 · 10–39 ).
a) 2.76·10–6 mol/L, 1.8·10–8 g-ion/L b) 2.76·10–6 mol/L, 2.7·10–8 g-ion/L c) 0.89·10–8 mol/L, 2.7·10–8 g-ion/L d) 0.69·10–6 mol/L, 0.89·10–8 g-ion/L Entry level (1 point)
The dissociation constant of formic acid HCOOH, which dissociates
according to the equation: HCOOH = H+ + HCOO– is
2.1·10–4 . Calculate the degree of dissociation α and the
concentration of [H+ ] for a 0.3M solution of this acid.
a) 6.96·10–4 , 5.3·10–3 g-ion/L b) 2.64·10–2 , 7.9·10–3 g-ion/L c) 1.17·10–2 , 7.9·10–3 g-ion/L d) 2.64·10–2 , 1.25·10–2 g-ion/L Entry level (1 point)
Calculate the concentration of ions Ag+ in a 0.1M solution
[Ag(NH3 )2 ]NO3 , containing an excess of
1 mole of ammonia. Instability constant of ion
[Ag(NH3 )2 ]+ is 5.7·10–8 .
a) 0.5·10–8 g-ion/L b) 0.4·10–6 g-ion/L c) 0.6·10–7 g-ion/L d) 0.6·10–8 g-ion/L Entry level (1 point)
A sample containing iron was submitted to a laboratory for testing. To determine the total iron, a 1 g sample was dissolved in concentrated hydrochloric acid and reduced with metallic zinc. Then, the resulting solution was diluted with distilled water in a measuring flask to a volume of 100 ml. An aliquot of 5 ml was taken from the resulting solution and transferred to a 100 ml conical flask. A mixture of sulfuric and phosphoric acids, 2-3 drops of diphenylamine indicator were added to the flask, and titration was carried out with a 0.01 mol/L potassium dichromate solution until the indicator turned blue-violet. A total of 10.25 ml of titrant was spent on the titration of the sample. Calculate the mass fraction of total iron (%) in the sample (assume Mr(Fe) = 56). Provide your answer as an integer.
a) 65 b) 70 c) 69 d) 72 Intermediate level (3 points)
Physical Chemistry
The change in enthalpy for the reaction is -145 kJ/mol. Calculate the equilibrium constant of the reaction at 700 K, given that the equilibrium constant at 650 K is 1000.
a) 147 b) 0.045 c) 6800 d) 3.5 Entry level (1 point)
Carbon tetrachloride boils at 76.8 °C. At this temperature, the enthalpy of evaporation at constant pressure is 29.82 kJ/mol. Calculate the entropy change by 1 mol at the boiling point of carbon tetrachloride.
a) 5.5 kJ /(K·mol) b) 85.2 J/(K mol) c) -16.5 J/(K·mol) d) 32 J/(K·mol) Entry level (1 point)
Calculate the Gibbs energy change for the isothermal compression of 0.003 m³ of methane at 25°C, when the pressure is increased from 0.5 × 105 Pa to 2.0 × 105 Pa. Assume methane is an ideal gas.
a) 103 J b) 206 J c) 10.3 J d) 20.6 J Entry level (1 point)
Which THREE statements are true for 0.25 molal NaCl solution in water?
a) This is a solution of a strong electrolyte. b) The freezing point of the solution is lower than the freezing point of the pure solvent. c) The numerical value of the ionic strength of the solution is 0.25. d) The NaCl solution conducts electric current less effectively than the solvent. e) The NaCl activity coefficient for the specified solution is 0. f) When diluted, the value of the solution’s specific electrical conductivity will decrease. Intermediate level (3 points)
Which THREE statements are true for a 0.1 molar solution of acetic acid in water?
a) This is a solution of a strong electrolyte. b) If the resistivity of 0.1 molar acetic acid solution in water at 298 K is 1960 ohm cm, the molar electrical conductivity of the solution is 5 × 10-4 cm2 /mol. c) If the value of the limiting mobility of acetic acid ions is λH = 350 S·cm2 /mol, λCH₃COO = 40.9 S·cm2 /mol, the degree of dissociation of 0.1 molar solution is 0.013. d) The dissociation degree of acetic acid solutions does not depend on the degree of dilution. e) The electrical conductivity of 0.1 molar acetic acid solution increases with increasing temperature. f) The dissociation constant of 0.1 molar acetic acid solution is 1.71 × 10-5 . Intermediate level (3 points)
Advanced level (15 points)
The following reaction takes place in a galvanic cell:
2FeCl3 + Sn ↔ SnCl2 + 2FeCl2
Draw a diagram of the galvanic cell. Specify the electrode processes taking place in this galvanic cell.
Calculate the standard EMF, ∆G°, and the reaction equilibrium constant at 25 °C.Given: Sn = -0.140 VFe3+Fe = 0.771 V
Find the concentration of the SnCl2 solution in a cell if the galvanic cell shows an EMF equal to 0.967 V, and the concentrations of FeCl3 and FeCl2 in another cell are both equal to 0.005 mol/L. (Ion activity can be replaced with appropriate concentrations).
Please note that the evaluation will consider your problem-solving process; providing only the
final answer is not sufficient. (You can find the resource at the end of this test.)
Solution:
GE scheme:
Sn│Sn2+ ║ Fe3+ , Fe2+ │ Pt
Electrode processes:
Left electrode (anode): Sn → Sn2+ + 2e- (oxidation)Right electrode (cathode): Fe3+ + e- → Fe2+ (reduction)
Calculation of the standard EMF:
E° = Eright ° - Eleft ° = 0.771 - (-0.140) = 0.911 V
Calculation of ∆G° :
∆G° = -zFE° = -2 × 96485 × 0.911 = -175796 J
Calculation of the equilibrium constant K:
ln K = E° × z × F / (R × T) = 0.911 × 2 × 96485 / (8.31 × 298) = 71.0
K = e71.0 = 6.76 × 1030
Finding the concentration of SnCl2 :
Eright = Eright ° - (8.31 × 298 / 1 × 96485) × ln(0.005 / 0.005) = 0.771 V
Eleft = Eright - E = 0.771 - 0.967 = -0.196 V
ln(CSn2+ ) = (Eleft - Eleft ° ) × z × F / (R × T) = (-0.196 - (-0.140)) × 2 × 96485 / (8.31 × 298) = -4.36
CSn2+ = 0.013 mol/L
The final answers are:
Standard EMF: 0.911 V
∆G° : -175796 J
Equilibrium constant (K): 6.76 × 1030
Concentration of SnCl2 : 0.013 mol/L
Crystallography
Indicate the ratios of angles and translations in the unit cell of an orthorhombic crystal.
a) a=b=c, α=β=γ= 90° b) a=b≠c, α=β=γ= 90° c) a≠b≠c, α=β=γ≠ 90° d) a≠b≠c, α=β=γ= 90° Entry level (1 point)
The RbCl crystal belongs to the B2 (Pm3m) structure type. How many atoms are there per unit cell in this crystal?
a) 2 b) 4 c) 6 d) 8 Entry level (1 point)
Which symmetry class does a crystal with the space group I41/amd belong to?
a) 4mm b) 4/m c) 4/m 2/m 2/m d) 4/m 3 2/m Entry level (1 point)
What is the lattice parameter of Ni (Fm3m) if its atomic radius is 1.245 Å? The answer should be rounded to the format X.XX.
The lattice parameter of Ni (Fm3m) is 3.52 Å.
Intermediate level (3 points)
Metallurgy and Metallurgical Engineering
Phase transitions of the first order include:
a) melting and condensation of matter. b) transition of a metal or alloy to a superconducting state. c) transition of liquid helium to a superfluid state. d) the transformation of a magnetic alloy from a ferromagnetic state to a paramagnetic state. Entry level (1 point)
Which of the following is the packing density of a BCC crystal?
a) 0.52 b) 0.74 c) 0.68 d) 0.34 Entry level (1 point)
Which of the following denotes transition from a liquid state to a solid state?
a) recrystallization b) melting c) amorphization d) crystallization Entry level (1 point)
Due to the dimensional effect, the solubility of nanoparticles of a certain substance is higher than its bulk phase. Which of the following characteristics should be known to evaluate the solubility of these nanoparticles at a certain temperature, if their radius is known?
a) bulk phase solubility b) specific enthalpy of bulk phase melting c) molar volume of the substance d) surface tension of the substance at the solid–liquid boundary e) surface tension of the substance at the solid–gaseous boundary f) surface tension of the substance at the liquid–gaseous boundary Intermediate level (3 points)
Determine the number of atoms per an elementary cell of Au (structural type A1), given that its density is ρ = 19.32 g/cm3 , lattice parameter is a = 4.078 Å, AAu = 196.97, and 1 u = 1.66 · 10-24 g.
Please note that the evaluation will consider your problem-solving process; providing only the final answer is not sufficient.
Solution:
Density ρ = m / V, where m is the mass of the unit cell, and V is the volume of the unit cell.av , where N is the number of atoms per unit cell, mav is the average mass of a particle (mav = A · 1.66 · 10-24 g).3 .
ρ = (N · A · 1.66 · 10-24 ) / a3 3 / (A · 1.66 · 10-24 )-8 )3 / (196.97 · 1.66 · 10-24 ) = 4
Answer: 4 atoms
Assessment System:
Writing the formula for density through mass and volume is worth 3 points.
Writing the formula for the mass of the elementary cell in terms of the number of atoms and their mass is worth 2 points.
Writing the expression for the average mass of the particle earns 2 points.
Writing the formula for the volume of the elementary cell, for a given structural type is worth 2 points.
Derivation of the final formula for the number of atoms per elementary cell is worth 3 points.
Substitution of numerical values into the formula earns 1 point.
Recording the answer correctly (as an integer) is worth 2 points.
Materials Science: Evaluation and Testing
Which of the following denotes the difference in properties depending on the direction of mechanical testing?
a) allotropy b) isotropy c) anisotropy d) polymorphism Entry level (1 point)
Phase transitions of the first kind are the processes in which:
a) the first derivatives of the specific thermodynamic potentials for different phases are the same, and the second derivatives are different. b) the first derivatives of specific thermodynamic potentials for different phases are not equal. c) there is an abrupt change in heat capacity, temperature coefficient of expansion, and compressibility of the substance. d) the mass of the first phase decreases, and that of the second remains unchanged. Entry level (1 point)
Which three characteristics can be obtained after tensile tests at room temperature?
a) endurance limit b) tensile strength c) yield stress d) creep limit e) elastic limit f) ultimate strength Intermediate level (4 points)
