N heptane pt t x1 cl2 light. Reactions confirming the relationship between various classes of inorganic substances. Typical errors in structural formulas

Pt/MOR/Al2O3 catalysts containing mordenite zeolite from 10 to 50 wt.% were prepared. Solutions of H2PtCl6 and Cl2 were used as a Pt precursor. Translucent method electron microscopy It was shown that the localization of platinum on the MOR/Al2O3 mixed support directly depends on the nature of the metal precursor. The catalysts were tested in the isomerization reaction of n-heptane. It has been shown that the best samples of catalysts provide the yield of target products - di- and trimethyl-substituted heptane isomers at a level of 21 wt.% at a temperature of 280 °C and the yield of stable C5+ catalyzate at a level of 79–82 wt.%. Catalysts can be used to improve environmental characteristics gasolines by using them in the process of isomerization of the 70–105 °C fraction of straight-run gasoline.

About the authors

M. D. Smolikov

Omsk State Technical University
Russia

V. A. Shkurenok

Institute for Problems of Hydrocarbon Processing SB RAS, Omsk
Russia

S. S. Yablokova

Institute for Problems of Hydrocarbon Processing SB RAS, Omsk
Russia

D. I. Kiryanov

Institute for Problems of Hydrocarbon Processing SB RAS, Omsk
Russia

E. A. Belopukhov

Institute for Problems of Hydrocarbon Processing SB RAS, Omsk
Russia

V. I. Zaikovsky

Russia
Institute of Catalysis named after. G.K. Boreskov SB RAS, Novosibirsk

A. S. Bely

Russia

Bibliography

1. Technical regulations Customs Union TR CU 013/2011 “On requirements for automobile and aviation gasoline, diesel and marine fuel, fuel for jet engines and fuel oil." Bely A.S., Smolikov M.D., Kiryanov D.I., Udras I.E. // Russian chemical journal. 2007. T. L1. No. 4. pp. 38-47.

2. Sitdikova A.V., Kovin A.S., Rakhimov M.N. // Oil refining and petrochemistry. 2009. No. 6. P. 3-11.

3. Pat. RF No. 2408659 dated July 20, 2009. Method of isomerization of light gasoline fractions containing C7-C8 paraffin hydrocarbons / Shakun A.N., Fedorova M.L. Zhorov Yu.M. Thermodynamics of chemical processes. M.: Chemistry, 1985.

4. Liu P., Zhang X., Yao Y., Wang J. // Applied Catalysis A: General. 2009. Vol. 371. P. 142-147.

5. Corma A., Serra J.M., Chica A. // Catalysis Today. 2003. Vol. 81. P. 495-506.

6. Nie Y., Shang S., Xu X., Hua W., Yue Y., Gao Z. // Applied Catalysis A: General. 2012. Vol. 433-434. P. 69-74.

7. Belopukhov E.A., Bely A.S., Smolikov M.D., Kiryanov D.I., Gulyaeva T.I. // Catalysis in industry.

8. No. 3. P. 37-43.

Unified State Exam in Chemistry

Analysis of results
solutions part 2

1. The OVR equations are given in an implicit (incomplete) form and
it is necessary to determine the substances missing in the scheme.
2. Typically three components enter into ORR reactions:
reducing agent, oxidizing agent and medium (in the same
sequences and are recorded).
3. If there is a medium, then there will definitely be water (acid →
water, alkali → water, water → alkali or alkali + water).
4. Ions are determined by the medium.
5. It is often necessary to know the existence of ions in different
media (Mn, Cr).
6. The most common reactions are the following
elements: S, Mn, Hal, N, Cr, P, C (in organic compounds).

Typical reducing agents

Neutral atoms and molecules: Al, Zn, Cr, Fe, H, C,
LiAlH4, H2, NH3, etc.
Negatively charged non-metal ions:
S2–, I–, Br–, Cl–, etc.
Positively charged metal ions in
lowest oxidation state: Cr2+, Fe2+, Cu+, etc.
Complex ions and molecules containing atoms in
state of intermediate oxidation state: SO32–,
NO2–, CrO2–, CO, SO2, NO, P4O6, C2H5OH, CH3CHO,
HCOOH, H2C2O4, C6H12O6, etc.
Electric current at the cathode.

Typical oxidizing agents

Neutral molecules: F2, Cl2, Br2, O2, O3, S, H2O2 and
etc.
Positively charged metal ions and
hydrogen: Cr3+, Fe3+, Cu2+, Ag+, H+, etc.
Complex molecules and ions containing atoms
metal able highest degree oxidation:
KMnO4, Na2Cr2O7, Na2CrO4, CuO, Ag2O, MnO2, CrO3,
PbO2, Pb4+, Sn4+, etc.
Complex ions and molecules containing atoms
nonmetal in a state of positive degree
oxidation: NO3–, HNO3, H2SO4(conc.), SO3, KClO3,
KClO, Ca(ClO)Cl, etc.
Electric current at the anode.

Wednesday

Acidic: H2SO4, less often HCl and
HNO3
Alkaline: NaOH or KOH
Neutral: H2O

Half-reactions of Mn and Cr

acidic environment: MnO4– + 8H+ + 5ē → Mn2+ + 4H2O
Mn+7 + 5ē → Mn+2
alkaline environment: MnO4– + ē → MnO42–
Mn+7 + ē → Mn+6
neutral medium: MnO4– + 2H2O + 3ē → MnO2 + 4OH–
Mn+7 + 3ē → Mn+4
acidic environment: Cr2O72– + 14H+ + 6ē → 2Cr3+ + 7H2O
2Cr+6 + 6ē → 2Cr+3
alkaline environment: Cr3+ + 8OH– – 3ē → CrO42+ + 4H2O
Cr+3 – 3ē → Cr+6

The most famous half-reactions of the reduction of oxidizing agents

O2 + 4ē → 2O−2;
O3 + 6ē → 3O−2;
F2 + 2ē → 2F−;
Cl2 + 2ē → 2Cl−;
S+6 + 2ē → S+4 (H2SO4 → SO2);
N+5 + ē → N+4 (concentrated HNO3 → NO2);
N+5 + 3ē → N+2 (diluted HNO3 → NO;
reactions with weak reducing agents);
N+5 + 8ē → N−3 (diluted HNO3 → NH4NO3;
reactions with strong reducing agents);
2O−1 + 2ē → 2O−2 (H2O2)

Part 2: Poorly learned question

30. Redox reactions.
write the reaction equation:


25.93% – completely coped with this task

30.

-3
+5
+4
Ca3P2 + ... + H2O → Ca3(PO4)2 + MnO2 + ... .
1) We determine the substances missing in the diagram and compose
electronic balance:
3 2P-3 – 16ē → 2P+5 oxidation
16 Mn+7 + 3ē → Mn+4 recovery

3Ca3P2 + 16KMnO4 + 8H2O = 3Ca3(PO4)2 + 16MnO2 + 16KOH
rebel
ok-tel
3) Determine the reducing agent and oxidizing agent

A typical example of errors in task 30

Due to the lack of systematic knowledge about the oxidizing agent, the student assigns oxidation states for all
elements.
It must be remembered that if an element (not a simple substance) has
index, then it must be placed before the element (in the form
coefficient). Hence the incorrect balance and, as a consequence, not
reaction is correct.
The oxidizing agent at the site of the process is not indicated.

30

Using the electronic balance method,
write the reaction equation:
HCHO + KMnO4 + ... → CO2 + K2SO4 + ...
+ ... .
Identify the oxidizing agent and
reducing agent.
29.1–65.1% – performance range
30.0% – completely completed the task

30

0
+7
+4
HCHO + KMnO4 + ... → CO2 + K2SO4 + ... + ...

5 C0 – 4ē → C+4
oxidation
4 Mn+7 + 5ē → Mn+2 recovery
2) We arrange the coefficients in the reaction equation:
5HCOH + 4KMnO4 + 6H2SO4 = 5CO2 + 2K2SO4 + 4MnSO4 + 11H2O
rebel
ok-tel

30

Using the electronic balance method,
write the reaction equation:
Ca(HS)2 + HNO3 (conc.) → ... + CaSO4 + NO2
+ ... .
Identify the oxidizing agent and the reducing agent.
26.3–57.7% – range of task completion C1
4.9% – completely coped with this task

30

-2
+5
+6
+4
Ca(HS)2 + HNO3 (conc.) → ... + CaSO4 + NO2 + ...
.
1) We create an electronic balance:
1
2S-2 – 16ē → 2S+6 oxidation
16 N+5 + ē → N+4
recovery
2) We arrange the coefficients in the reaction equation:
Ca(HS)2 + 16HNO3 (conc.) → H2SO4 + CaSO4 + 16NO2 + 8H2O
rebel
ok-tel
3) Determine the oxidizing agent and reducing agent

31 Reactions confirming the relationship
various classes inorganic substances
1. Draw the genetic relationship of inorganic substances.
2. Mark characteristic properties substances: acid-base and redox
(specific).
3. Pay attention to the concentrations of substances (if
indicated): solid, solution, concentrated
substance.
4. It is necessary to write down four reaction equations
(not diagrams).
5. As a rule, two reactions are ORR, for metals –
complexation reactions.

Part 3: Unlearned Question

31Reactions confirming the relationship between various
classes of inorganic substances.
Hydrogen sulfide was passed through bromine water.
The resulting precipitate was treated with hot
concentrated nitric acid. Stand out brown
the gas was passed through a solution of barium hydroxide. At
interaction of one of the formed salts with aqueous
a brown precipitate formed with a solution of potassium permanganate.
Write equations for the four reactions described.
5.02–6.12% – range full implementation tasks C2
5.02% – completely coped with this task

31

H2S
Br2(aq)
Solid HNO3 (conc.) Brown Ba(OH)2
gas
substance
to
Salt with KMnO4 anion
with AC Art. OK.
H2O
H2S (gas),
S (TV),
NO2 (gas),
Ba(NO2)2,
please
please
brown gas
salt with element
disproportional in variable st. OK.
Brown
sediment
MnO2 (sol.)
brown sediment

1) H2S + Br2 = S↓ + 2HBr
to
2) S + 6HNO3 = H2SO4 + 6NO2 + 2H2O
3) 2Ba(OH)2 + 4NO2 = Ba(NO3)2 + Ba(NO2)2 + 2H2O
4) Ba(NO2)2 + 4KMnO4 + 2H2O = 3Ba(NO3)2 + 4MnO2↓+ 4KOH

A typical example of errors in task 31

The second equation is written incorrectly - sulfur when heated
oxidizes to sulfuric acid.
The third equation is not equalized.

Solid lithium chloride heated with concentrated
sulfuric acid. The gas released was dissolved in
water. When the resulting solution interacts with
potassium permanganate formed a simple gas
yellow-green substance. When burning iron
the wires in this substance received salt. Salt
dissolved in water and mixed with carbonate solution
sodium Write equations for the four reactions described.
11.3–24.2% – range of task completion C2
2.7% – completely coped with this example

31

LiCl
H2SO4 (k)
Gas
soluble
in water
LiCl(tv),
salt
KMnO4
Gas
yellow-green
H2SO4 (conc.),
ok, whoa
Fe, to
Salt
soluble
in water
KMnO4,
ok
Na2CO3(solution)
Fe,
met., v-l
Gas, sediment
or water
Na2CO3 (solution)
salt sl. who-you
We write down possible reaction equations:
1) LiCl + H2SO4 = HCl + LiHSO4
2) 2KMnO4 + 16HCl = 2MnCl2 + 5Cl2 + 2KCl + 8H2O
3) 2Fe + 3Cl2 = 2FeCl3
4) 2FeCl3 + 3Na2CO3 + 3H2O = 2Fe(OH)3↓ + 6NaCl + 3CO2

31 Reactions confirming the relationship between various classes of inorganic substances

A mixture of nitric oxide (IV) and oxygen was passed through
potassium hydroxide solution. The resulting salt
dried and calcined. Balance received after
calcination of salt, dissolved in water and mixed with
solution
iodide
potassium
And
sulfur
acid.
The simple substance formed during this reaction
reacted with aluminum. Write the equations
the four reactions described.

31

NO2 + O2
KOH (solution)
KOH(solution),
alkali
Salt
to
HI + H2SO4(solution)
Solid
substance
(soluble in water)
KNO3,
KNO2,
term. undef. salt sol. salt, ok, v-l
Simple
substance
Al
HI,
Al
v-l
amph. meth
We write down possible reaction equations:
1) 4NO2 + O2 + 4KOH = 4KNO3 + 2H2O
to
Salt
2) 2KNO3 = 2KNO2 + O2
3) 2KNO2 + 2HI + 2H2SO4 = I2 + 2NO + 2K2SO4 + 2H2O
4) 3I2 + 2Al = 2AlI3

organic compounds
1.
2.
3.
4.
5.
6.
All classes of organic compounds studied in
school curriculum.
The chains are presented in an implicit form (by product or by
reaction conditions).
Particular attention must be paid to the flow conditions
reactions.
All reactions must be equalized (including ORR). No schemes
There shouldn't be any reactions!
If it is difficult to run the chain in the forward direction,
solve from the end of the chain or in fragments. Try anything
execute!
Write organic substances in the form of structural
formulas!

32 Reactions confirming the relationship
organic compounds
3H2
H2
[H]
CnH2n+2
alkanes
H2
+Hal2
HHal
CnH2n
alkenes
H2
2H2
CnH2n-2
alkadienes
kat
CnH2n-6
arenas
H2O
+H2O,
Hg2+, H+
[O]
H2O
CnH 2n+1Hal
halogenated HHal
CnH2n
cycloalkanes
CnH2n-2
alkynes
H2O
H2O
+HHal
H2
[O]
CnH 2n+1OH
alcohols
[H]
[O]
RCHO
aldehydes
(R)2CO
ketones
[H]
RCOOH
carboxylic acids
[O]
+H2O, H+ +R"OH
+RCOOH
+H2O, H+
RCOOR"
esters
24

On the structural formulas of organic compounds

When writing reaction equations, examinees must
use structural formulas of organic
substances (this indication is given in the task conditions).
Structural formulas can be presented on
different levels, without distorting the chemical meaning:
1) full or abbreviated structural formula
acyclic compounds;
2) schematic structural formula of cyclic
connections.
It is not allowed (even fragmentarily) to combine clause 2 and
3.
25

Structural formula

Structural formula - symbol chemical
composition and structure of substances using chemical symbols
elements, numeric and auxiliary characters (brackets, dashes, etc.).
full structural
H
H
H
C C
H
H H H
H
C
HH
H C C C O H
H H H
H C C C H
H
C
C
C
H
H
H
H
C
C C
H
H
H
H
abbreviated structural
CH
CH2 CH CH3
CH3 CH2 CH2 OH
HC
CH2
CH
HC
CH
H2C
CH2
CH
schematic structural
OH
26

Typical errors in structural formulas

27

Alternative reactions

C3H6
C3H6
Cl2, 500 oC
Cl2
CCl4, 0 oC
CH2CH
CH2Cl + HCl
CH2CH
CH3
Cl
Cl2
C3H6 light, > 100 oC
C3H6
Cl2
light
Cl
CH2 CH2
CH2
Cl
Cl
Cl+HCl

Alternative reactions

CH3CH2Cl + KOH
CH3CH2Cl + KOH
H2O
CH3CH2OH + KCl
alcohol
CH2 CH2 + H2O + KCl
CH3
Cl2
light
CH2Cl + HCl
CH3
Cl2
Fe
CH3+Cl
Cl
2CH3CH2OH
CH3CH2OH
H2SO4
140 oC
H2SO4
170 oC
(CH3CH2)2O + H2O
CH2 CH2 + H2O
CH3 + HCl

Typical errors in drawing up reaction equations

30

32 Reactions confirming the relationship
organic compounds.
Write the reaction equations using
which can be implemented as follows
transformations:
heptane
Pt, to
KMnO4
X1
KOH
X2
KOH, to
benzene
HNO3
H2SO4
X3
Fe, HCl


0.49–3.55% – range of complete completion of task C3
0.49% – completely coped with this task
X4

heptane
Pt, to
KMnO4
X1
KOH
KOH, to
X2
benzene
HNO3
H2SO4
X3
Fe, HCl
X4

1) CH3CH2CH2CH2CH2CH2CH3
2)
Pt, to
CH3 + 4H2
CH3 + 6KMnO4 + 7KOH
COOK + 6K2MnO4 + 5H2O
o
3)
4)
5)
COOK + KOH
+ HNO3
t
H2SO4
NO2 + 3Fe + 7HCl
16,32 % (36,68 %, 23,82 %)
+ K2CO3
NO2 + H2O
NH3Cl + 3FeCl2 + 2H2O

1)
2)
3)
4)
5)
Equations 2 and 5 are not composed correctly. Equation 3 is not correct.

A typical example of errors in task 32

2)
Permanganate ion (MnO4–) in an alkaline medium transforms into
manganate ion (MnO42–).
5)
In an acidic environment, aniline forms an ammonium salt -
V in this case Phenylammonium chloride.

A typical example of errors in task 32

2)
3)
It is not allowed to write a scheme or multi-stage reaction
(second reaction).
When writing reaction equations for organic compounds, you cannot
forget about inorganic substances - not as in the textbook, but as in
condition of the task (third equation).

32 Reactions confirming the relationship between organic
connections.


benzene
H2,Pt
X1
Cl2, UV
X2
cyclohexanol
H2SO4(conc.)
160 oС
O
X3
O
HOC(CH2)4COH
When writing reaction equations, use
structural formulas of organic substances.
3.16% – completely coped with this task

benzene
H2,Pt
X1
Cl2, UV
X2
cyclohexanol
H2SO4(conc.)
160 oС
O
X3
O
HOC(CH2)4COH
We write down the reaction equations:
1)
2)
3)
4)
Pt
+ 3H2
+Cl2
hv
Cl+KOH
OH
Cl+HCl
H2O
H2SO4 (conc.)
160 oC
OH + KCl
+ H2O
O
5) 5
+ 8KMnO4 + 12H2SO4
O
5HOC(CH2)4COH + 4K2SO4 + 8MnSO4 + 12H2O

A typical example of errors in task 32

The idea of ​​the structural formula has not been formed
cyclic compounds (second and third reactions).
The second equation (substitution reaction) is incorrect.
It is better to write conditions above the arrow.

A typical example of errors in task 32

Lack of attention to formulas (both cyclohexene and
and the formula of the dicarboxylic acid in the fifth reaction).

A typical example of errors in task 32

Cu
ethanol o
t
Cu(OH)2
X1
to
X2
Ca(OH)2
X3
to
X4
H2, cat.
propanol-2
Not paying attention to the conditions of the task: copper (II) oxide is not given,
and copper (as a catalyst in the dehydrogenation reaction).
Primary aldehydes are formed from aldehydes upon reduction.
alcohols.

A typical example of errors in task 32

Cu
ethanol o
t
Cu(OH)2
X1
to
X2
Ca(OH)2
X3
to
X4
H2, cat.
propanol-2
How do you get three carbon atoms from two, and one of them?
in the trivalent state.

X2
32 Reactions confirming
relationship between organic
connections
Write reaction equations that can be used to
carry out the following transformations:
X1
Zn
cyclopropane
ï ðî ï åí
HBr, to
KMnO4, H2O, 0 oC
X2
X3
propene
hut HBr
KMnO4, H2O, 0 oC
X4
When writing reaction equations, use
structural formulas of organic substances.
16.0–34.6% – range of task completion C3
3.5% - completely coped with this task
X3

32

X1
Zn
cyclopropane
HBr, to
X2
propene
KMnO4, H2O, 0 oC
X3
hut HBr
X4
We write down the reaction equations:
1) BrCH2CH2CH2Br + Zn → ZnBr2 +
2)
t°
+ HBr → CH3CH2CH2Br
3) CH3CH2CH2Br + KOH(alcohol solution) → CH3–CH=CH2 + H2O +KBr
4) 3CH3–CH=CH2 + 2KMnO4 + 4H2O → 3CH3 CH CH2 + 2KOH + 2MnO2
5) CH3 CH CH2 + 2HBr → CH3
OH OH
OH OH
CH CH2 + 2H2O
Br
Br

32 Reactions confirming the relationship of organic compounds

Write reaction equations that can be used to
carry out the following transformations:
potassium acetate
KOH, alloy
X1
CH3
C2H2
C act., to
X2
potassium benzoate
When writing reaction equations, use
structural formulas of organic substances.
14.6–25.9% – range of task completion C3
2.0% - completely coped with this task

32

potassium acetate
KOH, alloy
X1
C2H2
C act., to
CH3
X2
potassium benzoate
We write down the reaction equations:
t°
1) CH3COOK + KOH (solid) → CH4 + K2CO3
t°
2) 2CH4 → C2H2 + 3H2
C
, t°
Act.
3) 3C2H2 →
C6H6
AlCl3
4) C6H6 + СH3Cl →
C6H5–CH3 + HCl
5) C6H5–CH3 + 6KMnO4 + 7KOH → C6H5–COOK + 6K2MnO4 + 5H2O
or C6H5–CH3 + 2KMnO4 → C6H5–COOK + 2MnO2 + KOH + H2O

33. Calculation problems for solutions and
mixtures
1. Write down the equation(s) of the reaction(s).
2. Select an algorithm for solving the problem: using excess (or
impurity), the yield of the reaction product from theoretically
possible and determine the mass fraction (mass) of the chemical
compounds in the mixture.
3. There are only 4 stages of solving the problem.
4. In calculations, refer to reaction equations and use
corresponding mathematical formulas.
5. Don't forget to check your units of measurement.
6. If the amount of a substance is less than 1 mol, then it is necessary
Round to three decimal places.
7. Separate mass fractions and percentages in brackets or write
through union or.
8. Don't forget to write down the answer.

33

1. Calculations according to
equation
reactions
4. Finding
mass fraction
one of the products
reactions in solution
according to the equation
material
balance
2. Objectives
on the mixture
substances
33
3. Tasks on
“type of salt”
(definition
composition
product
reactions)
5. Finding
mass of one of
starting materials
according to the equation
material
balance

Part 2: Unlearned Question

Calculation of the mass (volume, amount of substance) of reaction products,
if one of the substances is given in excess (has impurities), if one of the
substances are given in the form of a solution with a certain mass fraction
dissolved substance. Mass or volume fraction calculations
the yield of the reaction product from the theoretically possible. Calculations
mass fraction (mass) chemical compound in the mixture.
44.8 liters (n.s.) of hydrogen chloride were dissolved in 1 liter of water. To that
the substance obtained as a result was added to the solution
reactions of calcium oxide weighing 14 g with excess
carbon dioxide. Determine the mass fraction of substances in
the resulting solution.
3.13% – completely coped with this task

44.8 liters (n.s.) of hydrogen chloride were dissolved in 1 liter of water. TO
to this solution was added a substance obtained in
as a result of the reaction of calcium oxide weighing 14 g with
excess carbon dioxide. Determine the mass
the proportion of substances in the resulting solution.
Given:
V(H2O) = 1.0 l
V(HCl) = 44.8 l
m(CaO) = 14 g
Solution:
CaO + CO2 = CaCO3
ω(CaCl2) – ?
Vm = 22.4 mol/l
M(CaO) = 56 g/mol
M(HCl) = 36.5 g/mol
2HCl + CaCO3 = CaCl2 + H2O + CO2

1) We calculate the amounts of reagent substances:
n=m/M
n(CaO) = 14 g / 56 g/mol = 0.25 mol
n(CaCO3) = n(CaO) = 0.25 mol
2) Calculate the excess and amount of the substance
hydrogen chloride:
n(HCl)tot. = V / Vm = 44.8 l / 22.4 l/mol = 2 mol
(in excess)
m(HCl) = 2 mol · 36.5 g/mol = 73 g
n(HCl)react. = 2n(CaCO3) = 0.50 mol

3) Calculate the amount of carbon dioxide and
calcium chloride:
n(HCl)res. = 2 mol – 0.50 mol = 1.5 mol
n(CO2) = n(CaCO3) = 0.25 mol
n(CaCl2) = n(CO2) = 0.25 mol
4) Calculate the mass of the solution and mass fractions
substances:
m(HCl)res. = 1.5 mol · 36.5 g/mol = 54.75 g
m(CaCO3) = 0.25 mol 100 g/mol = 25 g
m(CO2) = 0.25 mol 44 g/mol = 11 g
m(CaCl2) = 0.25 mol 111 g/mol = 27.75 g

Calculate the mass of the solution and mass fractions
substances:
m(solution) = 1000 g + 73 g + 25 g – 11 g = 1087 g
ω = m(in-va) / m(r-ra)
ω(HCl) = 54.75 g / 1087 g = 0.050 or 5.0%
ω(CaCl2) = 27.75 g / 1087 g = 0.026 or 2.6%
Answer: mass fraction of hydrochloric acid and calcium chloride in
the resulting solution is 5.0% and 2.6%
respectively.

Note. In the case when the answer
there is an error in the calculations in
one of the three elements (the second,
third or fourth), which led
to the wrong answer, score for
task performance is reduced only by
1 point.

C4
Calculation of mass (volume, amount of substance) of products
reactions if one of the substances is given in excess (has
impurities), if one of the substances is given in the form of a solution with
a certain mass fraction of the dissolved substance.
Calculations of mass or volume fraction of product yield
reactions from the theoretically possible. Mass calculations
proportion (mass) of a chemical compound in a mixture.
Phosphorus weighing 1.24 g reacted with 16.84 ml of a 97% solution of sulfuric acid (ρ = 1.8 g/ml) with
formation of orthophosphoric acid. For complete
To neutralize the resulting solution, a 32% sodium hydroxide solution (ρ = 1.35 g/ml) was added.
Calculate the volume of sodium hydroxide solution.
0% – completely coped with this task

2) We calculate the excess and quantity of reagent substances:
2P + 5H2SO4 = 2H3PO4 + 5SO2 + 2H2O
2 mol
5 mol
0.04 mol 0.1 mol
n=m/M
n = (V ρ ω)/M
n(P) = 1.24 g / 31 g/mol = 0.040 mol
n(H2SO4)tot. = (16.84 ml · 1.8 g/ml · 0.97) / 98 g/mol = 0.30 mol
(excess)
n(H3PO4) = n(P) = 0.04 mol
n(H2SO4)react. = 5/2n(P) = 0.1 mol
n(H2SO4)res. = 0.3 mol – 0.1 mol = 0.2 mol

3) Calculate the excess and amount of alkali substance:
H3PO4 + 3NaOH = Na3PO4 + 3H2O
1 mole
3 mol
0.04 mol 0.12 mol
n(NaOH)H3PO4 = 3n(H3PO4) = 3 0.04 mol = 0.12 mol
n(NaOH)tot. = 0.12 mol + 0.4 mol = 0.52 mol
4) Calculate the volume of alkali:
m=n·M
V = m / (ρ ω)
m(NaOH) = 0.52 mol 40 g/mol = 20.8 g
V(solution) = 65 g / (1.35 g/ml 0.32) = 48.15 ml

Calculation problems for solutions

A mixture of iron and aluminum powders reacts with
810 ml 10% sulfuric acid solution
(ρ = 1.07 g/ml). When interacting the same
mass of mixture with excess hydroxide solution
sodium, 14.78 liters of hydrogen (n.s.) were released.
Determine the mass fraction of iron in the mixture.
1.9% - completely coped with this task

1) Write down the reaction equations for metals
Fe + H2SO4 = FeSO4 + H2


2) We calculate the amounts of reagent substances:
n = m/M
n = (V ρ ω) / M n = V / Vm
n(H2SO4) = (810 g 1.07 g/ml 0.1) / 98 g/mol
= 0.88 mol
n(H2) = 14.78 l / 22.4 l/mol = 0.66 mol
n(Al) = 2/3n(H2) = 0.44 mol
2Al + 2NaOH + 6H2O = 2Na + 3H2
2 mol
3 mol
0,44
0,66

2) We calculate the amounts of reagent substances:
n(H2SO4 spent on the reaction with Al) = 1.5 n(Al) = 0.66
mole
n(H2SO4, spent on the reaction with Fe) =
= 0.88 mol – 0.66 mol = 0.22 mol
n(Fe) = n(H2SO4) = 0.22 mol
2Al + 3H2SO4 = Al2(SO4)3 + 3H2
0,44
0,66
Fe + H2SO4 = FeSO4 + H2
0,22
0,22
3) Calculate the masses of metals and their mixtures:
m(Al) = 0.440 mol 27 g/mol = 11.88 g
m(Fe) = 0.22 mol 56 g/mol = 12.32 g
m(mixture) = 11.88 g + 12.32 g = 24.2 g
4) Calculate the mass fraction of iron in the mixture:
ω(Fe) = 12.32 g / 24.2 g = 0.509 or 50.9%

Calculation problems for solutions

When dissolving 4.5 g partially
oxidized aluminum in excess solution
KOH produces 3.7 L(N) of hydrogen.
Determine the mass fraction of aluminum in
sample.

2Al + 2KOH + 6H2O = 2K + 3H2
2 mol
0.110 mol
3 mol
0.165 mol
Al2O3 + 2KOH + 3H2O = 2K
2) Calculate the amount of aluminum substance:
n = V / Vm
n(H2) = 3.7 L / 22.4 L/mol = 0.165 mol
n(Al) = 2/3n(H2) = 0.110 mol
3) Calculate the masses of aluminum and aluminum oxide:
m(Al) = n M = 0.110 mol 27 g/mol = 2.97 g
m(Al2O3) = m(mixtures) – m(Al) = 4.5 g – 2.97 g = 1.53 g
4) Calculate the mass fraction of aluminum in the mixture:
ω(Al) = mv-va / mmixture = 2.97 g / 4.5 g = 0.660 or 66.0%
– according to theory
- on practice

Problem (2008)

Hydrogen sulfide with a volume of 5.6 l (n.s.) reacted
without residue with 59.02 ml potassium hydroxide solution
with a mass fraction of 20% (ρ=1.186g/ml). Define
mass of salt obtained as a result of this
chemical reaction.
1. Type 3 “Salt Type”.
2. Excess and deficiency.
3. Determination of salt composition.

Problem (2008)

After 35 ml of 40% sodium hydroxide solution
pl. 1.43 g/ml missed 8.4 l
carbon dioxide (n.s.) Determine
mass fractions of substances in the resulting
solution.
1. Type 3 “Salt Type”.
2. Excess and deficiency.
3. Determination of salt composition.
4. Determination of the mass of reaction products - salts.

Problem (2009)

Magnesium weighing 4.8 g was dissolved in 200 ml of 12%
sulfuric acid solution (ρ=1.5 g/ml). Calculate
mass fraction of magnesium sulfate in the final
solution.
1. Type 4 “Finding the mass fraction of one of
reaction products in solution according to the equation
material balance".
2. Excess and deficiency.
3. Calculation of the mass fraction of a substance in solution.
4. Determination of the mass of the dissolved substance.

Problem (2010)

Aluminum carbide was dissolved in 380 g of solution
hydrochloric acid with a mass fraction of 15%.
The gas released took up a volume of 6.72 liters
(Well.). Calculate the mass fraction of hydrogen chloride in
the resulting solution.



3. Drawing up an equation to calculate the mass fraction
starting material

Challenge (2011)

Potassium nitrite weighing 8.5 g was added while heating
270 g ammonium bromide solution with mass fraction
12%. What volume (n.s.) of gas will be released in this case and
what is the mass fraction of ammonium bromide in
the resulting solution?
1.Type 5 “Finding the mass and mass fraction of one of
starting substances according to the material balance equation.”
2. Drawing up the reaction equation.
3. Finding the amount of a substance, their mass, volume.
4. Drawing up an equation to calculate the mass fraction
original substance.

Problem (2012)

Determine the mass of Mg3N2, completely
subjected to decomposition by water, if for
salt formation with hydrolysis products
it took
150 ml of 4% hydrochloric acid solution
density 1.02 g/ml.

Challenge (2013)

Determine the mass fractions (in%) of iron sulfate
and aluminum sulfide in the mixture, if during processing
25g of this mixture with water released a gas, which
reacted completely with 960g of 5%
copper sulfate solution.
1. Type 5 “Finding the mass and mass fraction of one of
starting substances according to the material balance equation.”
2. Drawing up reaction equations.
3. Finding the amount of a substance, their mass.
4. Determination of the mass fraction of the starting substances of the mixture.

Problem 2014 The gas obtained by reacting 15.8 g of potassium permanganate with 200 g of 28% hydrochloric acid was passed through 100 g of a 30% solution of sul

Challenge 2014
Gas obtained by interaction 15, 8
g potassium permanganate with 200 g 28% hydrochloric acid
acid, passed through 100 g of 30%
potassium sulfite solution. Define
mass fraction of salt in the resulting
solution

Problem (2015) A mixture of copper(II) oxide and aluminum with a total mass of 15.2 g was set on fire using a magnesium tape. After the reaction is completed, the resulting solid

Challenge (2015)
A mixture of copper(II) oxide and total aluminum
weighing 15.2 g was set on fire using
magnesium tape. After graduation
reaction resulting solid residue
partially dissolved in hydrochloric acid
with the release of 6.72 liters of gas (n.o.).
Calculate mass fractions (in %)
substances in the original mixture.

1) Reaction equations have been compiled: 3CuO + 2Al = 3Cu + Al2O3, Al2O3 + 6HCl = 2AlCl3 + 3H2O. 2Al + 6HCl = 2AlCl3 + 3H2 2) The amounts of hydrogen and aluminum substances remaining are calculated

1) Reaction equations are drawn up:
3CuO + 2Al = 3Cu + Al2O3,
Al2O3 + 6HCl = 2AlCl3 + 3H2O.
2Al + 6HCl = 2AlCl3 + 3H2
2) The amounts of hydrogen substance and
aluminum remaining after the reaction:
(H2) = 6.72 / 22.4 = 0.3 mol,
(remaining Al) = 2/3 0.3 = 0.2 mol.
3) The amount of copper(II) oxide was calculated,
reacted:
Let n(CuO) = x mol, then n(react. Al) = 2/3 x
mole.

m(CuO) + m(react. Al) = 15.2 – m(remaining Al) 80x + 27 * 2/3 x = 15.2 – 0.2 * 27 x = 0.1 4) Mass fractions calculated substances in the mixture: W(CuO) = 0.1 *80 / 15.2 *100% = 52.6%, W(Al) = 100% – 52.6% = 47.4%

m(CuO) + m(react. Al) = 15.2 –
m(remaining Al)
80x + 27 * 2/3 x = 15.2 – 0.2 * 27
x = 0.1
4) Mass fractions calculated
substances in the mixture:
W(CuO) = 0.1 *80 / 15.2 *100% =
52,6 %,
W(Al) = 100% – 52.6% = 47.4%.

2016 When a sample of sodium bicarbonate was heated, part of the substance decomposed. In this case, 4.48 liters (n.s.) of gas were released and 63.2 g of solid were formed

2016
When heating a sample of bicarbonate
sodium part of the substance has decomposed.
At the same time, 4.48 liters (n.s.) of gas were released and
63.2 g of solid was formed
anhydrous residue. To the received balance
added minimum volume
20% hydrochloric acid solution,
necessary for complete selection
carbon dioxide. Determine the mass fraction
final sodium chloride
solution.

2NaHCO3 = Na2CO3 + CO2 + H2O NaHCO3 + HCl = NaCl + CO2 + H2O Na2CO3 + 2HCl = 2NaCl + CO2 + H2O 2) The amount of substance of the compounds in the solid is calculated

1) The reaction equations are written:
2NaHCO3 = Na2CO3 + CO2 + H2O
NaHCO3 + HCl = NaCl + CO2 + H2O
Na2CO3 + 2HCl = 2NaCl + CO2 + H2O
2) The amount of compound substances in
hard
balance:
n(CO2) = V / Vm = 4.48 / 22.4 = 0.2 mol
n(Na2CO3) = n(CO2) = 0.2 mol
m(Na2CO3) = n ∙ M = 0.2 ∙ 106 = 21.2 g
m(NaHCO3 residue) = 63.2 – 21.2 = 42 g
n(NaHCO3 residue) = m / M = 42 / 84 = 0.5 mol

3) The mass of reacted hydrochloric acid and the mass of sodium chloride in the final solution were calculated: n(HCl) = 2n(Na2CO3) + n(NaHCO3 residue) = 0.2 ∙ 2 + 0.5 = 0.9 mol

m(HCl) = n ∙ M = 0.9 ∙ 36.5 = 32.85 g
m(HCl solution) = 32.85 / 0.2 = 164.25 g
n(NaCl) = n(HCl) = 0.9 mol
m(NaCl) = n ∙ M = 0.9 ∙ 58.5 = 52.65 g
4) The mass fraction of sodium chloride in the solution is calculated:
n(CO2) = n(Na2CO3) + n(NaHCO3 remainder) = 0.2 + 0.5 = 0.7 mol
m(CO2) = 0.7 ∙ 44 = 30.8 g
m(solution) = 164.25 + 63.2 – 30.8 = 196.65 g
ω(NaCl) = m(NaCl) / m(solution) = 52.65 / 196.65 = 0.268, or 26.8%

Problem (2016) As a result of heating 20.5 g of a mixture of magnesium oxide and magnesium carbonate powders, its mass decreased by 5.5 g. Calculate the volume of

Challenge (2016)
As a result of heating 20.5 g of the mixture
magnesium oxide and carbonate powders
magnesium, its mass decreased by 5.5
d. Calculate the volume of sulfuric solution
acids with a mass fraction of 28% and
density 1.2 g/ml, which
required
to dissolve the original mixture.

1) The reaction equations are written: MgCO3 = MgO + CO2 MgO + H2SO4 = MgSO4 + H2O MgCO3 + H2SO4 = MgSO4 + H2O + CO2 2) The amount of carbon dioxide released is calculated

1) The reaction equations are written:
MgCO3 = MgO + CO2
MgO + H2SO4 = MgSO4 + H2O
MgCO3 + H2SO4 = MgSO4 + H2O + CO2
2) Calculated the amount of substance released
carbon dioxide
gas, mass of magnesium carbonate and magnesium oxide in
starting mixture:
n(CO2) = 5.5 / 44 = 0.125 mol
n(MgCO3) = n(СO2) = 0.125 mol
m(MgCO3) = 0.125 84 = 10.5 g
m(MgO) = 20.5 – 10.5 = 10 g

3) The amount of magnesium oxide substance and the amount of sulfuric acid substance required to dissolve the mixture are calculated: n(MgO) = 10 / 40 = 0.25 mol n

3) The amount of magnesium oxide substance and
the amount of sulfuric acid required for
dissolving the mixture:
n(MgO) = 10 / 40 = 0.25 mol
n(H2SO4 for reaction with MgCO3) = 0.125 mol
n(H2SO4 for reaction with MgO) = 0.25 mol
n(H2SO4 total) = 0.125 + 0.25 = 0.375 mol
4) The volume of sulfuric acid solution is calculated:
V(H2SO4(solution)) = 0.375 98 / 1.2 0.28 = 109.4 ml

C5 Finding molecular
formulas of substances (until 2014)
1. Make up the reaction equation in general view, wherein
write substances in the form of molecular formulas.
2. Calculate the amount of a substance from a known value
mass (volume) of a substance, most often inorganic.
3. According to the stoichiometric ratios of the reactants
substances find the amount of organic substance
compounds with known mass.
4. Find molecular weight organic matter.
5. Determine the number of carbon atoms in the composition of the desired
substances, based on the general molecular formula and
calculated molecular weight.
6. Write down the found molecular weight of the organic
substances.
7. Don't forget to write down the answer.

Formula

Chemical formula - symbol
chemical composition and structure of substances using
characters chemical elements, numeric and
auxiliary characters (brackets, dashes, etc.).
Gross formula (true formula or empirical) –
reflects the composition (the exact number of atoms of each
element in one molecule), but not the structure of the molecules
substances.
Molecular formula (rational formula) –
formula that identifies groups of atoms
(functional groups) characteristic of classes
chemical compounds.
The simplest formula is a formula that reflects
certain content of chemical elements.
A structural formula is a type of chemical
formulas that graphically describe the location and
the order of bonding of atoms in a compound, expressed in
plane.

The solution to the problem will include three
sequential operations:
1. drawing up a chemical reaction diagram
and determination of stoichiometric
ratios of reacting substances;
2. calculation of the molar mass of the desired
connections;
3. calculations based on them, leading to
establishing the molecular formula
substances.

Part 2: Unlearned Question

When interacting with a limiting monobasic
carboxylic acid with bicarbonate
calcium, 1.12 liters of gas were released (n.o.) and
4.65 g of salt was formed. Write down the equation
reactions in general form and determine
molecular formula of the acid.
9.24–21.75% – range of complete completion of task C5
9.24% – completely coped with this task
25.0–47.62% – range of complete completion of task C5
in the second wave

2СnH2n+1COOH + Ca(HCO3)2 = (СnH2n+1COO)2Ca + 2CO2 + 2H2O
1 mole
2 mol
2) Calculate the amount of carbon dioxide and
salt:

n((СnH2n+1COO)2Ca) = 1/2n(СO2) = 0.025 mol
3) Determine the number of carbon atoms in the salt and
establish the molecular formula of the acid:
M ((СnH2n+1COO)2Ca) = (12n + 2n + 1 + 44) 2 + 40 = 28n +
130
M ((СnH2n+1COO)2Ca) = m / M = 4.65 g / 0.025 mol = 186
g/mol
28n + 130 = 186
n=2
The molecular formula of the acid is CH COOH

34. Finding the molecular formula of substances.
When interacting with a limiting monobasic carbonic acid
acid with magnesium carbonate released 1120 ml of gas (n.o.)
and 8.5 g of salt was formed. Write the reaction equation in
general view. Determine the molecular formula of the acid.
21.75% – completely coped with this task

1) Write down the general reaction equation:
2СnH2n+1COOH + MgCO3 = (СnH2n+1COO)2Mg + CO2 + H2O
1 mole
1 mole
2) Calculate the amount of carbon dioxide and salt:
n(CO2) = V / Vm = 1.12 l / 22.4 l/mol = 0.050 mol
n((СnH2n+1COO)2Mg) = n(СO2) = 0.050 mol
3) Determine the number of carbon atoms in the salt and establish
molecular formula of acid:
M ((СnH2n+1COO)2Mg) = (12n + 2n + 1 + 44) 2 + 24 = 28n + 114
M ((СnH2n+1COO)2Mg) = m / M = 8.5 g / 0.050 mol = 170 g/mol
28n + 114 = 170
n=2
The molecular formula of the acid is C2H5COOH

The reaction is not equalized. Although
this did not affect
mathematical calculations.
Transition from general
molecular formula to
the desired molecular
the formula is not true,
due to use
in practice mostly
gross formulas

A typical example of errors in task 34

Reaction
compiled with
using gross formulas.
Mathematical
part of the problem
solved correctly
(method
proportions).
The difference between
gross formula
and molecular
no formula
learned.

34. Finding the molecular formula of substances

During the oxidation of saturated monohydric alcohol
copper(II) oxide yielded 9.73 g of aldehyde, 8.65 g
copper and water.
Determine the molecular formula of the original
alcohol
88

Solution:
Given:
m(СnH2nO) = 9.73 g
m(Cu) = 8.65 g
СnH2n+2O – ?
1) We write down the general reaction equation and
We calculate the amount of copper substance:

0.135 mol
0.135 mol 0.135 mol
1 mole
1 mole 1 mole
n(Cu) = m / M = 8.65 g / 64 g/mol = 0.135 mol
89

Determine the molecular formula of the original alcohol.
СnH2n+2O + CuO = СnH2nO + Cu + H2O
1 mole
1 mole 1 mole
0.135 mol
0.135 mol 0.135 mol
2) Calculate the molar mass of the aldehyde:
n(Cu) = n(СnH2nO) = 0.135 mol
M(СnH2nO) = m / n = 9.73 g / 0.135 mol = 72 g/mol
90

3) Establish the molecular formula of the original alcohol from the formula
aldehyde:
M(СnH2nO) = 12n + 2n + 16 = 72
14n = 56
n=4
C4H9OH
Answer: the molecular formula of the original alcohol is C4H9OH.
91

34. Finding the molecular formula of substances (since 2015)

The solution to the problem will include four
sequential operations:
1. finding the amount of a substance by
chemical reaction (combustion products);
2. Determination of molecular formula
substances;
3. drawing up the structural formula of a substance,
based on the molecular formula and
qualitative reaction;
4. drawing up an equation for a qualitative reaction.

34.

When burning a sample of some organic compound mass
14.8 g yielded 35.2 g of carbon dioxide and 18.0 g of water. It is known that
the relative vapor density of this substance with respect to hydrogen is 37.
During the study chemical properties this substance
It has been established that when this substance interacts with copper oxide
(II) a ketone is formed.
Based on the data of the task conditions:
1) make the necessary calculations;
2) establish the molecular formula of the original organic
substances;
3) compose the structural formula of this substance, which
unambiguously reflects the order of bonding of atoms in its molecule;
4) write the equation for the reaction of this substance with copper (II) oxide.

34

Given:
m(CxHyOz) = 14.8 g
m(CO2) = 35.2 g
m(H2O) = 18 g
DH2 = 37
СхHyOz – ?
M(CO2) = 44 g/mol
M(H2O) = 18 g/mol
Solution:
1) a)
C → CO2
0.80 mol
0.80 mol
n(CO2) = m / M = 35.2 g / 44 g/mol = 0.80 mol
n(CO2) = n(C) = 0.8 mol
b)
2H → H2O
2.0 mol
1.0 mol
n(H2O) = 18.0 g / 18 g/mol = 1.0 mol
n(H) = 2n(H2O) = 2.0 mol

34

c) m(C) + m(H) = 0.8 12 + 2.0 1 = 11.6 g (oxygen available)
m(O) = 14.8 g – 11.6 g = 3.2 g
n(O) = 3.2 / 16 = 0.20 mol
2) Determine the molecular formula of the substance:
Mist(CxHyOz) = DH2 MH2 = 37 2 = 74 g/mol
x: y: z = 0.80: 2.0: 0.20 = 4: 10: 1
The calculated gross formula is C4H10O
Mcalc(C4H10O) = 74 g/mol
The true formula of the original substance is C4H10O

34
3) We compose the structural formula of the substance based on the true
formulas and qualitative reaction:
CH3 CH CH2 CH3
OH
4) We write down the equation for the reaction of a substance with copper (II) oxide:
CH3 CH CH2 CH3 + CuO
OH
to
CH3 C CH2 CH3 + Cu + H2O
OThe need for increased attention to
organizing targeted work to prepare for
one state exam in chemistry, which
involves systematic repetition of the studied material
and training in performing tasks of various types.
The result of the repetition work should be the reduction
into the knowledge system of the following concepts: substance, chemical
element, atom, ion, chemical bond,
electronegativity, oxidation state, mol, molar
mass, molar volume, electrolytic dissociation,
acid-base properties of a substance, redox properties, oxidation processes and
reduction, hydrolysis, electrolysis, functional
group, homology, structural and spatial isomerism. It is important to remember that mastering any concept
lies in the ability to identify its characteristic
signs, identify its relationships with other
concepts, as well as the ability to use this concept
to explain facts and phenomena.
Repetition and generalization of material is advisable
arrange according to the main sections of the chemistry course:
Theoretical basis chemistry
Inorganic chemistry
Organic chemistry
Methods of knowledge of substances and chemicals
reactions. Chemistry and life. Mastering the content of each section involves
mastery of certain theoretical
information, including laws, rules and concepts,
and also, most importantly, understanding them
relationships and boundaries of application.
At the same time, mastery conceptual apparatus course
chemistry is a necessary but not sufficient condition
successful completion of exam tasks
work.
Most jobs of CMM variants of a single
state exam in chemistry are directed,
mainly to test the ability to use
theoretical knowledge in specific situations. Examinees must demonstrate skills
characterize the properties of a substance based on their
composition and structure, determine the possibility
reactions between substances,
predict possible reaction products with
taking into account the conditions of its occurrence.
Also, to complete a number of tasks you will need
knowledge about the signs of studied reactions, rules
handling laboratory equipment and
substances, methods of obtaining substances in
laboratories and in industry. Systematization and generalization of the studied material in the process
repetitions should be aimed at developing the ability to highlight
The main thing is to establish cause-and-effect relationships between
individual elements of content, especially the relationship of composition,
structure and properties of substances.
There are still many questions that you should familiarize yourself with in advance.
every student who chooses this exam must.
This is information about the exam itself, about the features of its conduct, about
how you can check your readiness for it and how to
organize yourself when doing exam work.
All these questions should be the subject of the most careful
discussions with students. The following are posted on the FIPI website (http://www.fipi.ru)
normative, analytical, educational and methodological
information materials:
documents defining the development of the KIM Unified State Examination in Chemistry 2017
(codifier, specification, demo version appear by 1
September);
educational materials for members and chairmen
regional subject commissions to verify implementation
tasks with a detailed answer;
methodological letters from previous years;
educational computer program"Expert Unified State Exam";
training tasks from the open segment of the federal bank
test materials.

1. The structure of part 1 of the CMM has been fundamentally changed:
tasks with a choice of one answer are excluded; tasks
grouped into separate thematic blocks, in
each of which has tasks of both basic and
increased levels of difficulty.
2. Reduced total tasks from 40 (in 2016) to
34.
3. The rating scale has been changed (from 1 to 2 points) for completion
tasks basic level difficulties that test
mastering knowledge about genetic connection inorganic and
organic substances (9 and 17).
4 Maximum primary score for performing work in
overall will be 60 points (instead of 64 points in 2016).

Parts Number Job Type
work assignments and
th
level
difficulties
Maximum
th
primary
point
%
maximum
primary
points
behind
this part of the work from
general
maximum
primary score – 60
Part 1
29
Tasks with a short
answer
40
68,7%
Part 2
5
Tasks with
expanded
answer
20
31,3%
TOTAL
34
60
100%

Approximate time allocated to complete individual
assignments, tasks
is:
1) for each task of the first part 1 – 5 minutes;
2) for each task of the second part 3 – up to 10 minutes.
Total execution time
exam paper is
3.5 hours (240 minutes).