Each student taking the exam in chemistry should be prepared for the fact that 3 astronomical hours, or 180 minutes, are allotted for the completion of the examination work, which consists of three parts and includes 45 tasks. In official documents, this time is recommended to be distributed as follows:

• each task of part A - 2-3 minutes;
• each task of part B - up to 5 minutes;
• each task of part C - up to 10 minutes.

However, the teacher should encourage students to save time on the relatively easy parts A and B in order to use more slack for part C, which is the most difficult and therefore the most “expensive” in terms of points.

Part C (C1-C5) includes 5 high-level tasks with a detailed answer, tasks of increased complexity. Each task of this part is individual and non-standard.

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CHEMISTRY

Each USE student in chemistry should be prepared for the fact that the examination work about You, consisting of three parts and including 45 tasks, are given 3 astronomical hours, or 180 minutes. In official documents, this time is recommended to be distributed as follows:

• each task of part A - 2-3 minutes;
• each task of part B - up to 5 minutes;
• each task of part C - up to 10 minutes.

However, the teacher should encourage students to save time on the relatively easy parts A and B in order to use more slack for part C, which is the most difficult and therefore the most “expensive” in terms of points.

Part C (C1-C5) includes 5 high-level tasks with a detailed answer.

Tasks with a detailed answer provide for the simultaneous verification of the assimilation of several content elements from various content blocks.

In the exam paper in 2009. the following types of tasks with a detailed answer are presented:

• tasks that test the assimilation of the topic of redox reactions;
• tasks that test knowledge of the chemical properties of inorganic substances;
• tasks that test the assimilation of educational information about the relationship of various classes of substances (organic and inorganic);
• combined calculation problems;
• tasks to determine the molecular formula of a substance.

The third part test - 5 tasks of part C, - includes tasks of increased complexity. Each task of this part is individual and non-standard.

In task C1 it is proposed, using the method of electronic balance, to formulate an equation for any redox reaction and determine the oxidizing agent and reducing agent. Tasks C1 can be divided into three types:

1) the formulas of any substances are omitted on the right side of the equation

Example: P + HNO 3 → NO 2 + ...

KMnO 4 + H 2 S + H 2 SO 4 → MnSO 4 + S + ... + ...

K 2 Cr 2 O 7 + HCl → Cl 2 + KCl + … + …

2) the formulas of any substances are omitted in its left part

Example: KMnO 4 + KBr + ... → MnSO 4 + Br 2 + K 2 SO 4 + H 2 O

P 2 O 3 + H 2 Cr 2 O 7 + ... → H 3 PO 4 + CrPO 4

3) formulas of substances are omitted in both parts of the equation

Example: Cr 2 (SO 4) 3 + ... + NaOH → Na 2 CrO 4 + NaBr + ... + H 2 O

The maximum score for this task is 3 points (1st point is given for writing a balance, 2nd for writing an equation, 3 for determining an oxidizing agent and a reducing agent).

In task C2 given four or five substances, between which it is necessary to write four equations of reactions, and in this case it is necessary to use all the substances indicated in the task.

Example:

1. Aqueous solutions are given: iron (III) chloride, sodium iodide, sodium bichromate, sulfuric acid and cesium hydroxide. Give equations for four possible reactions involving the indicated substances.
2. Substances are given: sodium nitrate, white phosphorus, bromine, potassium hydroxide (solution). Give equations for four possible reactions involving the indicated substances.

This task is perhaps the most difficult of all the tasks of the USE test and tests knowledge of the chemical properties of inorganic substances. The maximum score in this task is 4 points (1 point is given for each correctly written reaction equation).

In task C3 it is necessary to carry out a chain of five transformations between organic substances, in which several links are missing.

Example: +Zn +HBr t ° +KMnO 4

1. CH 2 Br-CH 2 -CH 2 Br → X 1 → X 2 → propene → X 3 → 1,2-dibromopropane

H2O

H 2 O t ° KMnO 4 + H 2 O

1. Al 4 C 3 → X 1 → X 2 → ethanal X 3 → X 1

The maximum score in this task is 5 points (1 point is given for each correctly written reaction equation).

In task C4 it is necessary to calculate the mass (volume, amount of substance) of the reaction products, if one of the substances is given in excess and indicated in the assignment as a solution with a certain mass fraction of the solute or contains impurities. The maximum score for the correct completion of this task is 4 points (points are awarded for each intermediate action).

Example:

1. Sulfur oxide (IV) weighing 8 g was dissolved in 110 g of 8% sulfuric acid. What salt and in what quantity is formed if 10.6 g of potassium hydroxide is added to the resulting solution?
2. What mass of calcium carbonate must be added to 600 g of a solution of nitric acid with a mass fraction of 31.5%, so that it decreases to 10.5%?

In task C5 it is necessary to determine the molecular formula of the substance. The maximum score is 2 (points are awarded for each intermediate action).

Example:

1. In the interaction of 11.6 g of saturated aldehyde with an excess of copper (II) hydroxide, a precipitate of 28.8 g is formed upon heating. Derive the molecular formula of the aldehyde.
2. When burning 9 g of the limiting secondary amine, 2.24 liters of nitrogen and 8.96 liters of carbon dioxide were released. Determine the molecular formula of the amine.

Thus, for the correct execution of part C, you can score 18 points (slightly more than 27% of the maximum possible).

The maximum possible number of primary scores for the USE test in chemistry in 2009 is 66.

Analysis of the performance of tasks Part C

In 2009, the percentage of graduates who started to complete part C of the USE test in chemistry was 90.2%. The generalized results of completing the tasks of part C are presented in table 1.

Table 1

The results of the high-level assignments (part C) of the examination work of the USE 2009

Mandatory minimum content section education	Task designation in work	Content elements and activities to be checked	Task difficulty level	Maximum Points	Average primary score (main exam)
Chemical reaction		Redox reactions.	Tall		1,65
Substance		Reactions confirming the relationship of various classes of inorganic substances.	Tall		1, 05
		Reactions confirming the relationship between various classes of hydrocarbons and oxygen-containing organic compounds.	Tall		1,25
Knowledge and application of substances and chemical reactions		Calculations: mass (volume, amount of substance) of the reaction product, if one of the substances is given in excess, if one of the substances is given as a solution with a certain mass fraction of the dissolved substance.	Tall		1,55
		Finding the molecular formula of a substance according to qualitative and quantitative analysis	Tall		1,15

The average completion rate for Part C items in 2009 was 36.94%,

Typical mistakes when performing task C1:

• inability to determine the substance that determines the medium of the redox reaction solution (for example, water);
• inability to choose an oxidizing agent and a reducing agent among compounds with a variable degree of oxidation (for example, in the interaction of potassium nitrite and potassium permanganate);
• the inability to predict the reduction products of typical oxidizing agents (potassium permanganate, iodine, potassium nitrite) and the oxidation products of reducing agents (manganese dioxide) in various media, as well as the possibility of the participation of water molecules in these processes;
• inability to predict the oxidizing (reducing) properties of elements with an intermediate oxidation state in specific processes (for example, the chromium element in chromium (III) oxide).

This can be explained by the fact that these topics are studied in detail only in the profile course of chemistry. In the basic course, these issues are addressed in an introductory plan.

Tasks C2 tested knowledge of the properties and genetic relationship of the main classes of inorganic compounds.

With task C2, in general, less than a third of the graduates coped, which can be explained by the complexity of the task.Typical difficulties in completing this task were:

• inability to analyze the possibility of the interaction of substances (simple and complex) from the standpoint of their belonging to certain classes of inorganic compounds, as well as from the standpoint of the possibility of redox reactions;
• ignorance of the specific properties of halogens, phosphorus and their compounds, acids - oxidizing agents, amphoteric oxides and hydroxides, reducing properties of sulfides and halides.

Task C3 was completed by less than a quarter of the graduates. This is due to the complication of tasks of this type.Typical mistakes when performing task C3:

• ignorance of the conditions for the occurrence of chemical reactions, the genetic connection of classes of organic compounds;
• ignorance of the mechanisms, nature and conditions of reactions involving organic substances, properties and formulas of organic compounds;
• inability to predict the properties of an organic compound on the basis of ideas about the mutual influence of atoms in a molecule;
• ignorance of redox reactions (for example, with potassium permanganate).

Task C4 was a combined calculation task. More than a third of the graduates completed the task.

Under the conditions of tasks of this type, the following actions were combined:

• calculations according to the equation, when one of the substances is given as a solution with a certain mass fraction of the dissolved substance;
• calculations when one of the reactants is given in excess;
• determination of the mass of a solute in a solution;
• calculations according to the equations of successive reactions.

Most often, students are allowed errors :

• when determining the mass of the solution without taking into account the mass of the evolved gas or precipitate;
• when determining the mass fraction of a solute in a solution obtained by mixing solutions with different mass fractions of a solute;
• when determining the amount of substances entering the reaction.

Tasks C5 – finding the molecular formula of a substance according to the data of qualitative and quantitative analysis.

The problem was solved by more than half of the graduates. Many students were able to correctly perform the first action - to find the simplest ratio of moles of atoms in a compound, but could not proceed to determine the true formula.

The difficulty was caused by the task involving the definition of the molecular formula, if the combustion products are known - the volume of carbon dioxide and the mass of nitrogen and water.

Familiarization of graduates with the technology of assessment of tasks of Part C

Part C assignments are checked by experienced expert teachers, unlike parts A and B, which are checked using a computer. Therefore, when preparing answers to the tasks of part C, it is important, if possible, not to use abbreviations in words and write down the solution of problems as fully as possible.

You can complete the solution of any task of part C from any link, each of which has its own price of 1 point. In this case, graduates will score a certain number of points from the maximum provided by the test for the complete and correct completion of the task. For example, almost every examinee will be able to identify the oxidizing agent and reducing agent in problem C1 or write down the reaction equation for problem C4, thereby providing himself with 1 point for each action.

In other words, they need to complete all the fragments that they can complete for each task of part C.

The teacher needs to bring to the attention of students that when developing assessment criteria, the features of checking the assimilation of the elements of the content of all five tasks with a detailed answer included in the examination paper are taken into account. It also takes into account the fact that the wording of the examinees' answers can be either very general, streamlined and not specific, or too short and not sufficiently reasoned. Close attention is also paid to the distribution of the text of the original response to equivalent content elements, estimated at one point. This takes into account the inevitability of a gradual increase in the difficulty of obtaining each subsequent score for a correctly formulated content element.

Thus, when compiling a scale for evaluating computational problems, the multivariance of the ways to solve them is taken into account, and, consequently, the presence in the answer of its main stages and results indicated in the evaluation criteria. A common feature of the assessment of all tasks with a detailed answer, the teacher emphasizes, is the need to fix the conditions for the implementation of a given chemical reaction in the answers.

Let us illustrate what has been said on examples of evaluating certain types of tasks with a detailed answer used in the KIMs of the Unified State Examination.

Exercise.

SO 2 + K 2 Cr 2 O 7 + ... → K 2 SO 4 + ... + H 2 O

	Points
Response elements: │ S +4 - 2 e → S +6 2 │ Cr +6 + 3 e → Cr +3 3SO 2 + K 2 Cr 2 O 7 + H 2 SO 4 \u003d K 2 SO 4 + Cr 2 (SO 4) 3 + H 2 O 3) it is indicated that sulfur in the +4 oxidation state (sulfur dioxide due to sulfur in the +4 oxidation state) is a reducing agent, and chromium in the +6 oxidation state (or potassium dichromate due to chromium in the +6 oxidation state) is an oxidizing agent
Maximum score

Let us illustrate the evaluation by experts on the example of the original work of a graduate.

	Points
C1. 3SO 2 + K 2 Cr 2 O 7 + H 2 SO 4 \u003d K 2 SO 4 + Cr 2 (SO 4) 3 + H 2 O S +4 - 2 e → S +6 6 3 2Cr +6 + 6 e → Cr +3 2 1
The answer is correct, but not complete: an electronic balance has been drawn up, the coefficients in the reaction equation are placed, but the oxidizing agent and reducing agent are not indicated
Maximum score

It will be helpful if the teacher asks students to complete a similar task and then evaluate this performance in accordance with the proposed assessment criteria.

For example.

Using the electron balance method, write the equation for the reaction:

P + HNO 3 +… → NO +…

Determine the oxidizing agent and reducing agent.

	Points
Response elements: 1) an electronic balance has been drawn up: 3 P 0 - 5 e → P +5 2 N +5 + 3 e → N +2 2) the coefficients are placed in the reaction equation: 3 P + 5HNO 3 +2 H 2 O \u003d 5 NO + 3 H 3 PO 4 3) it is indicated that phosphorus in oxidation state 0 is a reducing agent, and nitrogen in oxidation state +5 (or nitric acid due to nitrogen in oxidation state +5) is an oxidizing agent
The answer is correct and complete, includes all the above elements
The answer contains an error in only one of the elements
There are two errors in the response.
All elements of the answer are written incorrectly
Maximum score

When working out this stage of preparing graduates for the Unified State Examination -2010, you can use the original of one of the works of graduates of schools in the Moscow region.

It is easy to see that this work deserves only 1 point, since, despite the compilation of the electronic balance, it does not indicate which element (substance) is an oxidizing agent and which is a reducing agent. Also, the graduate in his work did not place the coefficients in the reaction equation.

Typical errors of part C (2006-2007)

Task C1.

Typical mistakes: when determining possible products, the reaction medium, starting materials are not taken into account. For example:

P + HNO 3 → P 2 O 5 + ... - nitric acid, even concentrated, always contains water, phosphorus oxide interacts vigorously with water - can it be formed in an aquatic environment? Of course not, the right product is H 3PO4.

K 2 Cr 2 O 7 + ... H 2 SO 4 → ... + Cr (OH) 3 + ... - chromium (III) hydroxide - a base, albeit an amphoteric one, can it be obtained in an acidic environment? Or Cr oxide 2O3 ? Of course not, the right product is Cr 2 (SO 4 ) 3 .

An offensive mistake - everything seems to be correct, but the oxidizing agent is not specified, as a result, the score is lost. Or are the letters “o” - “v” written, and figure out what a person meant by this: “oxidizer” or “oxidation”?

Task C2.

Common Mistake #1: Reacting Metals with Nitric Acid - The vast majority of participants write: Me + HNO 3 →… + H 2 .

When nitric acid reacts with reducing agents, the nitrate ion is reduced.

Typical mistake #2: The possibility of OVR occurring along with exchange reactions is not taken into account, for example:

CuS + HNO 3 → Cu(NO 3 ) 2 + H 2 S. - Nitric acid, as already mentioned, is an oxidizing agent, sulfur in the oxidation state (-2) is a strong reducing agent, therefore, it is not an exchange reaction that occurs, but an OVR:

CuS + HNO 3 → Cu(NO 3 ) 2 + H 2 SO 4 + NO 2 + H 2 O.

Or: Fe 2 O 3 + HI → FeI 3 + H 2 O. - Iron (+3) is an oxidizing agent, iodide ion is a good reducing agent, so the real process can be expressed by the scheme: Fe 2 O 3 + HI → FeI 2 + I 2 + H 2 O.

Offensive mistakes: the reaction scheme is correct, but the coefficients are not placed. If he couldn’t, then there’s nothing to be done, and if it’s due to inattention, then it’s a shame, points are lost.

Typical mistake No. 2: Simplified reaction equations are written that do not take into account media, without indicating inorganic products: CH 3 CHO + Ag 2 O → CH 3 COOH + 2Ag - the reaction takes place in the presence of an excess of ammonia, which, of course, reacts with an acid, the product is a salt:

CH 3 CHO + Ag 2 O + NH 3 → CH 3 COONH 4 + 2Ag; or more precisely like this:

CH 3 CHO + 2OH → CH 3 COONH 4 + 3NH 3 + 2Ag

Or when oxidized with permanganate, it is written: C 6 H 5 CH 3 + [O] → C 6 H 5 COOH - without considering what happened to permanganate, what other products are formed ....

Common Mistake #3: Lack of odds.

C2 Substances are given: sulfur, potassium hydroxide, nitric acid, phosphoric acid. Write equations for four possible reactions between these substances.

(Other formulations of the answer are allowed that do not distort its meaning)	Points
Response elements:
The answer is correct and complete, includes all the above elements
All elements of the answer are written incorrectly
Maximum score

Given substances: magnesium, concentrated sulfuric acid, nitrogen, ammonium chloride.

(Other formulations of the answer are allowed that do not distort its meaning)	Points
Response elements: 1) 3Mg + N 2 = Mg 3 N 2 2) Mg + 2NH 4 Cl \u003d MgCl 2 + 2NH 3 + H 2 3) 2NH 4 Cl (solid) + H 2 SO 4 (conc.) \u003d (NH 4) 2 SO 4 + 2HC1 4) 4Mg + 5H 2 SO 4 (conc.) = 4MgSO 4 + H 2 S + 4H 2 O
The answer is correct and complete, includes all the above elements
Correctly written 3 reaction equations
Correctly written 2 reaction equations
Correctly written one reaction equation
All elements of the answer are written incorrectly
Maximum score

Substances given: lead(11) sulfide, sodium sulfite, hydrogen peroxide, concentrated sulfuric acid. Write the equations for the four possible reactions between these substances.

(Other formulations of the answer are allowed that do not distort its meaning)	Points
Response elements: Four equations of possible reactions involving the indicated substances are written: 1) PbS + 4H 2 O 2 \u003d PbSO 4 + 4H 2 O 2) PbS + H 2 SO 4 = H 2 S + PbSO 4 3) Na 2 SO 3 + H 2 O 2 = Na 2 SO 4 + H 2 O 4) Na 2 SO 3 + H 2 SO 4 = Na 2 SO 4 + SO 2 + H 2 O
The answer is correct and complete, includes all the above elements
Correctly written 3 reaction equations
Correctly written 2 reaction equations
Correctly written one reaction equation
All elements of the answer are written incorrectly
Maximum score

Substances are given: potassium sulfite, hydrogen sulfide, sulfuric acid, potassium permanganate solution.

Write the equations for the four possible reactions between these substances.

(Other formulations of the answer are allowed that do not distort its meaning)	Points
Response elements: Four equations of possible reactions involving the indicated substances are written: 1) 2KMnO 4 + 3H 2 S \u003d 2MnO 2 + 3S + 2KOH + 2H 2 O 2) 3K 2 SO 3 + 2KMnO 4 + H 2 O \u003d 2MnO 2 + 3K 2 SO 4 + 2KOH 3) 5K 2 SO 3 + 2KMnO 4 + 3H 2 SO 4 \u003d 2MnSO 4 + 6K 2 SO 4 + 3H 2 O 4) K 2 SO 3 + H 2 SO 4 = K 2 SO 4 + SO 2 + H 2 O
The answer is correct and complete, includes all the above elements
Correctly written 3 reaction equations
Correctly written 2 reaction equations
Correctly written one reaction equation
All elements of the answer are written incorrectly
Maximum score

Given substances: bromine, hydrogen sulfide, sulfur dioxide, concentrated nitric acid.

Write the equations for the four possible reactions between these substances.

(Other formulations of the answer are allowed that do not distort its meaning)	Points
Response elements: Four equations of possible reactions involving the indicated substances are written: 1) H 2 S + Br 2 \u003d 2HBr + S 2) 2H 2 S + SO 2 \u003d 3S + 2H 2 O 3) H 2 S + 2HNO 3 \u003d SO 2 + 2NO 2 + S + 2H 2 O 4) SO 2 + 2HNO 3 (conc.) = H 2 SO 4 + 2NO 2
The answer is correct and complete, includes all the above elements
Correctly written 3 reaction equations
Correctly written 2 reaction equations
Correctly written one reaction equation
All elements of the answer are written incorrectly
Maximum score

Substances given: copper, iron(III) chloride, concentrated nitric acid, sodium sulfide.

Write the equations for the four possible reactions between these substances.

(Other formulations of the answer are allowed that do not distort its meaning)	Points
Response elements: Four equations of possible reactions involving the indicated substances are written: 1) 2FeС1 3 + Cu \u003d CuCl 2 + 2FeCl 2 2) Na 2 S + 4HNO 3 \u003d 2NaNO 3 + 2NO 2 + S + 2H 2 O 3) 2FeС1 3 + 3Na 2 S = 2FeS + S + 6NaCl 4) Cu + 4HNO 3 (conc.) \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
The answer is correct and complete, includes all the above elements
Correctly written 3 reaction equations
Correctly written 2 reaction equations
Correctly written one reaction equation	1
All elements of the answer are written incorrectly	0
Maximum score	4

Cl2 KOH, alcohol С act, 650° KMnO4 , H2 SO4

ethene → X1 → X2 → X3 → toluene →X4

(Other formulations of the answer are allowed that do not distort its meaning)	Points
Response elements: 1) C2 H4 + Сl2 → CH2 Cl-CH2 Cl C2 H5 OH(X1 =C2 H4 Cl2 ) 2)CH2 Cl-CH2 Cl + 2KOH → HC≡CH + 2KS1 + 2H2 O C act, 650° (X2 = C2 H2 ) 3)3C2 H2 → C6 H6 (X3 = C6 H6 ) AlCl3 , t° 4) C6 H6 + CH3 C1 → C6 H3 -CH3 + HC1 5) 5C6 H5 -CH3 + 6KMnO4 + 9H2 SO4 = 6MnSO4 + 3K2 SO4 + 5C6 H5 -COOH + 14H2 O (X5 =C6 H5 -COOH)
The answer is correct and complete, includes all the above elements	5
	4
Correctly written 3 reaction equations	3
Correctly written 2 reaction equations	2
Correctly written one reaction equation	1
All elements of the answer are written incorrectly	0
Maximum score	5

Write the reaction equations that can be used to carry out the following transformations:

Correct Answer Content and Grading Instructions (Other formulations of the answer are allowed that do not distort its meaning)	Points
Response elements: The reaction equations corresponding to the transformation scheme are written:
The answer is correct and complete, includes all the above elements	5
Correctly written 4 reaction equations	4
Correctly written 3 reaction equations	3
Correctly written 2 reaction equations	2
Correctly written one reaction equation	1
All elements of the answer are written incorrectly	0
Maximum score	5

To the solution obtained by adding 20 g of potassium hydride to 500 ml of water was added 100 ml of a 32% hydrochloric acid solution (density 1.16 g/ml). Determine the mass fractions of substances in

the resulting solution.

Response elements:

KH + H2 O = H2 + KOH

KOH + HC1 = KS1 + H2 O

m (r-ra HC1) \u003d p V \u003d 1.16 100 \u003d 116 (g)

m(HCl) \u003d m (p-pa HCl) w \u003d 116 0.32 \u003d 37.12 (g)

n(HCl) = m(HCl): M(HCl)= 37,12: 36,5 = 1.02 (mol)

n(KOH)= n(KH) \u003d m: M \u003d 20: 40 \u003d 0.5 (mol)excess HCl

n(KCl) = n(KOH) = 0.5 (mol)

m(KCl) \u003d M n \u003d 74.5 0.5 \u003d 37.25 (g)

n(H2 ) = n(KH) = 0.5 (mol);m(H2 ) = M n = 2 0.5 = 1 (d)

n(ex.HC1) = 1.02 - 0.5 = 0.52 (mol)

m (ex. HC1) \u003d M n \u003d 36.5 0.52 \u003d 18.98 (g)

m(solution) = m(KH) + m(H2 O) + m(p-pa HCl) - m(H2 ) =

20 + 500 + 116 - 1 = 635 (g)

w (KCl) \u003d m (KCl) : m (solution) \u003d 37.25: 635 \u003d 0.059, or 5.9%

w(HCl) = m(ex.HC1): m(solution) = 18.98: 635 = 0.03, or 3%

27.2 g of a mixture of calcium and aluminum carbides was treated with acid, 11.2 liters of a mixture of gases (at n.o.) were obtained. Determine the volume fraction of acetylene in the mixture.

Content of the correct answer

(Other formulations of the answer are allowed that do not distort its meaning)

Response elements:

CaS2 + 2HC1 = CaCl2 + C2 H2 M(CaS2 ) = 64 g/mol

Al4 C3 + 12HC1 = 4A1C13 + 3CH4 M(A14 With3 ) = 144 g/mol

n(СаС2 ) = n(С2 H2 ) = X n(А14 With3 ) = y n(CH4 ) = Zu

n(CH4 + C2 H2 ) = V: Vn, = 11.2: 22.4 = 0.5 (mol)

x + 3y = 0.5

=> x = 0.2; y = 0.1

64x + 144y = 27.2

φ(C2 H2 ) = V(C2 H2 ): V(CH4 +C2 H2 ) = n(C2 H2 ): n(CH4 +C2 H2 ) =

0.2: 0.5 = 0.4, or 40%

The vapor density of organic matter in terms of oxygen is 1.875. When burning 15 g of this substance, 16.8 liters of carbon dioxide (at n.o.) and 18 g of water are formed. Determine the composition of the organic

substances.

Response elements:

M(СхНУОz) = D М(O2 ) = 1.875 32 = 60 (g/mol)

n(CxHyOz) \u003d m: M \u003d 15: 60 \u003d 0.25 (mol)

n(CO2 ) = V: Vm\u003d 16.8: 22.4 \u003d 0.75 (mol) => n (C) \u003d 0.75 (mol)

n(N2 O) \u003d m: M \u003d 18: 18 - 1 (mol) => n (H) \u003d 2 (mol)

n(CxHyOz) : n(C) : n(H) = 0.25: 0.75: 2 = 1: 3: 8 = > x = 3; y = 8

M(S3 H8 Oz)=12 3 + 1 8 + 16 Z

44 + 16 z = 60=> z=l

Composition of organic matter C3 H8 O

For the complete neutralization of a solution containing 18.5 g of the limiting monobasic carboxylic acid, 50 g of a 20% sodium hydroxide solution was used. Determine the composition of the acid.

Content of the correct answer

(Other formulations of the answer are allowed that do not distort its meaning)

Response elements:

1) m(NaOH) - m(p-pa) w(NaOH) = 50 0.2 = 10 (g)

n(NaOH) \u003d m: M \u003d 10: 40 \u003d 0.25 (mol)

CnH2n+1COOH + NaOH = CnH2n+1COONa + H2 O

n(CnH2n+1COOH) = n(NaOH) = 0.25 (mol)

2) M(CnH2n+1COOH) \u003d m: n \u003d 18.5: 0.25 \u003d 74 (g / mol)

12n + 2n + 1 + 12 + 16 2 + 1 = 74 => n = 2

Composition of acid C2 H5 COOH

In the tasks of part C, the most difficult were those where it was necessary

show knowledge of the following reactions:

- interaction of complex salts K3 [A1(OH)6 ], K3[Cr (OH)6 ] with weak acids (H2 S, N2 O + CO2 , N2 O + SO2 ) or acid solutions of salts strongly hydrolyzed by cation (FeCl3 , А1С13 , СrСl3 );

-reactions involving H2 O2 as an oxidizing agent (with H2 S, SO2 , TO3 [Cr(OH)6 ]);

- decomposition of KClO3 ;

- the interaction of salt solutions formed by a weak base and a weak acid (CrCl3 and K2 SO3 , А1С13 and Na2 S);

- the interaction of an acid and its average salt with the formation of an acid salt (K2 SO3 + H2 O + CO2 ; Na2 S+H2 S);

- interaction of phosphorus with concentrated sulfuric and nitric acids;

-characterizing properties of amphoteric oxides (including the fusion of aluminum oxide with sodium carbonate to form sodium metaaluminate and carbon dioxide);

- interaction of chlorine with alkalis in the cold and when heated;

- interaction of iron with nitric acid at different degrees of its dilution;

-characterizing the properties of concentrated sulfuric and nitric acids as oxidizing agents in reactions not only with metals, but also with non-metals, and with complex substances;

- Wurtz reaction;

- interaction of an alcohol solution of alkali with halogen-substituted alkanes;

- alkylation of amines;

- alkylation of benzene and its homologues;

- obtaining acetaldehyde by catalytic oxidation of ethylene.

Typical mistakes when passing the exam in chemistry

Chemistry teacher MOUSOSH №9 Shapkina Zh.A.

The Unified State Exam in Chemistry, as an experiment, has been held in some regions of the Russian Federation since 2002. During this time, there has been an increase in the number of participants. So, if in 2002 5320 graduates from 10 regions of the Russian Federation took part in the exam, then in 2003 23778 graduates from 34 regions took the exam, in 2004 - 28069 graduates from 50 regions, and in 2006 - 30389 graduates from 58 regions.

The average score scored by the participants is 49% (from 2002 to 2006). The number of graduates who scored 100 points increased from 3 people in 2003 to 39 people in 2006.

Conducting the Unified State Examination in Chemistry for a number of years allows us to analyze the typical mistakes made by graduates when completing assignments.

One of the significant shortcomings of the USE is the inability to familiarize applicants with the mistakes made during the performance of the test tasks of parts A and B, which does not allow them to be analyzed in detail, deprives applicants of the legal right to appeal and creates tension among parents and students. This situation is exacerbated by the fact that the control and measuring materials are not developed enough, many questions are posed incorrectly, and there are errors in the proposed answers.

Let's comment on some test tasks.

In periods, the reducing properties of atoms of chemical elements with an increase in their serial number:

1) weaken, 2) increase, 3) do not change, 4) change periodically.

Answer 1 is given as correct. At the same time, it is known that the reducing properties of atoms of chemical elements with an increase in their serial number weaken in a period, and in periods they change periodically. So the correct answer would be 4.

Acetic aldehyde reacts with each of two substances: 1) an ammonia solution of silver oxide ( I ) and oxygen; 2) copper hydroxide ( II ) and calcium oxide; 3) hydrochloric acid and silver; 4) sodium hydroxide and hydrogen.

Answer 1 is considered correct. However, it is known that acetaldehyde easily reacts with aldol condensation in the presence of alkalis, and forms primary alcohols with hydrogen, so answer 4 is also correct.

Phenol interacts with solutions: A) Cu ( Oh ) 2 ; B) FeCl 3 ; AT) H 2 SO 4 ; G) Br 2 ; D)[ Ag ( NH 3 ) 2 ] Oh ; E) Na 2 CO 3

The correct answer is B, D, F, but phenol does not react with sodium carbonate, since it exhibits weaker acidic properties than carbonic acid. So the correct answer is B, D.

Some tasks include questions that are not provided for by school programs. For example, in task C1, it was proposed to write the equations for reactions occurring at the anode and cathode, and the general equations for the electrolysis of aqueous solutions not only of salts, which is included in the school curriculum, but also of acids and alkalis, which is not included in the program. The average percentage of completion of the task on the topic "Electrolysis" is 40. Typical mistakes in completing this task are as follows: graduates confused the signs of the electric charge of the cathode and anode; we did not take into account the sequence of discharge of particles present in the solution at the cathode and anode, including depending on their concentration; did not indicate (or indicated incompletely) the conditions of electrolysis - mixing, the presence or absence of a diaphragm, temperature, concentration; could not write the overall equation of the electrolysis process.

Some tasks that more than 75% of students failed to complete had an unusual wording for them. For example, questions on the topic "Hydrolysis".

Establish a correspondence between the composition of salt and the reaction of the environment of its aqueous solution.

Salt: 1) potassium nitrate, 2) aluminum sulfate, 3) potassium sulfide, 4) sodium orthophosphate.

Medium: A) acidic, B) neutral, C) alkaline.

Almost all students completed this task.

Establish a correspondence between the composition of salt and the type of its hydrolysis. Salt formula: 1) BeSO 4 , 2) KNO 3 , 3) Pb ( NO 3 ) 2 , 4) CuCl 2 Type of hydrolysis: A) by cation, B) by anion, C) by cation and anion.

Only 23.3% of students coped with the task in this formulation, since in the school curriculum they do not study such names of hydrolysis types as “by cation”, “by anion”. A very common mistake is to count HF strong acid.

In assignments for correspondence, it must be remembered that the answer under the same letter can be used several times, i.e. the same letter is the correct answer to several questions.

Many mistakes were made in the answers to questions containing negation. Students forget to take into account negation. For example:

Zinc oxide does not react with 1) HCl , 2) NaOH , 3) H 2 O , 4) H 2 SO 4

In the tasks of part B, attention should be paid to students' knowledge of the physical properties of substances, their application, production in industry and in the laboratory. Since graduates often find it difficult to answer such simple questions as “Does this substance have a color or smell?”

In the tasks of part B, another reason for errors appears due to the replacement of letters with numbers. This is an indication of the coefficients, not the numbers of the correct answers.

In task B3, it is necessary to establish a correspondence between the starting substances and the sum of all coefficients in the full or reduced ionic equation. One of the typical mistakes is that many students forget to take into account the coefficient 1, which is not written in the equations. Another common mistake is that when moving from a full ionic equation to a reduced one, students forget that coefficients can also be reduced if they can all be divided by the same number.

For example:

Establish a correspondence between the starting substances and the sum of all coefficients in the reduced ionic reaction equation. Starting substances: A) Al 2 ( SO 4 ) 3 + KOH , B) Ba ( Oh ) 2 + HNO 3 , AT) Zn ( Oh ) 2 + HCl , G) MgCl 2 + Na 2 CO 3 .Sum of coefficients: 1)3, 2)4, 3)5, 4)6.

The correct answer is 3141, not 5363. You need to know that numbers can be repeated in the answers.

Problems are also caused by tasks with multiple choice, for example:

The reagents for carbon dioxide and ethylene are solutions of: 1) potassium permanganate, 2) nitric acid, 3) calcium hydroxide, 4) sodium chloride, 5) copper sulfate ( II ), 6) hydrogen chloride. Answer ...

It is not known how many digits should be in the answer, and it is desirable to write in the answer all the necessary ones, and not write anything extra. The second feature is that the numbers must be specified in ascending order. If you first write, for example, "24", and then think and attribute "1", then the answer "241" will be considered incorrect, even if "124" is correct.

The calculation problems of part B are not too difficult, but many errors are made during rounding.

Task C1 - OVR. Many errors are probably due to inattention: students, having written the equation correctly, forget to indicate the oxidizing agent and lose the score.

Tasks C2 and C3 are aimed at testing students' knowledge of the relationship between inorganic and organic substances (chains of transformations) and included 5 content elements: 5 reaction equations indicating the conditions for their occurrence. The maximum score for completing this task is 5. Some tasks included transformations of chromium and iron compounds, the study of which is not provided for by the school curriculum.

Write the reaction equations that can be used to carry out the following transformations:

Cr2S3 X 1 > K 2 CrO 4 X 1 X 2 > KCrO 2

(average score was 0.3 out of 5);

K 2 Cr 2 O 7 X > K 3 [ Cr ( Oh ) 6 ] X > KCrO 2 X

(mean score was 0.4);

KFeO 2 X 1 X 2 X 1 > Na 2 FeO 4 X 3

(average score - only 0.1);

Fe 3 O 4 > Fe ( NO 3 ) 3 X 1 X 2 X 3 > K 2 FeO 4

(average score - 0.3).

The tasks in which there is no scheme for the relationship of substances turned out to be very difficult, for example:

P X 1 X 2 X 4 X 5

The average score for this task was 0.2 out of 5 possible, which is not surprising, since all the substances sought are encrypted. Despite the fact that the reactions are quite simple and well-known, an error at any stage, especially at the first one, does not leave students with a chance to complete the task as a whole.

Task C3 - a chain of transformations of organic substances. Typical mistakes in this task are as follows: correctly indicating the main product of the reaction, the student does not indicate side substances, does not place the coefficients. The reaction conditions are not taken into account when determining their products. So, during the hydrolysis of esters in an alkaline medium, free acids are indicated as products, and acids are also indicated during the oxidation of aldehyde in the “silver mirror” reaction, although this reaction proceeds in an excess of ammonia solution and its products are ammonium salts.

Typical errors in task C4 (combined task) are errors in the nomenclature: the examiner does not understand the difference between nitrate-nitrite-nitride, carbonate-carbide, phosphate-phosphide, chlorate-chlorite-chloride, etc.

There are many errors in the equations for the reactions of copper with nitric acid, chlorine with alkali, decomposition of nitrates, chlorates.

Many errors are caused by the inability to take into account all the substances in a given system, in a given solution. So, having determined, in one of the tasks, that nitric acid remained in excess, schoolchildren “forget” about it when sodium hydroxide is added to the solution. Or, finding the mass of the solution, they do not take into account that a precipitate has fallen out of it.

There are also errors in calculations according to the reaction equations, in the analysis of the excess-lack of the reagent.

In task C5, the definition of the formula of a substancecaused significant difficulties for the examiners. These difficulties, firstly, are due to the fact that some of the proposed tasks for finding the formulas of substances included an element of the solution unfamiliar to students. In particular, it was required to establish the true formula by selection, without having data on the molar mass of the substance.

With the complete combustion of gaseous organic matter that does not contain oxygen, 4.48 l (N.O.) of carbon dioxide, 1.8 g of water and 4 g of hydrogen fluoride were released. Set the molecular formula of the burned compound, calculate its volume and mass.

As a result, the examiners found only the simplest formula, but could not determine the true one.

In tasks on combustion products, hydrogen is lost, which is in the composition of hydrogen halides.

Sometimes - an incorrect transition from the amount of substance of the combustion product to the amount of substance of the element: n (H 2 O) n (H).

Hydrogen is often lost by over-rounding in calculations.

Using the relative density of nitrogen, hydrogen, oxygen, sometimes the student "forgets" that the molecules of these gases are diatomic.

In order to improve the quality of passing the exam, it is necessary to give students a few tips.

For Part A:

1) plan 2-3 rounds of work on questions. On the first lap, everything is too difficult to skip. On the second - think, on the third - guess.

2) Working on a difficult question, decide if cheat sheet #1 (Periodic Table) can be used to answer? Cheat Sheet #2 (Solubility Table)? Cheat sheet No. 3 (Series of metal stresses)?

3) If you see several correct answers in the proposed answers, then first re-read the question, did you understand it correctly, did you miss the negative? Do you confuse what is possible in principle with what is done in practice? Then choose the most typical, most obvious option.

Don't forget that part A tests knowledge of the most obvious things.

If in the question “what metal interacts with water” there are options “iron”, “sodium”, “aluminum”, then remember that pipes and pans are still not made of sodium.

4) In the proposed options, you do not see a single correct one, which means you need to re-read the question, did you understand it correctly, did you miss the negation? If this does not help, remember that there are exceptions to many rules in chemistry. Are there any special properties of the presented substances? Special conditions for reactions?

For Part B:

1) At the first stage of work, write down the answers to the questions in the text of the assignment, in special plates or fields. Only after final checking, transfer them to the answer sheet.

2) A task with a short free answer is considered to be completed correctly if the sequence of numbers (number) is correctly indicated.

3) For a complete correct answer to tasks B1-B8, 2 points are given, for a correct incomplete answer - 1 point, for an incorrect answer or its absence - 0 points.

For part C:

1) The difficulty coefficient of the tasks in part C is large, so 1 point in part C can be worth several points in part A, so you should try to do at least something in part C.

2) Try to present this something as legibly as possible.

Part C on the exam in chemistry begins with task C1, which involves compiling a redox reaction (already containing part of the reagents and products). It is worded like this:

C1. Using the electron balance method, write an equation for the reaction. Determine the oxidizing agent and reducing agent.

Often applicants believe that this task does not require special preparation. However, it contains pitfalls that prevent you from getting a full score for it. Let's see what to pay attention to.

Theoretical information.

Potassium permanganate as an oxidizing agent.

+ reducing agents
in an acidic environment	in a neutral environment	in an alkaline environment
(salt of the acid involved in the reaction)		Manganate or, -

Dichromate and chromate as oxidizing agents.

(acidic and neutral environment), (alkaline environment) + reducing agents always turns out
acidic environment	neutral environment	alkaline environment
Salts of those acids that participate in the reaction:		in solution or melt

Increasing the oxidation states of chromium and manganese.

+ very strong oxidizing agents (always regardless of the medium!)
, salts, hydroxo complexes	+ very strong oxidizing agents: a), oxygen-containing salts of chlorine (in an alkaline melt) b) (in alkaline solution)	Alkaline environment: formed chromate
, salt	+ very strong oxidizing agents in an acidic environment or	Acid environment: formed dichromate or dichromic acid

- oxide, hydroxide, salts	+ very strong oxidizing agents: , oxygen-containing salts of chlorine (in melt)	Alkaline environment: manganate
- salt	+ very strong oxidizing agents in an acidic environment or	Acid environment: Permanganate - manganese acid

Nitric acid with metals.

- no hydrogen is released, nitrogen reduction products are formed.

The more active the metal and the lower the acid concentration, the further nitrogen is reduced.
Nonmetals + conc. acid	Inactive metals (to the right of iron) + dil. acid	Active metals (alkaline, alkaline earth, zinc) + conc. acid	Active metals (alkali, alkaline earth, zinc) + medium dilution acid	Active metals (alkaline, alkaline earth, zinc) + very dil. acid
Passivation: do not react with cold concentrated nitric acid:
do not react with nitric acid at any concentration:

Sulfuric acid with metals.

- diluted sulfuric acid reacts like an ordinary mineral acid with metals to the left of the voltage series, while hydrogen is released;
- when reacting with metals concentrated sulfuric acid no hydrogen is released, sulfur reduction products are formed.

Inactive metals (to the right of iron) + conc. acid Nonmetals + conc. acid	Alkaline earth metals + conc. acid	Alkali metals and zinc + concentrated acid.	Dilute sulfuric acid behaves like a normal mineral acid (like hydrochloric acid)
Passivation: do not react with cold concentrated sulfuric acid:
do not react with sulfuric acid at any concentration:

Disproportionation.

Disproportionation reactions are reactions in which the same the element is both an oxidizing agent and a reducing agent, both raising and lowering its oxidation state:

Disproportionation of non-metals - sulfur, phosphorus, halogens (except fluorine).

Sulfur + alkali 2 salts, metal sulfide and sulfite (reaction occurs during boiling)	and
Phosphorus + alkali phosphine and salt hypophosphite(reaction proceeds at boiling)	and
Chlorine, bromine, iodine + water (without heating) 2 acids, Chlorine, bromine, iodine + alkali (without heating) 2 salts, and and water	and
Bromine, iodine + water (when heated) 2 acids, Chlorine, bromine, iodine + alkali (when heated) 2 salts, and and water	and

Disproportionation of nitric oxide (IV) and salts.

+ water 2 acids, nitric and nitrogenous + alkali 2 salts, nitrate and nitrite	and
	and
	and

Activity of metals and non-metals.

To analyze the activity of metals, either the electrochemical series of metal voltages or their position in the Periodic Table is used. The more active the metal, the easier it will donate electrons and the better it will be as a reducing agent in redox reactions.

Electrochemical series of voltages of metals.

Features of the behavior of some oxidizing and reducing agents.

a) oxygen-containing salts and acids of chlorine in reactions with reducing agents usually turn into chlorides:

b) if substances participate in the reaction in which the same element has a negative and positive oxidation state, they occur in the zero oxidation state (a simple substance is released).

Required skills.

1. Arrangement of oxidation states.
   It must be remembered that the degree of oxidation is hypothetical the charge of an atom (i.e. conditional, imaginary), but it should not go beyond common sense. It can be integer, fractional, or zero.

   Exercise 1: Arrange the oxidation states of the substances:

2. Arrangement of oxidation states in organic substances.
   Remember that we are only interested in the oxidation states of those carbon atoms that change their environment in the redox process, while the total charge of the carbon atom and its non-carbon environment is taken as 0.

   Task 2: Determine the oxidation state of the carbon atoms circled along with the non-carbon environment:

   2-methylbutene-2: - =

   acetone:

   acetic acid: -

3. Do not forget to ask yourself the main question: who donates electrons in this reaction, and who accepts them, and what do they turn into? So that it does not work that electrons arrive from nowhere or fly away to nowhere.

   Example:

   In this reaction, one must see that potassium iodide can be only reducing agent, so potassium nitrite will accept electrons, lowering its degree of oxidation.
   Moreover, under these conditions (dilute solution) nitrogen goes from to the nearest oxidation state.

4. Drawing up an electronic balance is more difficult if the formula unit of a substance contains several atoms of an oxidizing or reducing agent.
   In this case, this must be taken into account in the half-reaction by calculating the number of electrons.
   The most common problem is with potassium dichromate, when it goes into the role of an oxidizing agent:

   These deuces cannot be forgotten when calling, because they indicate the number of atoms of a given type in the equation.

   Task 3: What coefficient should be put before and before

   
   

   Task 4: What coefficient in the reaction equation will stand in front of magnesium?

5. Determine in which medium (acidic, neutral or alkaline) the reaction takes place.
   This can be done either about the products of the reduction of manganese and chromium, or by the type of compounds that were obtained on the right side of the reaction: for example, if in the products we see acid, acid oxide- it means that this is definitely not an alkaline environment, and if metal hydroxide precipitates, it is definitely not acidic. And of course, if on the left side we see metal sulfates, and on the right - nothing like sulfur compounds - apparently, the reaction is carried out in the presence of sulfuric acid.

   Task 5: Determine the environment and substances in each reaction:

6. Remember that water is a free traveler, it can both participate in a reaction and be formed.

   Task 6:Which side of the reaction will the water be on? What will the zinc go to?

   Task 7: Soft and hard oxidation of alkenes.
   Add and equalize the reactions, after placing the oxidation states in organic molecules:

   (cold solution)

   (aqueous solution)

7. Sometimes a reaction product can be determined only by compiling an electronic balance and understanding which particles we have more:

   Task 8:What other products will be available? Add and equalize the reaction:

8. What are the reactants in the reaction?
   If the schemes we have learned do not give an answer to this question, then we need to analyze which oxidizing agent and reducing agent in the reaction are strong or not?
   If the oxidizer is of medium strength, it is unlikely that it can oxidize, for example, sulfur from to, usually oxidation only goes up to.
   Conversely, if is a strong reducing agent and can recover sulfur from up to , then only up to .

   Task 9: What will the sulfur turn into? Add and equalize the reactions:

   (conc.)

9. Check that there is both an oxidizing agent and a reducing agent in the reaction.

   Task 10: How many other products are in this reaction, and which ones?

10. If both substances can exhibit the properties of both a reducing agent and an oxidizing agent, it is necessary to consider which of them more active oxidant. Then the second one will be the restorer.

    Task 11: Which of these halogens is the oxidizing agent and which is the reducing agent?

11. If one of the reactants is a typical oxidizing agent or a reducing agent, then the second one will “do his will”, either by donating electrons to the oxidizing agent, or by accepting them from the reducing agent.

    Hydrogen peroxide is a substance with dual nature, in the role of an oxidizing agent (which is more characteristic of it) passes into water, and as a reducing agent - passes into free gaseous oxygen.

    Task 12: What role does hydrogen peroxide play in each reaction?

The sequence of arrangement of the coefficients in the equation.

First put down the coefficients obtained from the electronic balance.
Remember that you can double or reduce them only together. If any substance acts both as a medium and as an oxidizing agent (reducing agent), it will have to be equalized later, when almost all the coefficients are arranged.
Hydrogen is equalized penultimately, and we only check for oxygen!

Take your time counting the oxygen atoms! Remember to multiply rather than add indices and coefficients.
The number of oxygen atoms on the left and right sides must converge!
If this does not happen (provided that you count them correctly), then there is a mistake somewhere.

Possible mistakes.

1. Arrangement of oxidation states: check each substance carefully.
   Often mistaken in the following cases:

   a) oxidation states in hydrogen compounds of non-metals: phosphine - oxidation state of phosphorus - negative;
   b) in organic substances - check again whether the entire environment of the atom is taken into account;
   c) ammonia and ammonium salts - they contain nitrogen always has an oxidation state;
   d) oxygen salts and acids of chlorine - in them chlorine can have an oxidation state;
   e) peroxides and superoxides - in them, oxygen does not have an oxidation state, it happens, and in - even;
   f) double oxides: - in them, metals have two different oxidation states, usually only one of them is involved in the transfer of electrons.

   Task 14: Add and equalize:

   Task 15: Add and equalize:

2. The choice of products without taking into account the transfer of electrons - that is, for example, in the reaction there is only an oxidizing agent without a reducing agent, or vice versa.

   Example: free chlorine is often lost in a reaction. It turns out that electrons came to manganese from outer space...

3. Incorrect products from a chemical point of view: a substance that interacts with the environment cannot be obtained!

   a) in an acidic environment, metal oxide, base, ammonia cannot be obtained;
   b) in an alkaline environment, acid or acid oxide will not be obtained;
   c) an oxide, let alone a metal that reacts violently with water, is not formed in an aqueous solution.

   Task 16: Find in reactions erroneous products, explain why they cannot be obtained under these conditions:

Answers and solutions to tasks with explanations.

Exercise 1:

Task 2:

2-methylbutene-2: - =

acetone:

acetic acid: -

Task 3:

Since there are 2 chromium atoms in the dichromate molecule, they donate 2 times more electrons - i.e. 6.

Task 4:

Since in a molecule two nitrogen atoms, this deuce must be taken into account in the electronic balance - i.e. before magnesium it should be coefficient .

Task 5:

If the environment is alkaline, then phosphorus will exist in the form of salt- potassium phosphate.

If the medium is acidic, then phosphine turns into phosphoric acid.

Task 6:

Since zinc is amphoteric metal, in alkaline solution it forms hydroxocomplex. As a result of the arrangement of the coefficients, it turns out that water must be present on the left side of the reaction:

Task 7:

Electrons give away two atoms in an alkene molecule. Therefore, we must take into account general the number of electrons donated by the whole molecule:

(cold solution)

Please note that out of 10 potassium ions, 9 are distributed between two salts, so alkali will turn out only one molecule.

Task 8:

In the balance sheet process, we see that 2 ions have 3 sulfate ions. This means that in addition to potassium sulfate, another sulfuric acid(2 molecules).

Task 9:

(permanganate is not a very strong oxidizing agent in solution; note that water passes during adjustment to the right!)

(conc.)
(concentrated nitric acid is a very strong oxidizing agent)

Task 10:

Don't forget that manganese accepts electrons, wherein chlorine should give them away.
Chlorine is released in the form of a simple substance.

Task 11:

The higher the non-metal in the subgroup, the more active oxidizing agent, i.e. Chlorine is the oxidizing agent in this reaction. Iodine passes into the most stable positive oxidation state for it, forming iodic acid.

Task 12:

(peroxide is an oxidizing agent, because the reducing agent is )

(peroxide is a reducing agent, because the oxidizing agent is potassium permanganate)

(peroxide is an oxidizing agent, since the role of a reducing agent is more characteristic of potassium nitrite, which tends to turn into nitrate)

The total charge of a particle in potassium superoxide is . Therefore, he can only give.

(water solution) (acid environment)

For strong students

Many tasks of the unified state examination in chemistry contain serious errors or inaccuracies, so that they do not have a solution at all or allow several correct answers. Such tasks are mostly based on the "paper" idea of ​​chemical reactions. Faced with such questions, strong students who know chemistry well experience great difficulties. There is no one to ask the question, since the teacher who is on duty at the exam does not himself know what the author of the problem had in mind. What to do in this situation?

In this section, we will analyze several chemistry assignments from 2003 and try to determine what the authors had in mind.

Exercise 1. In a vessel containing 156 g of water, 46 g of sodium were placed. Determine the mass fraction of sodium hydroxide in the resulting solution.

2Na + 2H 2 O \u003d 2NaOH + H 2,

and then do the following: n (NaOH) = n (Na) = 46 / 23 = 2 mol; m(NaOH) = 2 x 40 = 80 g. m(r-ra) = m(H 2 O) + m(Na)- m(H 2) \u003d 156 + 46 - 2 \u003d 200 g. w (NaOH) \u003d 80 / 200 \u003d 0.4 \u003d 40%.

Actually: if you put the indicated amount of sodium in water, then an explosion of such force will occur that there will be no solution left, and there will be no mass fraction. In addition, there will be no one to count it. This task is a typical example of "paper" chemistry, and harmful.

Task 2. When propene reacts with hydrogen chloride, the following is formed:

1) 1-chloropropane

2) 2-chloropropane

3) 2-chloropropene

Actually: in this reaction, a mixture of two substances is formed - 1-chloropropane and 2-chloropropane, and the second substance predominates in the mixture. Strictly speaking, there are two correct answers: (1) and (2). It must be understood that Markovnikov's rule is not a law, it does not have absolute force. This rule speaks only about the preferred direction of the reaction.

Task 3. Chlorination of butane produces:

1) 1-chlorobutane

2) 2-chlorobutane

3) 1,2-dichlorobutane

4) 3-chlorobutane

What were the authors thinking? The correct answer is (2), which follows from the fact that the C-H bond energy at the secondary carbon atom is less than at the primary one, and therefore radical chlorination occurs first at the tertiary, and then at the secondary carbon atom.

Actually: chlorination of alkanes is not a regioselective reaction, in this reaction a mixture of substances is formed, and it is necessary to take into account not only the binding energy, but also the number of hydrogen atoms of each type. Chlorination of alkanes always results in a complex mixture of substances. There are three correct answers in this task: 1), 2) and 3).

Task 4. The substance formed during the oxidation of isopropylbenzene is called __________.

What were the authors thinking? If isopropylbenzene C 6 H 5 CH (CH 3) 2 is oxidized with potassium permanganate in an acidic medium, then benzoic acid C 6 H 5 COOH is formed. Apparently, this is the correct answer from the point of view of the authors.

Actually: even in this reaction, CO 2 is formed. In addition, the reaction products of the oxidation of isopropylbenzene depend on the conditions. If oxygen is used as an oxidizing agent, then phenol and acetone are formed (cumene method). There are at least three more correct answers in this problem: carbon monoxide (IV), phenol and acetone.

Task 5. The decomposition products of ammonium nitrate are:

NH 4 NO 3 \u003d N 2 O + 2H 2 O.

Actually: decomposition products of ammonium nitrate depend on the conditions. At a higher temperature (about 700 o C), nitric oxide (I) decomposes into simple substances, so the decomposition equation takes the form:

2NH 4 NO 3 \u003d 2N 2 + O 2 + 4H 2 O.

Then the correct answers are A, D.

Task 6. The oxidation state of chlorine in the KClO 3 molecule is

Actually: the KClO 3 molecule does not exist, because in solid form, potassium chlorate consists of ions, but in liquid and gaseous form it does not exist. KClO 3 is a substance of non-molecular structure. This error is editorial and does not lead to incorrect answers. Such errors are quite common.

Task 7. Establish a correspondence between the reagents and the ion-molecular reaction equation.

Actually: in reactions (3) and (4), not only the interaction of sulfate ions with ions of an alkaline earth metal occurs, but also a neutralization reaction occurs simultaneously. Strictly speaking, no ion-molecular equation corresponds to reactions (3) and (4) in the right column.

Task 8. What acid is found in natural fats?

2) C 17 H 35 COOH

4) NH 2 CH 2 COOH

Actually: fats are esters, they do not contain acids, but there are residues of acids. This is not a mistake, but rather an inaccuracy. She is not fatal.

Task 9. The strongest basic properties are:

1) ethylamine

2) trimethylamine

3) phenylamine

4) dimethylamine

What were the authors thinking? They believed that the basic properties of saturated amines increase in the series: primary< вторичные < третичные. Этого можно было бы ожидать, так как три углеводородных радикала увеличивают электронную плотность на атоме азота сильнее, чем два. Подразумевается правильный ответ 2) – триметиламин.

Actually: Contrary to popular belief, tertiary saturated amines are weaker bases than secondary and even primary ones. This is due, in particular, to spatial effects: three radicals impede the access of reagents to the nitrogen atom. Strictly speaking, the correct answer is 4), dimethylamine. The difference in basicity of secondary and tertiary amines is small and may be the subject of study in universities, but not in general education schools.

Task 10. When methanol was heated with an amount of 0.5 mol of substance with an excess of potassium bromide, bromomethane was obtained with a mass of 38 g and a practical yield of ______%.

CH 3 OH® CH 3 Br.

n practical (CH 3 Br) = 38/95 = 0.4 mol. Product yield: h (CH 3 Br) = 0.4 / 0.5 = 0.8 = 80%.

Actually: methanol does not react with potassium bromide without the addition of a strong acid. In addition, the Russian language suffers greatly here - from the text of the task it follows that methanol is heated not with a burner, but with a quantity of substance.

Task 11. Indicate a compound in which all bonds are covalent polar

What were the authors thinking? The task lists four salts. Three of them contain metal atoms and are clearly ionic. It seems that the authors believed that ammonium chloride contains only covalent bonds. They meant the correct answer was 2) - NH 4 Cl.

Actually: NH 4 Cl - ionic crystals. True, one of the two NH 4 + ions contains covalent polar bonds. There isn't a single correct answer here.

Task 12. Specify the carbohydrate that dissolves copper(II) hydroxide to form a bright blue solution and enters into the "silver mirror" reaction

1) maltose

2) sucrose

3) glucose

Actually: maltose is a reducing disaccharide, it also reacts with a silver mirror and dissolves copper (II) hydroxide. There are two correct answers in this task - 1) and 3).

Task 13. How will the rate of the CaO + CO 2 ® CaCO 3 reaction be affected by a 3-fold increase in carbon dioxide pressure?

1) speed increases by 3 times

2) speed increases by 9 times

3) speed decreases by 3 times

4) speed does not change

What were the authors thinking? Formally applying the law of mass action, they believed that this reaction is of the first order in CO 2, therefore, increasing the pressure by 3 times will increase the reaction rate by 3 times. Their correct answer is 1).

Actually: this reaction is heterogeneous, and heterogeneous reactions rarely have an integer order, since the rate of reaction is affected by the rate of diffusion and adsorption on the surface of the solid. The order of heterogeneous reactions can even depend on the degree of fineness of the solid! In assignments for the law of mass action, only elementary reactions can be given. There is no correct answer here at all.

Task 14. Reacts most rapidly with hydrochloric acid:

Actually: the rate of interaction of a metal with an acid depends not only on the nature of the metal, but also on other factors, such as the degree of grinding of the metal, the concentration of the acid, the presence of an oxide film, etc. Thus, iron powder will dissolve faster in acid than a zinc granule, although zinc is a more active metal. The task is formulated in such a way that there is no single correct answer.

Task 15. The sum of the coefficients in the reaction equation for the complete combustion of propane is:

C 3 H 8 + 5O 2 \u003d 3CO 2 + 4H 2 O.

The sum of the coefficients in this equation is 13, the correct answer is 3).

Actually: all problems based on the absolute values ​​of stoichiometric coefficients are incorrect. It is not the coefficients themselves that make sense, but only their ratio. For example, Fe + 2HCl does not mean that two moles of hydrogen chloride are involved in the reaction, but that the amount of hydrogen chloride is 2 times the amount of iron. There are two correct answers in this task - 3) and 4), because both propane combustion equations:

C 3 H 8 + 5O 2 \u003d 3CO 2 + 4H 2 O

2C 3 H 8 + 10O 2 \u003d 6CO 2 + 8H 2 O

are equally correct.

Task 16. The volume of hydrogen released during the interaction of 146 g of hydrochloric acid with 2 mol of zinc is _______ liters.

Zn + 2HCl \u003d ZnCl 2 + H 2.

Further, the authors identified hydrochloric acid (solution) and the individual substance HCl: n (HCl) = 146 / 36.5 = 4 mol, which corresponds to the amount of zinc. n (H 2) \u003d n (HCl) / 2 \u003d 2 mol, V(H 2) \u003d 2 × 22.4 \u003d 44.8 liters.

Actually: hydrochloric acid is not an individual substance, but a solution. The individual substance is hydrogen chloride. The correct answer cannot be given here, since the concentration of hydrochloric acid is not indicated.

Task 17. Establish a correspondence between the compound formula and the sequence of hybridizations of its carbon atoms.

What were the authors thinking? Carbon atoms in a double bond have sp 2 - hybridization, with triple - sp, and if all bonds are single - sp 3 . Thus, the intended correspondence: 1 - B, 2 - D, 3 - D, 4 - A.

Actually: in cumulated dienes, the carbon atom bonded to two double bonds is in the state sp-hybridization. This is not included in the condition. Paragraph 4 must correspond to the sequence sp 2 –sp–sp 2. In addition, the Russian language suffers again: hybridization is a phenomenon that does not have a plural. There are no "hybridizations", but there are "types of hybridization".

Task 18. The product of complete hydrolysis of starch is:

1) a-glucose

2) b-glucose

3) fructose

Actually: during the hydrolysis of starch, an equilibrium mixture of a-glucose, b-glucose and a linear form of glucose is formed. Thus, there are two correct answers: 1) and 2).

Task 19. During the electrolysis of the NaOH melt, the following is released at the anode:

Actually: anodic process equation:

4OH – – 4 e® O 2 + 2H 2 O

There are two correct answers here: 3) and 4).

Task 20. From 319 g of a 37.3% hot solution of calcium chloride, 33.4 g of a precipitate stood out on cooling. What is the mass fraction of salt in the remaining solution?

What were the authors thinking? Judging by the round answer that we will now receive, the following solution was supposed. Mass of CaCl 2 in the final solution: m(CaCl 2) \u003d 319 × 0.373 - 33.4 \u003d 85.6 g. Mass of solution: m(solution) \u003d 319 - 33.4 \u003d 285.6 g. w (CaCl 2) \u003d 85.6 / 285.6 \u003d 0.3 \u003d 30%.

Actually: when the CaCl 2 solution is cooled, the CaCl 2 × 6H 2 O crystalline hydrate will precipitate. The correct solution takes into account the mass content of anhydrous salt in the crystalline hydrate: m(CaCl 2) \u003d 319 × 0.373 - 33.4 × (111/219) \u003d 102.1 g. Mass of solution: m(solution) \u003d 319 - 33.4 \u003d 285.6 g. w (CaCl 2) \u003d 102.1 / 285.6 \u003d 0.357 \u003d 35.7%.

What can you advise in a situation when you are faced with an incorrect task? There is no one to prove the incorrectness: the answers are checked by a computer in which the author's answers are stored. Therefore, in order to get a high mark, first of all try to guess what the author meant. Give the answer he had in mind, then write it down and distribute it on the Internet to future generations of students who will be writing chemistry tests.

A graduate of school No. 1284 listens attentively to the last instructions before the final test. He knows that in the geography exam you have to use a map and a ruler. This will help you avoid mistakes.

PHOTO: Anna Ivantsova

According to the teacher in chemistry of Lyceum No. 1580 at the Moscow State Technical University named after N.E. Bauman, candidate of pedagogical sciences, USE expert Irina Yakunina, the unified state exam in chemistry has undergone some changes in recent years. For example, in the first part of the exam (there are three in total), questions were removed in which you can guess the correct answer by selection.

Questions have become more correct. The student should now understand what is being asked. Guessing the correct answer is almost impossible, - says Irina Yakunina.

The expert also noted that most of the students' mistakes in the exam in chemistry are caused by inattention. This is especially true for the first part of the exam.

This also happens with well-prepared children. Often mistakes are due to the fact that the student hastened to answer or was simply overconfident. But poorly prepared graduates make mistakes, because they do not see the pitfalls that may arise in the issue, - said Irina Yakunina.

For this reason, it is so important for students to read assignments carefully. And if necessary, several times in a row. It is worth noting that the first part of the exam is not appealable, therefore, it will not work to get an extra score lost as a result of an unfortunate mistake.

Also in the first part there are errors in tasks related to the chemistry of elements.

This is one of the most difficult areas of the subject, there are a lot of exceptions, so not all students demonstrate a high level of knowledge, - says Yakunina.

A student of school No. 1284 takes a trial Unified State Examination in Literature, which was also attended by journalists and public figures

Most of the errors occur when in tasks you need to answer questions about which substances, organic and inorganic, react with others presented in the list.

The main difficulty for graduates is with organic substances - many of them have ambiguous reaction products. Therefore, it is important to properly prepare before the exam, to repeat all possible exceptions to the rules, - says Irina Yakunina.

In the second part of the exam in chemistry this year, the calculation tasks in inorganic chemistry were complicated. Now, to answer the question, one often needs to solve algebraic equations. Today, a student who plans to take an exam in chemistry must also have a high level of knowledge in mathematics.

The student should be well able to involve mathematics in chemical problems for a more rational and faster solution, - said Yakunina.

Difficulties may arise in the last part. This year, graduates will have to solve tasks to determine the formula of organic matter in a changed situation.

A student can remember, for example, alkenes with one double bond, and on the exam he will come across a cyclic alkene. And if the child is not used to applying knowledge in a changed situation, then he will solve the task by one point, because he will recognize the molecular formula, but he will no longer be able to offer the correct structural formula and draw up the correct equation, - says Irina Yakunina.

Therefore, students should practice in adapting knowledge in a given situation. In some cases, you can contact a chemistry tutor for this.