Typical mistakes when taking the exam in chemistry. Task C1 on the exam in chemistry. Features, advice, recommendations. You will have three official cheat sheets for the exam. And you need to understand them

Typical mistakes when passing the exam in chemistry

Chemistry teacher MOSOSH # 9 Shapkina Zh.A.

The Unified State Examination 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 5,320 graduates from 10 regions of the Russian Federation took part in the exam, then in 2003 23,778 graduates from 34 regions took the exam, in 2004 - 28,069 graduates from 50 regions, and in 2006 - 30,389 graduates from 58 regions.

The average score of 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 USE in chemistry for a number of years makes it possible to analyze the typical mistakes made by graduates when completing assignments.

One of the significant drawbacks of the Unified State Exam is the impossibility of acquainting applicants with mistakes made during the performance of test items in parts A and B, which does not allow for their detailed analysis, deprives applicants of the legal right of appeal and creates tension among parents and students. This situation is exacerbated by the fact that control and measuring materials are not developed enough, many questions are posed incorrectly, there are errors in the proposed answers.

Let's comment on some of the 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 shown as correct. At the same time, it is known that the reducing properties of atoms of chemical elements with an increase in their ordinal number weaken in the period, and in the periods they change periodically. Thus, the correct answer is 4.

Acetic aldehyde reacts with each of two substances: 1) 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 to be correct. However, it is known that acetaldehyde in the presence of alkalis easily reacts with aldol condensation, and forms primary alcohols with hydrogen, therefore answer 4 is also correct.

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

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

Some assignments include questions that are not included in the school curriculum. For example, in task C1, it was proposed to write the equations of the 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 are not included in the program. The average percentage of completion of the assignment on the topic "Electrolysis" is 40. Typical mistakes when completing this assignment are as follows: graduates confused the signs of the electric charge of the cathode and anode; the order of discharge of particles present in the solution at the cathode and anode was not taken into account, including depending on their concentration; did not indicate (or indicated incompletely) the electrolysis conditions - stirring, presence or absence of a diaphragm, temperature, concentration; could not write the summary equation of the electrolysis process.

Some of the tasks that more than 75% of the students did not cope with had an unusual wording for them. For example, questions on the topic "Hydrolysis".

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

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

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

Almost all students completed the task in this formulation.

Establish a correspondence between the composition of the 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 such names of hydrolysis types as "by cation", "by anion" are not studied. A very common mistake is to count HF strong acid.

In the matching tasks, remember 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 the questions containing denial. Students forget to consider denial. 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, and not the numbers of the correct answer options.

In task B3, it is necessary to establish a correspondence between the starting substances and the sum of all the coefficients in the complete or abbreviated ionic equation. One of the common mistakes is that many students forget to include a factor of 1, which is not written in the equations. Another common mistake is that when going from a full ionic equation to an abbreviated one, students forget that you can also reduce the coefficients if they can all be divided by the same number.

For example:

Establish a correspondence between the starting materials and the sum of all the coefficients in the abbreviated ionic reaction equation. Starting materials: A) Al 2 ( SO 4 ) 3 + KOH , B) Ba ( OH ) 2 + HNO 3 , V) 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 the numbers can be repeated in the answers.

Tasks with multiple choice also cause difficulties, for example:

Reagents for carbon dioxide and ethylene are solutions: 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 advisable to write all the necessary ones in the answer, and not write anything superfluous. The second feature is that the numbers must be indicated in ascending order. If you first write, for example, "24", and then think and assign "1", then the answer "241" will be considered incorrect, even if "124" is correct.

The computational problems in Part B are not too difficult, but many rounding errors are made.

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

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

Write down the reaction equations with which you can carry out the following transformations:

Cr 2 S 3 NS 1 > K 2 CrO 4 X 1 X 2> KCrO 2

(the average score was 0.3 out of 5);

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

(the average score was 0.4);

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

(the average score is 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).

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

R 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 required substances are encrypted. Despite the fact that the reactions are quite simple and well-known, an error at any stage, especially at the first, does not leave students 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 reaction product, the student does not indicate by-substances, does not place coefficients. The conditions of the reactions are not taken into account when determining their products. Thus, in the hydrolysis of esters in an alkaline medium, free acids are indicated as products, and acids are also indicated in 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 the C4 task (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 substances in a given system, in a given solution. So, having determined, in one of the tasks, that nitric acid remained in excess, the 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.

Errors are also encountered in calculations using the reaction equations, in the analysis of excess-deficiency of the reagent.

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

With the complete combustion of gaseous organic matter, which does not contain oxygen, 4.48 liters (n.u.) of carbon dioxide, 1.8 g of water and 4 g of hydrogen fluoride were released. Establish 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 for combustion products, hydrogen is lost, which is in the composition of hydrogen halides.

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

Hydrogen is often lost when over-rounding in calculations.

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

In order to improve the quality of the exam, it is necessary to give students some advice.

For part A:

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

2) When working on a difficult question, decide if you can use Cheat Sheet # 1 (Periodic Table) to answer? Cheat Sheet # 2 (Solubility Table)? Cheat Sheet # 3 (Metal Stress Row)?

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

Remember, Part A tests knowing the most obvious things.

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

4) In the proposed options, you do not see a single correct one, which means that you need to re-read the question, did you understand it correctly, did you miss the negation? If that doesn't work, remember that there are exceptions to many of the rules in chemistry. Are there any special properties of the substances presented? Special conditions for carrying out the 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 you have finalized it, transfer them to the answer sheet.

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

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

For part C:

1) The coefficient of difficulty for 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 something as legibly as possible.

A graduate of school number 1284 listens attentively to the last instructions before the final test. He knows: on the exam in geography, you need to use a map and a ruler. This will help you avoid mistakes.

According to the chemistry teacher of the Lyceum No. 1580 at the Moscow State Technical University named after N.E. Bauman, candidate of pedagogical sciences, examiner of the Unified State Exam Irina Yakunina, the unified state examination 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 using the selection method.

The questions became more correct. Now the student should understand what he is being asked about. It is almost impossible to guess the correct answer, - says Irina Yakunina.

The expert also noted that most of the mistakes of students on 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 trained graduates make mistakes, because they do not see the pitfalls that may arise in the question, - said Irina Yakunina.

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

Also, in the first part, errors occur 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 in it, so not all students demonstrate a high level of knowledge, says Yakunina.

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

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

The main difficulty for graduates is precisely 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 USE in chemistry this year, the computational problems in inorganic chemistry were complicated. Now, to answer a question, you often need 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 good at using mathematics in chemical problems for a more rational and quick solution, - said Yakunina.

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

The student may 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 recognizes the molecular formula, but he will no longer be able to offer the correct structural formula and make the correct equation, - says Irina Yakunina.

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

Each examiner in chemistry must be prepared for the fact that 3 astronomical hours, or 180 minutes, are allotted for the examination work, which consists of three parts and includes 45 tasks. In official documents, it is recommended to distribute this time 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 advise students to save time on the relatively easy parts A and B in order to use more time reserve when completing 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.

Download:

Preview:

CHEMISTRY

Everyone who takes the exam in chemistry must be prepared for the fact that to perform an examination slave O you, consisting of three parts and including 45 tasks, are given 3 astronomical hours, or 180 minutes. In official documents, it is recommended to distribute this time 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 advise students to save time on the relatively easy parts A and B in order to use more time reserve when completing 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 different content blocks.

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

• tasks that check the assimilation of the topic of redox reactions;
• tasks that test knowledge of the chemical properties of inorganic substances;
• tasks that check the assimilation of educational information about the relationship of various classes of substances (organic and inorganic);
• combined computational tasks;
• 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 draw up an equation for any redox reaction and determine the oxidizing agent and reducing agent. C1 tasks can be roughly divided into three types:

1) missing formulas of any substances 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 in the left part of it are missing

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

Р 2 O 3 + H 2 Cr 2 O 7 +… → H 3 PO 4 + CrPO 4

3) formulas of substances are missing in both sides of the equation

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

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

In task C2 given four or five substances, between which it is necessary to write four reaction equations, 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 dichromate, sulfuric acid and cesium hydroxide. Give the equations of four possible reactions involving the indicated substances.
2. Substances are given: sodium nitrate, white phosphorus, bromine, potassium hydroxide (solution). Give the equations of four possible reactions involving the indicated substances.

This task is, perhaps, the most difficult of all the tasks of the USE test and tests the 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

H 2 O

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 a substance) of the reaction products if one of the substances is given in excess and is indicated in the task in the form of a solution with a certain mass fraction of a solute or contains impurities. The maximum score for the correct execution 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 should be added to 600 g of nitric acid solution 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. The interaction of 11.6 g of saturated aldehyde with an excess of copper (II) hydroxide upon heating forms a precipitate weighing 28.8 g. Derive the molecular formula of the aldehyde.
2. During the combustion of 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 points for the USE test in chemistry 2009 is 66.

Analysis of the fulfillment of tasks of Part C

In 2009, the percentage of graduates who started taking part C of the USE test in chemistry was 90.2%. The generalized results of the tasks of Part C are presented in Table 1.

Table 1

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

The average rate of completion of tasks in part C 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 oxidation state (for example, in the interaction of potassium nitrite and potassium permanganate);
• inability to predict the reduction products of typical oxidants (potassium permanganate, iodine, potassium nitrite) and oxidation products of reducing agents (manganese dioxide) in various environments, as well as the possibility of water molecules participating 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 a specialized chemistry course. The basic course covers these issues in an introductory plan.

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

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

• inability to analyze the possibility of 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 - oxidizers, amphoteric oxides and hydroxides, the reducing properties of sulfides and halides.

Less than a quarter of graduates completed task C3. This is due to the increasing complexity of tasks of this type.Typical mistakes when performing task C3:

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

Task C4 there was a combined computational problem. More than a third of the graduates completed the task.

In the context of tasks of this type, the following actions were combined:

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

Most often, students are allowed errors:

• when determining the mass of the solution without taking into account the mass of the released gas or sediment;
• 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 that react.

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

More than half of the graduates solved the problem. Many students were able to correctly perform the first step - to find the simplest ratio of moles of atoms in a compound, but could not move on to determining the true formula.

A problem was caused by the problem of determining 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

The items in Part C are checked by experienced expert teachers, in contrast to Parts A and B, which are checked using a computer. Therefore, it is important, when filling out the answers to the tasks of Part C, if possible, not to use abbreviations in words and to write down the solution of the problems as fully as possible.

You can perform the solution to any task of Part C from any link, each of which has its own price of 1 point. In this case, graduates will gain 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 examiner will be able to determine the oxidizing agent and reducing agent in task C1, or write down the reaction equation for task C4, thereby providing himself with 1 point for each action.

In other words, it is necessary to complete all the fragments that they can complete for each task of Part C.

The teacher needs to bring to the attention of the students that when developing the assessment criteria, the peculiarities of checking the assimilation of the elements of the content of all five tasks with a detailed answer included in the examination work are taken into account. It takes into account the fact that the wording of the examiners' 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 answer itself according 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 point for a correctly formulated element of content.

So, when compiling a scale for evaluating computational problems, the multivariance of ways to solve them is taken into account, and, therefore, the presence in the answer of its main stages and results indicated in the evaluation criteria. The teacher emphasizes that a common feature of assessing all tasks with a detailed answer 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 with examples of evaluating certain types of tasks with a detailed answer used in the CMMs of the USE.

Exercise.

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

Let us illustrate the assessment of it by experts using the example of the original work of a graduate.

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 electronic balance method, write the reaction equation:

P + HNO 3 +… → NO +…

Determine the oxidizing and reducing agent.

When working out this stage of preparing graduates for the Unified State Exam-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 contain an indication of which element (substance) is an oxidizing agent and which one 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 and 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 reacts vigorously with water - can it form in an aqueous medium? Of course not, the correct product is H 3 PO 4.

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

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

Task C2.

Typical mistake # 1: the interaction of metals with nitric acid - the overwhelming majority of participants write: Me + HNO 3 →… + H 2.

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

Typical mistake # 2: The possibility of OVR proceeding 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 takes place, 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, and the coefficients are not placed. If you could not, then nothing can be done, and if from inattention, then it's a shame, points are lost.

Typical mistake # 2: Simplified reaction equations are written that do not take into account media, without specifying inorganic products: CH 3 CHO + Ag 2 O → CH 3 COOH + 2Ag - the reaction proceeds 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 the permanganate, what other products are formed….

Typical mistake # 3: lack of coefficients.

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

Substances are given: magnesium, concentrated sulfuric acid, nitrogen, ammonium chloride.

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

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

Write the equations for four possible reactions between these substances.

Substances are given: bromine, hydrogen sulfide, sulfur dioxide, concentrated nitric acid.

Write the equations for four possible reactions between these substances.

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

Write the equations for four possible reactions between these substances.

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

ethene → X1 → X2 → X3 → toluene → X4

Write down the reaction equations with which you can carry out the following transformations:

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 fraction of substances in

the resulting solution.

Response elements:

KH + H2 O = H2 + KOH

KOH + HC1 = KC1 + H2 O

m (solution HC1) = p V = 1.16 100 = 116 (g)

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

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

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

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

m (KCl) = M n = 74.5 0.5 = 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) = M n = 36.5 0.52 = 18.98 (g)

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

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

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

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

27.2 g of a mixture of calcium and aluminum carbides was treated with an acid, and 11.2 liters of a mixture of gases were obtained (at normal conditions). 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:

CaC2 + 2HC1 = CaCl2 + C2 H2 M (CaC2 ) = 64 g / mol

Al4 C3 + 12HC1 = 4A1C13 + 3CH4 M (A14 WITH3 ) = 144 g / mol

n (CaC2 ) = n (С2 H2 ) = X n (A14 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 oxygen vapor density of organic matter is 1.875. When 15 g of this substance is burned, 16.8 liters of carbon dioxide (under normal conditions) and 18 g of water are formed. Determine the composition of the organic

substances.

Response elements:

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

n (CxHyOz) = m: M = 15: 60 = 0.25 (mol)

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

n (H2 O) = m: M = 18: 18 - 1 (mol) => n (H) = 2 (mol)

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

M (C3 H8 Oz) = 12 3 + 1 8 + 16 Z

44 + 16 z = 60=> z = l

Composition of organic matter C3 H8 O

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

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) = m: M = 10: 40 = 0.25 (mol)

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

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

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

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

Acid composition 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, H2 O + CO2 , H2 О + SO2 ) or acidic solutions of salts strongly hydrolyzed at the cation (FeCl3 , A1C13 , СrСl3 );

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

- decomposition of КСlO3 ;

-interaction of solutions of salts formed by a weak base and a weak acid (СrСl3 and K2 CO3 , A1C13 and Na2 S);

- the interaction of an acid and its average salt with the formation of an acid salt (K2 CO3 + H2 O + CO2 ; Na2 S + H2 S);

- interaction of phosphorus with concentrated sulfuric and nitric acids;

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

- interaction of chlorine with alkalis in the cold and during heating;

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

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

- Wurz reaction;

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

- alkylation of amines;

- alkylation of benzene and its homologues;

- obtaining acetaldehyde by catalytic oxidation of ethylene.

Part C on the exam in chemistry begins with the task C1, which involves the preparation of a redox reaction (which already contains part of the reagents and products). It is formulated as follows:

C1. Using the electronic balance method, write the reaction equation. Determine the oxidizing 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 figure out what to look for.

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 that participates in the reaction)		Manganat or, -

Dichromate and chromate as oxidizing agents.

(acidic and neutral medium), (alkaline medium) + reducing agents always works
acidic environment	neutral environment	alkaline environment
Salts of those acids that are involved in the reaction:		in solution, or in melt

Increased oxidation states of chromium and manganese.

Nitric acid with metals.

- no hydrogen released, nitrogen reduction products are formed.

Sulfuric acid with metals.

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

Disproportionation.

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

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

Disproportionation of nitric oxide (IV) and salts.

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 are used. The more active the metal, the easier it will donate electrons and the better it will be a reducing agent in redox reactions.

Electrochemical series of metal voltages.

Features of the behavior of some oxidizing and reducing agents.

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

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

Required skills.

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

   Exercise 1: Arrange the oxidation states in 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 during the redox reaction, while the total charge of the carbon atom and its non-carbon environment is taken as 0.

   Assignment 2: Determine the oxidation state of the boxed carbon atoms along with the non-carbon environment:

   2-methylbutene-2: - =

   acetone:

   acetic acid: -

3. Do not forget to ask yourself the main question: who in this reaction gives up electrons, and who accepts them, and into what do they go? So that it doesn't work out that electrons come from nowhere or fly away to nowhere.

   Example:

   In this reaction, it should be seen that potassium iodide can be only a reducing agent, so potassium nitrite will accept electrons, lowering its oxidation state.
   Moreover, under these conditions (diluted solution) nitrogen goes from to the nearest oxidation state.

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

   The same deuces cannot be forgotten when equalizing, because they indicate the number of atoms of a given type in the equation.

   Assignment 3: What ratio should be put before and before

   
   

   Assignment 4: What coefficient in the reaction equation will stand before 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- this means that it is definitely not an alkaline medium, and if a metal hydroxide precipitates, it is definitely not acidic. Well, 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.

   Assignment 5: Determine the medium and substances in each reaction:

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

   Assignment 6:Which side of the reaction will the water end up in? What will the zinc transfer to?

   Assignment 7: Soft and hard oxidation of alkenes.
   Add and equalize the reactions, having previously arranged the oxidation states in organic molecules:

   (cold solution)

   (water solution)

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

   Assignment 8:What other products will you get? Add and equalize the reaction:

8. What are the reagents in the reaction?
   If the schemes we have learned do not give the answer to this question, then it is necessary to analyze which oxidizing and reducing agents in the reaction are strong or not very strong?
   If the oxidizing agent is of medium strength, it is unlikely that it can oxidize, for example, sulfur from to, usually oxidation only proceeds to.
   Conversely, if is a strong reducing agent and can restore sulfur from to, then only to.

   Quest 9: What will the sulfur go into? Add and equalize the reactions:

   (conc.)

9. Check that the reaction contains both an oxidizing agent and a reducing agent.

   Quest 10: How many other products are there 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 oxidizing agent. Then the second will be a restorer.

    Quest 11: Which of these halogens is an oxidizing agent and which is a reducing agent?

11. If one of the reagents is a typical oxidizing agent or reducing agent, then the second will "do his will", either giving electrons to the oxidizing agent, or accepting 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 in the role of a reducing agent - passes into free gaseous oxygen.

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

The sequence of placing the coefficients in the equation.

First put down the coefficients obtained from the electronic balance.
Remember that you can double or shrink 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 placed.
The penultimate equals hydrogen, and we only check for oxygen!

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

Possible mistakes.

1. Allocation of oxidation states: check each substance carefully.
   They are often mistaken in the following cases:

   a) the oxidation state in hydrogen compounds of non-metals: phosphine - the oxidation state of phosphorus - negative;
   b) in organic substances - check again if 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 chlorine acids - 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.

   Quest 14: Add and equalize:

   Quest 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 flew to manganese from space ...

3. Products that are incorrect 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, an acid or acidic oxide will not be obtained;
   c) oxide or, moreover, metal, which react violently with water, are not formed in an aqueous solution.

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

Answers and solutions to tasks with explanations.

Exercise 1:

Assignment 2:

2-methylbutene-2: - =

acetone:

acetic acid: -

Assignment 3:

Since there are 2 chromium atoms in a dichromate molecule, they give up 2 times more electrons, i.e. 6.

Assignment 4:

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

Assignment 5:

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

If the medium is acidic, then phosphine is converted to phosphoric acid.

Assignment 6:

Since zinc - amphoteric metal, in an alkaline solution it forms hydroxo complex... As a result of placing the coefficients, it is found that water must be present on the left side of the reaction:

Assignment 7:

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

(cold solution)

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

Assignment 8:

In the process of drawing up the balance sheet, we see that there are 3 sulfate ions for 2 ions... This means that in addition to potassium sulfate, more sulphuric acid(2 molecules).

Quest 9:

(permanganate is not a very strong oxidizing agent in solution; note that water goes over in the process of equalizing to the right!)

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

Quest 10:

Don't forget that manganese accepts electrons, wherein chlorine must give them away.
Chlorine is released as a simple substance.

Quest 11:

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

Quest 12:

(peroxide is an oxidizing agent, since a reducing agent is)

(peroxide is a reducing agent, since 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 particle charge in potassium superoxide is. Therefore, he can only give.

For strong students

Many tasks of the unified state exam in chemistry contain serious errors or inaccuracies, so that they have no solution at all or admit several correct answers. Most of such tasks are based on the "paper" concept of chemical reactions. Faced with such questions, strong chemistry students have great difficulty. There is no one to ask a question, since the teacher who is on duty at the exam himself does not 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. A vessel containing 156 g of water was charged with 46 g of sodium. Determine the mass fraction of sodium hydroxide in the resulting solution.

2Na + 2H 2 O = 2NaOH + H 2,

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

Actually: if the specified amount of sodium is placed 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 interacts with hydrogen chloride, the following forms:

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 here: (1) and (2). It must be understood that Markovnikov's rule is not a law, it has no absolute force. This rule speaks only of the predominant direction of the reaction.

Task 3. Chlorination of butane produces:

1) 1-chlorobutane

2) 2-chlorobutane

3) 1,2-dichlorobutane

4) 3-chlorobutane

What the authors were thinking about. Under the correct answer is meant (2), which follows from the fact that the energy of the C - H bond at the secondary carbon atom is less than at the primary one, and therefore radical chlorination first of all occurs 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. This task has three correct answers: 1), 2) and 3).

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

What the authors were thinking about. If isopropylbenzene C 6 H 5 CH (CH 3) 2 is oxidized with potassium permanganate in an acidic medium, 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, 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. Decomposition products of ammonium nitrate are:

NH 4 NO 3 = N 2 O + 2H 2 O.

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

2NH 4 NO 3 = 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 non-molecular substance. This error is editorial, it 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 alkaline earth metal ions occurs, but also a neutralization reaction. Strictly speaking, reactions (3) and (4) in the right column do not correspond to any ionic-molecular equation.

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. It is not fatal.

Task 9. It exhibits the strongest basic properties:

1) ethylamine

2) trimethylamine

3) phenylamine

4) dimethylamine

What the authors were thinking about. They believed that the main properties of saturated amines increase in the following order: 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 limit amines are weaker bases than secondary and even primary ones. This is due, in particular, to spatial effects: three radicals hinder the access of reagents to the nitrogen atom. Strictly speaking, the correct answer is 4), dimethylamine. The difference in the basicity of secondary and tertiary amines is small and can be the subject of study in universities, but not in general education schools.

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

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 adding a strong acid. In addition, the Russian language suffers greatly here - from the text of the assignment it follows that methanol is heated not by a burner, but by the amount of substance.

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

What the authors were thinking about. 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 2) - NH 4 Cl.

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

Task 12. Indicate a carbohydrate that dissolves copper (II) hydroxide to form a bright blue solution and reacts with a "silver mirror"

1) maltose

2) sucrose

3) glucose

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

Task 13. How will an increase in the pressure of carbon dioxide by 3 times affect the reaction rate of CaO + CO 2 ® CaCO 3?

1) speed increases 3 times

2) speed increases 9 times

3) the speed is reduced by 3 times

4) the speed does not change

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

Actually: this reaction is heterogeneous, and heterogeneous reactions rarely have a whole order, since the rate of reaction is influenced by the rate of diffusion and adsorption on the surface of a solid. The order of heterogeneous reactions can even depend on the degree of grinding 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 fastest 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, for example, 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 unambiguous 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 = 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. In this task, there are two correct answers - 3) and 4), because both propane combustion equations:

C 3 H 8 + 5O 2 = 3CO 2 + 4H 2 O

2C 3 H 8 + 10O 2 = 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 = 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) = n (HCl) / 2 = 2 mol, V(H 2) = 2 × 22.4 = 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 formula of a compound and the sequence of hybridizations of its carbon atoms.

What the authors were thinking about. Carbon atoms with a double bond have sp 2 - hybridization, with triple - sp, and if all connections are single - sp 3. Thus, the intended correspondence is: 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 provided in the condition. Item 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 plurality. 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 NaOH melt, the following is released at the anode:

Actually: equation of the anodic process:

4OH - - 4 e® O 2 + 2H 2 O

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

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

What the authors were thinking about. Judging by the round answer that we will now receive, the following solution was assumed. The mass of CaCl 2 in the final solution: m(CaCl 2) = 319 × 0.373 - 33.4 = 85.6 g. Mass of solution: m(solution) = 319 - 33.4 = 285.6 g. w (CaCl 2) = 85.6 / 285.6 = 0.3 = 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) = 319 × 0.373 - 33.4 × (111/219) = 102.1 g. Mass of solution: m(solution) = 319 - 33.4 = 285.6 g. w (CaCl 2) = 102.1 / 285.6 = 0.357 = 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 the computer, which contains the author's answers. Therefore, to get a high score, first try to guess what the author meant. Give the answer he implied, and then write down this assignment and distribute it on the Internet for future generations of students who are going to write tests in chemistry.