Structure

Each version of the VPR in chemistry contains 15 tasks of various types and levels of difficulty.

The work contains 4 tasks of an increased level of complexity (their serial numbers: 9, 10, 13, 14).

The tasks included in the work can be conditionally distributed into four content blocks: “Theoretical foundations of chemistry”, “Inorganic chemistry”, “Organic chemistry”, “Methods of knowledge in chemistry. Experimental fundamentals of chemistry. Chemistry and life".

Explanations for grading assignments

Correct completion of the task with serial number 3 is scored 1 point.

Correct completion of each of the remaining tasks of a basic level of complexity is assessed with a maximum of 2 points. If there is one error or incomplete answer, 1 point is given. The remaining answer options are considered incorrect and are scored 0 points.

Assessment of tasks of an increased level of complexity is carried out on the basis of an element-by-element analysis of student responses. The maximum score for a correctly completed task is 3 points. Long answer tasks can be completed by students in different ways. Therefore, the sample solutions given in the evaluation criteria should be considered only as one of the possible answer options.

Average general education

Line UMK V.V. Lunin. Chemistry (10-11) (basic)

Line UMK V.V. Lunin. Chemistry (10-11) (U)

Line UMK N. E. Kuznetsova. Chemistry (10-11) (basic)

Line UMK N. E. Kuznetsova. Chemistry (10-11) (in-depth)

VPR in chemistry. 11th grade

The test includes 15 tasks. 1 hour 30 minutes (90 minutes) is allotted to complete the chemistry work.

Write down your answers to the questions in the space provided. If you write down an incorrect answer, cross it out and write a new one next to it.

When performing work you are allowed to use:

• Periodic table chemical elements DI. Mendeleev;
• table of solubility of salts, acids and bases in water;
• electrochemical voltage series of metals;
• non-programmable calculator.

When completing assignments, you can use a draft. Entries in draft will not be reviewed or graded.

We advise you to complete the tasks in the order in which they are given. To save time, skip a task that you cannot complete immediately and move on to the next one. If you have time left after completing all the work, you can return to the missed tasks.

The points you receive for completed tasks are summed up. Try to complete as many tasks as possible and score the most points.

We wish you success!

From your chemistry course you know the following methods for separating mixtures: sedimentation, filtration, distillation (distillation), magnetic action, evaporation, crystallization.

In Fig. 1-3 show examples of using some of the listed methods.

Determine which of the methods for separating the mixtures shown in the figure can be used for separation:

1. cereals and iron filings caught in it;
2. water and salts dissolved in it.

Write down the figure number and the name of the corresponding method of separating the mixture in the table.

Solution

1.1. The separation of a mixture of cereals and iron filings is based on the property of iron to be attracted by a magnet. Figure 3.

1.2. The separation of a mixture of water and dissolved salts occurs during distillation. Water, when heated to boiling point, evaporates and, cooling in a water refrigerator, flows into a pre-prepared vessel. Figure 1.

The figure shows a diagram of the distribution of electrons over energy levels an atom of some chemical element.

Based on the proposed scheme, complete the following tasks:

1. write down the symbol of the chemical element to which this atomic model corresponds;
2. write down the period number and group number in the Periodic Table of Chemical Elements D.I. Mendeleev, in which this element is located;
3. determine whether the simple substance that forms this element is a metal or non-metal.

Write your answers in the table.

Solution

The figure shows a diagram of the structure of an atom:

Where is shown a kernel having a certain positive charge(n), and electrons rotating around the nucleus on the electron layers. Based on this, they are asked to name this element, write down the number of the period and group in which it is located. Let's figure it out:

1. Electrons rotate on three electron layers, which means the element is in the third period.
2. There are 5 electrons rotating on the last electron layer, which means the element is located in the 5th group.

Task 3

Periodic table chemical elements D.I. Mendeleev is a rich repository of information about chemical elements, their properties and the properties of their compounds. For example, it is known that with an increase in the atomic number of a chemical element, the basic nature of the oxide in periods decreases, and in groups it increases.

Considering these patterns, arrange the following elements in order of increasing the basicity of the oxides: Na, Al, Mg, B. Write the symbols of the elements in the desired sequence.

Answer: ________

Solution

As is known, the sum of protons in the nucleus of an atom is equal to the atomic number of the element. But the number of protons is not indicated to us. Since an atom is an electrically neutral particle, the number of protons (positively charged particles) in the nucleus of an atom is equal to the number of electrons (negatively charged particles) orbiting the nucleus of the atom. The total number of electrons rotating around the nucleus is 15 (2 + 8 + 5), therefore, the atomic number of the element is 15. Now all that remains is to look at D.I. Mendeleev’s periodic table of chemical elements and find number 15. This is P (phosphorus). Since phosphorus has 5 electrons in its last electron shell, it is a non-metal; metals on the last layer have from 1 to 3 electrons.

Given 4 elements from the periodic system of Mendeleev: Na, Al, Mg, B. They must be arranged so that the basicity of the oxides formed by them increases. In answering this VPR question, it is necessary to remember how metallic properties change in periods and groups of the periodic table.

In periods from left to right, metallic properties decrease and non-metallic properties increase. Consequently, the basicity of the oxides decreases.

In groups and main subgroups, metallic properties increase from top to bottom. Consequently, the basicity of their oxides increases in the same order.

Now let's look at the elements given to us. Two of them are in the third group; these are B and Al. Aluminum is lower in the group than boron, therefore its metallic properties are more pronounced than boron. Accordingly, the basicity of aluminum oxide is more pronounced.

Al, Na and Mg are located in the 3rd period. Since metallic properties decrease from left to right in a period, the basic properties of their oxides also decrease. Taking all this into account, we can arrange these elements in the following order:

Task 4

The table below shows some characteristics of covalent and ionic species chemical bond.

Using this information, determine the type of chemical bond: 1) in calcium chloride (CaCl 2); 2) in a hydrogen molecule (H 2).

1. In calcium chloride _____________
2. In a hydrogen molecule _____________

Solution

In the next question, it is necessary to determine which type of chemical bond is characteristic of CaCl 2, and which of H 2. This table has a hint:

Using it, we can determine that CaCl 2 is characterized by an ionic type of bond, since it consists of a metal atom (Ca) and non-metal atoms (Cl), and for H 2 it is covalent nonpolar, since this molecule consists of atoms of the same element - hydrogen.

Complex inorganic substances can be conditionally distributed, that is, classified, into four classes, as shown in the diagram. In this diagram, enter the missing names of two classes and two formulas of substances that are representatives of the corresponding classes.

Solution

The following task tests knowledge of the main classes of inorganic substances.

You need to fill in the empty cells in the table. In the first two cases, formulas of substances are given; it is necessary to classify them into a certain class of substances; in the last two, on the contrary, write formulas for representatives of these classes.

CO 2 is a complex substance consisting of atoms of various elements. One of which is oxygen. He is in second place. It's an oxide. The general formula of oxides is RO, where R is a specific element.

RbOH – belongs to the class of bases. Common to all bases is the presence of an OH group, which is connected to the metal (the exception is NH 4 OH, where the OH group is connected to the NH 4 group).

Acids are complex substances consisting of hydrogen atoms and an acid residue.

Therefore, the formulas of all acids begin with hydrogen atoms, followed by an acid residue. For example: HCl, H 2 SO 4, HNO 3, etc.

And lastly, write the formula for salt. Salts are complex substances consisting of metal atoms and an acid residue, for example NaCl, K 2 SO 4.

To complete tasks 6-8, use the information contained in this text

Phosphorus(V) oxide (P 2 O 5) is formed when phosphorus is burned in air and is a white powder. This substance is very active and reacts with water, releasing a large amount of heat, so it is used as a desiccant for gases and liquids, and a water-removing agent in organic syntheses.

The product of the reaction of phosphorus(V) oxide with water is phosphoric acid (H 3 PO 4). This acid exhibits all the general properties of acids, for example, it interacts with bases. Such reactions are called neutralization reactions.

Salts of phosphoric acid, such as sodium phosphate (Na 3 PO 4), are widely used. They are included in detergents and washing powders and are used to reduce water hardness. At the same time, the entry of excess amounts of phosphates into water bodies with wastewater contributes to the rapid development of algae (water blooms), which makes it necessary to carefully monitor the phosphate content in wastewater and natural waters. To detect the phosphate ion, you can use the reaction with silver nitrate (AgNO 3), which is accompanied by the formation of a yellow precipitate of silver phosphate (Ag 3 PO 4)

Task 6

1) Write an equation for the reaction of phosphorus with oxygen.

Answer: ________

2) On what property of phosphorus(V) oxide is its use as a drying agent based?

Answer: ________

Solution

In this task, you need to create an equation for the reaction of phosphorus with oxygen and answer the question why the product of this reaction is used as a drying reagent.

We write the reaction equation and set the coefficients: 4 P + 5 O 2 = 2 P 2 O 5

Phosphorus oxide is used as a drying agent for its ability to remove water from substances.

Task 7

1) Write a molecular equation for the reaction between phosphoric acid and sodium hydroxide.

Answer: ________

2) Indicate what type of reaction (compound, decomposition, substitution, exchange) the interaction of phosphoric acid with sodium hydroxide is.

Answer: ________

Solution

In the seventh task, you need to create an equation for the reaction between phosphoric acid and sodium hydroxide. In order to do this, it is necessary to remember that this reaction refers to exchange reactions, when complex substances exchange their constituent parts.

H 3 P.O. 4 + 3 NaOH = Na 3 P.O. 4 + 3 H 2 O

Here we see that hydrogen and sodium in the reaction products have swapped places.

Task 8

1) Write a shortened ionic equation for the reaction between solutions of sodium phosphate (Na 3 PO 4) and silver nitrate.

Answer: ________

2) Indicate the sign of this reaction.

Answer: ________

Solution

Let us write the reaction equation in abbreviated ionic form between solutions of sodium phosphate and silver nitrate.

In my opinion, you first need to write the reaction equation in molecular form, then arrange the coefficients and determine which of the substances leaves the reaction medium, that is, precipitates, is released as a gas, or forms a poorly dissociating substance (for example, water). The solubility table will help us with this.

Na 3 P.O. 4 + 3 AgNO 3 = Ag 3 P.O. 4 + 3 NaNO 3

An arrow standing next to silver phosphate, pointing downwards, indicates that this compound is insoluble in water and precipitates, therefore it does not undergo dissociation and is written in the form of a molecule in the ionic equations of the reaction. Let's write the complete ionic equation for this reaction:

Now let's cross out the ions that moved from the left side of the equation to the right without changing their charge:

3Na + + PO 4 3– + 3Ag + + 3NO 3 – = Ag 3 PO 4 + 3Na + + 3NO 3 –

We write out everything that is not crossed out in the abbreviated ionic equation:

PO 4 3– + 3Ag + = Ag3PO4

Task 9

A scheme of the redox reaction is given.

Mn(OH) 2 + KBrO 3 → MnO 2 + KBr + H 2 O

1. Make an electronic balance for this reaction.

Answer: ________

2. Identify the oxidizing agent and reducing agent.

Answer: ________

3. Arrange the coefficients in the reaction equation.

Answer: ________

Solution

The next task asks you to explain the redox process.

Mn(OH) 2 + KBrO 3 → MnO 2 + KBr + H 2 O

In order to do this, we write next to the symbol of each element its oxidation state in a given compound. Do not forget that in total all oxidation states of a substance are equal to zero, since they are electrically neutral. The oxidation state of atoms and molecules consisting of the same substance is also zero.

Mn 2+ (O 2– H +) 2 + K + Br 5+ O 3 2– → Mn 4+ O 2 2– + K + Br – + H 2 + O 2 –

Mn 2+ (O 2– H +) 2 + K+Br 5+ O 3 2– → Mn 4+ O 2 2– + K + Br – + H 2 + O 2 –

Mn 2+ –2e → Mn 4+ The process of electron loss is oxidation. In this case, the oxidation state of the element increases during the reaction. This element is a reducing agent; it reduces bromine.

Br 5+ +6e → Br – The process of accepting electrons is reduction. In this case, the oxidation state of the element decreases during the reaction. This element is an oxidizing agent; it oxidizes manganese.

An oxidizing agent is a substance that accepts electrons and is reduced (the oxidation state of the element decreases).

A reducing agent is a substance that gives up electrons and is oxidized (the oxidation state of the element decreases). At school it is written as follows.

The number 6, which appears after the first vertical line, is the least common multiple of the numbers 2 and 6 - the number of electrons given up by the reducing agent and electrons accepted by the oxidizing agent. We divide this figure by the number of electrons donated by the reducing agent and get the number 3, it is placed after the second vertical line and is a coefficient in the equation of the redox reaction, which is placed in front of the reducing agent, that is, manganese. Next, we divide the number 6 by the number 6 – the number of electrons accepted by the oxidizing agent. We get the number 1. This is the coefficient that is placed in the equation of the redox reaction in front of the oxidizing agent, that is, bromine. We enter the coefficients into the abbreviated equation, and then transfer them to the main equation.

3Mn(OH) 2 + KBrO 3 → 3MnO 2 + KBr + 3H 2 O

If necessary, we set other coefficients so that the number of atoms of the same element is the same. At the end, we check the number of oxygen atoms before and after the reaction. If their number is equal, then we did everything right. In this case, it is necessary to put a coefficient of 3 in front of the water.

The transformation scheme is given:

Cu → CuCl 2 → Cu(OH) 2 → Cu(NO 3) 2

Write molecular reaction equations that can be used to carry out these transformations.

Solution

We solve the transformation scheme:

Cu→ CuCl 2 → Cu(OH) 2 → Cu(NO 3 ) 2

1) Cu + Cl 2 = CuCl 2 – I draw your attention to the fact that copper does not interact with hydrochloric acid, since it ranks in the series of metal voltages after hydrogen. Therefore, one of the main reactions. Interaction directly with chlorine.

2) CuCl 2 + 2 NaOH = Cu(OH) 2 + 2 NaCl– exchange reaction.

3) Cu(OH) 2 + 2 HNO 3 = Cu(NO 3 ) 2 + 2 H 2 O–Copper hydroxide is a precipitate, so nitric acid salts are not suitable for obtaining copper nitrate from it.

Establish a correspondence between the name of an organic substance and the class/group to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number.

Write down the selected numbers in the table under the corresponding letters.

1. Methanol is an alcohol. Titles monohydric alcohols end in -ol, so A2.

2. Acetylene is an unsaturated hydrocarbon. This trivial name is given here. According to systematic nomenclature, it is called ethin. Choose B4.

3. Glucose is a carbohydrate, a monosaccharide. Therefore we choose B1.

In the proposed schemes chemical reactions insert the formulas of the missing substances and place the coefficients where necessary.

1) C 6 H 6 + Br 2		C 6 H 5 –Br + ...

2) CH 3 CHO + … → CH 3 CH 3 OH

Solution

It is necessary to insert the formulas of the missing substances and, if necessary, arrange the coefficients:

1) C 6 H 6 + Br 2 ⎯AlBr 3 → C 6 H 5 –Br + HBr Substitution reactions are characteristic of benzene and its homologues, therefore in this reaction bromine replaces the hydrogen atom in benzene and bromobenzene is obtained.

2) CH 3 CHO + H 2 → CH 3 CH 2 OH The reaction of the reduction of acetaldehyde to ethyl alcohol.

Acetic acid is widely used in the chemical and food industries. Aqueous solutions acetic acid (food additive E260) are used in household cooking, in canning, as well as for the production of medicinal and aromatic substances. The latter include numerous esters acetic acid, such as propyl acetate.

Calculate how many grams of propyl acetate (CH 3 COOC 3 H 7) can be obtained from the reaction of 300 g of acetic acid (CH 3 COOH) with 1-propanol (C 3 H 7 OH) with a 100% practical yield. Write down the equation for the reaction and a detailed solution to the problem.

Answer: ________

Task. Let's write down a brief statement of the problem:

m(CH 3 SOOS 3 H 7) = ?

1. The problem statement states that acetic acid reacted with a mass of 300 g. Let us determine the number of moles in 300 g of it. To do this, we will use a magic triangle, where n is the number of moles.

We substitute the numbers: n = 300 g: 60 g/mol = 5 mol. Thus, acetic acid reacted with propyl alcohol with an amount of substance of 5 mol. Next, we determine how many moles of CH 3 COOC 3 H 7 are formed from 5 moles of CH 3 COOH. According to the reaction equation, acetic acid reacts in an amount of 1 mol, and 1 mol of ether is also formed, since there are no coefficients in the reaction equation. Therefore, if you take an acid in an amount of 5 mol, then you will also get 5 mol of ether. Since they react in a 1:1 ratio.

Well, all that remains is to calculate the mass of 5 moles of ether using this triangle.

Substituting the numbers, we get: 5 mol · 102 g/mol = 510 g.

Answer: mass of ether = 510 g.

Acetylene is used as a fuel in gas welding and metal cutting, as well as a raw material for the production of vinyl chloride and other organic substances. In accordance with the diagram below, create equations for reactions characteristic of acetylene. When writing reaction equations, use the structural formulas of organic substances.

Solution

Carry out the transformations characteristic of acetylene according to the given scheme.

I would like to say that acetylene is an unsaturated hydrocarbon that has 2 π-bonds between carbon atoms, therefore it is characterized by addition, oxidation, and polymerization reactions at the site of rupture of π-bonds. Reactions can occur in two stages.

Ringer's solution is widely used in medicine as a regulator of water-salt balance, a plasma substitute and other blood components. To prepare it, 8.6 g of sodium chloride, 0.33 g of calcium chloride and 0.3 g of potassium chloride are dissolved in 1 liter of distilled water. Calculate the mass fraction of sodium chloride and calcium chloride in the resulting solution. Write down a detailed solution to the problem.

Answer: ________

Solution

To solve this problem, we write down its brief condition:

m(H 2 O) = 1000 g. m(CaCl 2) = 0.33 g. m(KCl) = 0.3 g. m(NaCl) = 8.6 g.	Since the density of water is equal to one, 1 liter of water will have a mass equal to 1000 grams. Next, to find the mass fraction as a percentage of the solution, we will use the magic triangle, m(in-va) – mass of substance; m(solution) – mass of solution; ω – mass fraction of a substance in percent in a given solution. Let us derive a formula for finding ω% in solution. It will look like this:
ω% (NaCl solution)

In order to immediately move on to finding the percentage mass fraction of a NaCI solution, we must know two other values, that is, the mass of the substance and the mass of the solution. We know the mass of the substance from the problem statement, and the mass of the solution must be found. The mass of the solution is equal to the mass of water plus the mass of all salts dissolved in water. The formula for calculation is simple: m(in-va) = m(H 2 O) + m(NaCl) + m(CaCl 2) + m(KCl), adding all the values, we get: 1000 g + 8.6 g. + 0.3 g + 0.33 g = 1009.23 g. This will be the mass of the entire solution.

Now we find the mass fraction of NaCl in the solution:

Similarly, we calculate the mass of calcium chloride:

We substitute the numbers and get:

Answer:ω% in NaCl solution = 0.85%; ω% in CaCl 2 solution = 0.033%.

The CD in chemistry for grade 11 consists of fifteen tasks. 11 of them relate to basic level difficulty, and only 4 – to advanced. Tasks are divided into 4 blocks:

In order to write a paper, students are given 90 minutes, that is, 2 lessons. During the chemistry test, eleventh graders are allowed to have the following things with them:

1. Calculator (non-programmable)
2. Periodic table of D. I. Mendeleev
3. Electrochemical voltage series of metals
4. Solubility table

Rating system

In total, you can score 33 points for the work. There is no clearly defined scale for converting them into grades - this is done at the discretion of the management of the educational institution.

Examples of tasks with scoring and explanations

Task 1

The first task opens a block of numbers that test students’ knowledge in the area theoretical foundations chemistry. At the beginning, the topic of the task is set - for example, methods for separating mixtures or methods for collecting gas (they are listed). There are 3 pictures (without captions) illustrating what is discussed in the text - for example, three ways to separate mixtures. Next, you need to correlate the figure number with the provisions of the table indicated in the assignment, and also indicate what this figure illustrates (for example, a method or method). The table might look like this:

If the entire table is filled out correctly, the student receives 2 points for this task. If there is an error in one of its elements - 1 point, and if there are 2 or more errors - 0 points.

Task 2

The second task tests the knowledge of eleventh-graders in such aspects of chemistry as the composition of the atom and the structure of its electron shell. The condition contains a drawing that shows a structural model or a diagram of the distribution of electrons among the levels of an element - for example, like this:

It is necessary to answer three questions: write the serial number of the element, the number of the period and group in which it is located, and also determine whether the simple substance formed by the element belongs to metals or non-metals.

If all the answers are correct, 2 points are given, with one error - 1 point, with two or more - 0.

Task 3

This number involves working with the periodic system of D.I. Mendeleev, knowledge of its laws and properties of elements. A list of 4 elements is given - for example, Si, O, N, P or Si, Al, S, Cl. You need to arrange them according to the condition - for these examples this is a decrease in the radii of atoms and an increase in the acidic properties of higher oxides - and write them down in the answer in the correct order. For a correct answer the student receives 1 point, for an incorrect answer - 0.

Task 4

The fourth VPR task in chemistry is related to the structure of chemical substances and their properties. A table is provided that displays the basic properties of substances of molecular and ionic structure. Next, you need to determine what structure the two given substances have - for example, iodine and carbon monoxide or baking soda and acetylene. If the structure of both substances is determined correctly, the eleventh grader receives 2 points for this task, if only one - 1 point, and if the entire answer is incorrect - 0 points.

Task 5

The fifth task opens a block of numbers related to inorganic chemistry. It is related to the classification inorganic compounds. A table is provided; in its first two columns the formulas of substances are written, but the classes to which they belong are omitted, and in the next two - vice versa. It might look like this:

You need to complete the table with the missing elements. If this is done correctly, the answer is scored 2 points, if one mistake is made - 1 point, if two or more - 0 points.

Further in the text of the work there is text about a chemical substance - for example, aluminum sulfate or ammonia. The text talks about methods of its preparation, appearance, use in life and industry, basic properties and reactions. Tasks 6-8, which include two questions each, are completed based on this text. For each of tasks 6, 7 and 8 you can get a maximum of 2 points - if the answer is completely correct. If there is 1 error, 1 point is given, and if the answer is incorrect, 0 points.

Tasks 6-8

All these tasks are built on the same principle - in the first part you need to create an equation for any reaction involving the substance (or its derivatives) that is mentioned in the text. All components of the reaction are indicated, and other details of the answer are also mentioned - for example, that the equation should be abbreviated or that it should represent the reaction that occurs before the formation of a precipitate.

For example, if the text is about aluminum sulfate, the first parts of the questions look like this:

• Write a molecular equation for the reaction of producing aluminum sulfate from aluminum oxide and sulfuric acid.
• Write a molecular equation for the reaction between aluminum sulfate and sodium hydroxide to form a precipitate.
• Write a shortened ionic equation for the reaction between aluminum hydroxide and sulfuric acid.

In the second part, questions are asked related to the written equation - about the type of reaction, about its characteristics, about the properties of the resulting substance. For our example they look like this:

• Describe the signs of an ongoing reaction.
• Explain why the resulting precipitate dissolves in excess alkali.
• What type of reactions (combination, decomposition, substitution, exchange) does this interaction belong to?

Task 9

The ninth issue of the VPR in chemistry, which belongs to an increased level of complexity, tests students’ ability to work with redox reactions - compiling their electronic balances, arranging coefficients and indicating which substance is an oxidizing agent and which acts as a reducing agent. A reaction scheme is given - for example:

Fe(OH)2 + NaBrO + H2O → Fe(OH)3 + NaBr

CH4 + NO2 → CO2 + NO + H2O

The task consists of 3 parts. In the first, you need to create an electronic balance, in the second, you need to indicate the reducing agent/oxidizing agent, and in the third, you need to set the coefficients. If all this is done correctly, the answer is scored 3 points, if the student made an error in one part of the answer - 2 points, in two parts - 1 point, and if the entire answer is incorrect - 0 points.

Task 10

The tenth task is somewhat easier than the previous one, although it is also a task of increased difficulty. It gives a chemical chain, usually including three equations - for example:

K2CO3 → CaCO3 → CO2 → NaHCO3

Na2O → NaOH → Na2CO3 → Na2SO4

It is required to create reaction equations. If all three of them are written correctly, the eleventh grader receives 3 points; if only two are correct, 2 points; if only one is correct, 1 point; and if everything is incorrect, 0.

Task 11

The eleventh task opens the block of tasks on organic chemistry. In it, you need to establish a correspondence between the formula of a substance and its name, or between the name of a substance and the class/group to which it belongs. Three names or three formulas are given, indicated by letters, but there are four positions that need to be matched, so simply choosing the answer will be more difficult. The answer is recorded in the table; if there are no errors, 2 points are given, if there is one error - 1 point, if two or three - 0 points.

Task 12

The conditions for this task provide two reaction schemes in which one substance is missing. You need to insert the missing substances, and, if necessary, arrange the coefficients. Task examples:

HBr → CH3–CH2–Br + H2O

CH3CH2OH + HCl → .................... + H2O

If the missing elements are written correctly, the answer is scored 2 points, with one error - 1 point. For an incorrect answer or no answer, the student does not receive points for this task.

Task 13

The thirteenth task is a task, and not the simplest one - it belongs to an increased level of complexity. Most often it is necessary to calculate the mass of a substance; sometimes the practical and theoretical yield and deficiency/excess are given. You need to write an equation for the reaction described in the condition and a detailed solution. Here are examples of conditions:

• Calculate how many grams of ethyl acetate can be obtained from 600 g of acetic acid with a 100% practical yield.
• Calculate the mass of calcium chloride, which is formed when an excess hydrochloric acid solution reacts with calcium hydroxide weighing 370 g.

If everything is done correctly - 3 points are given, if there is one error - 2 points, two errors - 1 point, three or more - 0.

Task 14

In the penultimate VPR task in chemistry, the complexity of which is also considered increased, you need to create 3 equations for reactions with an organic substance. Unlike task number 10, here it is not the result of the reactions that is given, but both of their components.

The condition looks, for example, like this:

If all three equations are correct, the answer is scored 3 points, if only two are correct - 2 points, only one - 1 point, none - 0.

Task 15

The fifteenth task involves solving a problem. Most often you need to calculate the mass or mass fraction. This problem is easier than the one that needs to be solved in issue 13. Examples:

• To prepare the marinade, the cookbook recommends dissolving 20 g of salt, 30 g of sugar and 10 g of acetic acid in 500 ml of water. Calculate the mass fraction of salt and acetic acid in the resulting marinade.
• To increase productivity, it is recommended to spray green onions weekly with a 0.2% solution of ammonium nitrate. Calculate the mass of ammonium nitrate and the mass of water required to prepare 500 g of such a solution.

This task belongs to the “chemistry and life” block, therefore the task conditions describe practical application the resulting substances. For a correct answer, the student receives 2 points; if there is one error in the answer, -1 point; in other cases, no points are awarded for the task.

VPR All-Russian Testing Work - Chemistry 11th grade

Explanations for the All-Russian sample test work

When familiarizing yourself with a sample test work, you should keep in mind that the tasks included in the sample do not reflect all the skills and content issues that will be tested as part of the all-Russian test work. A complete list of content elements and skills that can be tested in the work is given in the codifier of content elements and requirements for the level of training of graduates for the development of an all-Russian test in chemistry. The purpose of the sample test work is to give an idea of ​​the structure of the all-Russian test work, the number and form of tasks, and their level of complexity.

Instructions for performing the work

The test includes 15 tasks. 1 hour 30 minutes (90 minutes) is allotted to complete the chemistry work.
Formulate your answers in the text of the work according to the instructions for the assignments. If you write down an incorrect answer, cross it out and write a new one next to it.
When performing work, you are allowed to use the following additional materials:
– Periodic table of chemical elements D.I. Mendeleev;
– table of solubility of salts, acids and bases in water;
– electrochemical series of metal voltages;
– non-programmable calculator.
When completing assignments, you can use a draft. Entries in draft will not be reviewed or graded.
We advise you to complete the tasks in the order in which they are given. To save time, skip a task that you cannot complete immediately and move on to the next one. If you have time left after completing all the work, you can return to the missed tasks.
The points you receive for completed tasks are summed up. Try to complete as many tasks as possible and score the most points.
We wish you success!

1. From your chemistry course you know the following methods for separating mixtures: settling, filtration, distillation (distillation), magnetic action, evaporation, crystallization. Figures 1–3 show examples of the use of some of the listed methods.

Which of the following methods for separating mixtures can be used for purification:
1) flour from iron filings that got into it;
2) water from inorganic salts dissolved in it?
Write down the figure number and the name of the corresponding method of separating the mixture in the table.

iron filings are attracted by a magnet

During distillation, after condensation of water vapor, salt crystals remain in the vessel

2. The figure shows a model of the electronic structure of an atom of some chemicalelement.

Based on the analysis of the proposed model, complete the following tasks:
1) identify the chemical element whose atom has such an electronic structure;
2) indicate the period number and group number in the Periodic Table of Chemical Elements D.I. Mendeleev, in which this element is located;
3) determine whether the simple substance that forms this chemical element is a metal or non-metal.
Write your answers in the table.
Answer:

N; 2; 5 (or V); non-metal

to determine a chemical element, you should count the total number of electrons, which we see in figure (7)

taking the periodic table, we can easily determine the element (the number of electrons found is equal to the atomic number of the element) (N-nitrogen)

after this we determine the group number (vertical column) (5) and the nature of this element (non-metal)

3. Periodic table of chemical elements D.I. Mendeleev– a rich repository of information about chemical elements, their properties and the properties of their compounds, about the patterns of changes in these properties, about methods of obtaining substances, as well as about their location in nature. For example, it is known that with an increase in the atomic number of a chemical element in periods, the radii of atoms decrease, and in groups they increase.
Considering these patterns, arrange the following elements in order of increasing atomic radii: N, C, Al, Si. Write down the designations of the elements in the required sequence.

Answer: ____________________________

N → C → Si → Al

4. The table below lists characteristic properties substances that have a molecular and ionic structure.

Using this information, determine what structure the substances nitrogen N2 and table salt NaCl have. Write your answer in the space provided:

1) nitrogen N2 ________________________________________________________________
2) table salt NaCl ___________________________________________________

nitrogen N2 – molecular structure;
table salt NaCl – ionic structure

5. Complex inorganic substances can be conditionally distributed, that is, classified, into four groups, as shown in the diagram. In this diagram for each of the four groups, fill in the missing names of the groups or chemical formulas of substances (one example of formulas) belonging to this group.

The names of the groups are written down: bases, salts;
formulas of substances of the corresponding groups are written down

CaO, bases, HCl, salts

Read the following text and complete tasks 6–8.

The food industry uses the food additive E526, which is calcium hydroxide Ca(OH)2. It is used in the production of: fruit juices, baby food, pickled cucumbers, table salt, confectionery and sweets.
It is possible to produce calcium hydroxide on an industrial scale by mixing calcium oxide with water, this process is called quenching.
Calcium hydroxide is widely used in the production of building materials such as whitewash, plaster and gypsum mortars. This is due to his ability interact with carbon dioxide CO2 contained in the air. The same property of calcium hydroxide solution is used to measure the quantitative content carbon dioxide in the air.
A useful property of calcium hydroxide is its ability to act as a flocculant that purifies wastewater from suspended and colloidal particles (including iron salts). It is also used to increase the pH of water, since natural water contains substances (e.g. acids), causing corrosion in plumbing pipes.

1. Write a molecular equation for the reaction to produce calcium hydroxide, which
mentioned in the text.

2. Explain why this process is called quenching.
Answer:__________________________________________________________________________

________________________________________________________________________________

1) CaO + H 2 O = Ca(OH) 2
2) When calcium oxide interacts with water, a large amount is released
amount of heat, so the water boils and hisses, as if it hits a hot coal, when the fire is extinguished with water (or “this process is called quenching because as a result slaked lime is formed”)

1. Write a molecular equation for the reaction between calcium hydroxide and carbon dioxide
gas, which was mentioned in the text.
Answer:__________________________________________________________________________

2. Explain what features of this reaction allow it to be used for detection
carbon dioxide in the air.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) Ca(OH) 2 + CO 2 = CaCO 3 ↓ + H 2 O
2) As a result of this reaction, an insoluble substance is formed - calcium carbonate, cloudiness of the original solution is observed, which allows us to judge the presence of carbon dioxide in the air (qualitative
reaction to CO 2)

1. Write an abbreviated ionic equation for the reaction mentioned in the text between
calcium hydroxide and hydrochloric acid.
Answer:__________________________________________________________________________

2. Explain why this reaction is used to increase the pH of water.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) OH – + H + = H 2 O (Ca(OH)2+ 2HCl = CaCl2 + 2H2O)
2) The presence of acid in natural water causes low pH values ​​of this water. Calcium hydroxide neutralizes the acid and pH values ​​increase

The pH scale exists from 0-14. from 0-6 - acidic environment, 7 - neutral environment, 8-14 - alkaline environment

9. A diagram of the redox reaction is given.

H 2 S + Fe 2 O 3 → FeS + S + H 2 O

1. Make an electronic balance for this reaction.
Answer:__________________________________________________________________________

2. Identify the oxidizing agent and reducing agent.
Answer:__________________________________________________________________________

3. Arrange the coefficients in the reaction equation.
Answer:__________________________________________________________________________

1) An electronic balance has been compiled:

2Fe +3 + 2ē → 2Fe +2	2		1
		2
S -2 – 2ē → S 0	2		1

2) It is indicated that sulfur in the oxidation state –2 (or H 2 S) is a reducing agent, and iron in the oxidation state +3 (or Fe 2 O 3) is an oxidizing agent;
3) The reaction equation has been compiled:
3H 2 S + Fe 2 O 3 = 2FeS + S + 3H 2 O

10. The transformation scheme is given:

Fe → FeCl 2 → Fe(NO 3) 2 → Fe(OH) 2

Write molecular reaction equations that can be used to carry out
the indicated transformations.
1) _________________________________________________________________________
2) _________________________________________________________________________
3) _________________________________________________________________________

The reaction equations corresponding to the transformation scheme are written:
1) Fe + 2HCl = FeCl 2 + H 2
2) FeCl 2 + 2AgNO 3 = Fe(NO 3) 2 + 2AgCl
3) Fe(NO 3) 2 + 2KOH = Fe(OH) 2 + 2KNO 3
(Other equations that do not contradict the conditions for specifying equations are allowed
reactions.)

11. Establish a correspondence between the formula of an organic substance and the class/group, to which this substance belongs: for each position indicated by a letter, select the corresponding position indicated by a number.

Write down the selected numbers in the table under the corresponding letters.
Answer:

A	B	IN

1. C3H8 - CnH2n+2 - alkane
2. C3H6 - CnH2n-alkene
3. C2H6O - CnH2n+2O- alcohol

12. In the proposed schemes of chemical reactions, insert the formulas of the missing substances and arrange the coefficients.

1) C 2 H 6 + ……………..… → C 2 H 5 Cl + HCl
2) C 3 H 6 + ……………..… → CO 2 + H 2 O

1) C 2 H 6 + Cl 2 → C 2 H 5 Cl + HCl
2) 2C 3 H 6 + 9O 2 → 6CO 2 + 6H 2 O
(Fractional odds are possible.)

13. Propane burns with low levels of toxic emissions into the atmosphere Therefore, it is used as an energy source in many areas, for example in gas lighters and for heating country houses.
What volume of carbon dioxide (CO) is produced when 4.4 g of propane is completely burned?
Write down a detailed solution to the problem.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) The equation for the propane combustion reaction has been compiled:
C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O
2) n(C 3 H 8) = 4.4/44 = 0.1 mol
n(CO 2) = 3n(C 3 H 8) = 0.3 mol
3) V(O 2) = 0.3 22.4 = 6.72 l

14. Isopropyl alcohol is used as a universal solvent: it is included in household chemicals, perfumes and cosmetics, and windshield washer fluids for cars. In accordance with the diagram below, create reaction equations for the production of this alcohol. When writing reaction equations, use the structural formulas of organic substances.

1) _______________________________________________________
2) _______________________________________________________
3) _______________________________________________________

The reaction equations corresponding to the scheme are written:

(Other reaction equations that do not contradict the conditions for specifying reaction equations are allowed.)

15. In medicine, a saline solution is a 0.9% solution of sodium chloride in water. Calculate the mass of sodium chloride and the mass of water required to prepare 500 g of saline solution. Write down a detailed solution to the problem.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) m(NaCl) = 4.5 g
2) m(water) = 495.5 g

m(solution) = 500g m(salt) = x

x/500 * 100%= 0.9%

m(salt) = 500* (0.9/100)= 4.5 g

© 2017 Federal service for supervision in the field of education and science of the Russian Federation

On April 27, 2017, for the first time, the All-Russian testing work of the VPR in chemistry in 11 grades was held in approbation mode.

Official website of VPR (StatGrad)- vpr.statgrad.org

Options for VPR in chemistry, grade 11, 2017

№	Download answers (evaluation criteria)
Option 11	answers
Option 12	answers
Option 13	answers
Option 14	answers
Option 15	variant 15 answer
Option 16	variant 16 answer
Option 17	variant 17 answer
Option 18	variant 18 answer

To get acquainted with approximate options works on the official website of FIPI, posted demo options with answers and descriptions.

Samples of VPR in chemistry grade 11 2017 (demo version)

The test includes 15 tasks. 1 hour 30 minutes (90 minutes) is allotted to complete the chemistry work.

When performing work, you are allowed to use the following additional materials:

– Periodic table of chemical elements D.I. Mendeleev;

– table of solubility of salts, acids and bases in water;

– electrochemical series of metal voltages;

– non-programmable calculator.

The structure and content of the all-Russian test work of the VPR in chemistry

Each version of the VPR contains 15 tasks of various types and levels of difficulty. The options include tasks of various formats.

These tasks differ in the required form of recording the answer. So, for example, the answer can be: a sequence of numbers, symbols; words; formulas of substances; reaction equations.

The work contains 4 tasks of an increased level of complexity (their serial numbers: 9, 10, 13, 14). These tasks are more complex, since their implementation requires the complex use of the following skills:

– draw up reaction equations confirming the properties of substances and/or the relationship of various classes of substances, and the electronic balance of the redox reaction;

Explain how the properties and methods of obtaining substances are determined by their composition and structure;

– simulate a chemical experiment based on its description.

When completing assignments, you can use a draft. Entries in draft will not be reviewed or graded.