In 2018, VPR in chemistry in grade 11 was carried out by the decision of the educational organization, for graduates who did not choose this subject for passing the exam.

The 2018 versions will come in handy to prepare for the 2019 CDF.

VPR in chemistry grade 11 2018 options + answers

The verification work in chemistry is conditionally divided into four content blocks: “Theoretical Foundations of Chemistry”, “Inorganic Chemistry”, “Organic Chemistry”, “Methods of Learning Chemistry. Experimental foundations of chemistry. Chemistry and Life.

Graduates taking part in the writing of the VLOOKUP in chemistry must demonstrate a basic level of knowledge of the subject:

write an equation for a chemical reaction
simulate a chemical experiment based on its description,
explain the conditionality of the properties and methods of obtaining substances by their composition and structure.

Test work for grade 11 includes 15 tasks of various levels of complexity.

1.5 hours (90 minutes) are given to complete the entire work.

The tasks included in the test work check mastery
graduates with certain skills and methods of action that
meet the requirements for the level of training of graduates.

Table for converting the scores of the VLOOKUP of chemistry to the assessment

Secondary general education

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

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

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

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

VPR in chemistry. Grade 11

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

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

When performing work, it is allowed to use:

Periodic system of chemical elements D.I. Mendeleev;
table of solubility of salts, acids and bases in water;
electrochemical series of voltages of metals;
non-programmable calculator.

When completing assignments, you can use a draft. Draft entries 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 the task that you can't complete right away and move on to the next one. If after completing all the work you have time left, you can return to the missed tasks.

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

We wish you success!

From the course of chemistry, you know the following methods for separating mixtures: settling, filtering, distillation (distillation), magnet action, evaporation, crystallization.

On fig. 1-3 show examples of the use of some of these methods.

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

cereals and iron filings that got into it;
water and salts dissolved in it.

Record in the table the number of the figure and the name of the corresponding method for separating the mixture.

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. Separation of a mixture of water and dissolved salts occurs during distillation. Water, when heated to the boiling point, evaporates and, cooling in a water cooler, flows into a pre-prepared vessel. Picture 1.

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

Based on the proposed scheme, complete the following tasks:

write down the symbol of the chemical element to which the given model of the atom corresponds;
write down the period number and group number in the Periodic system of chemical elements D.I. Mendeleev, in which this element is located;
determine whether the simple substance that forms this element belongs to metals or non-metals.

Record your answers in a table.

Solution

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

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

Electrons rotate on three electron layers, which means that the element is in the third period.
On the last electron layer, 5 electrons rotate, which means that the element is located in the 5th group.

Task 3

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

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

Answer: ________

Solution

As you know, the sum of protons in the nucleus of an atom is equal to the ordinal number of the element. But the number of protons is not given 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) revolving around the atom's nucleus. The total number of electrons revolving around the nucleus is 15 (2 + 8 + 5), therefore, the serial number of the element is 15. Now it remains to look into the periodic system of chemical elements of D. I. Mendeleev and find the number 15. This is P (phosphorus). Since phosphorus has 5 electrons in the last electron layer, it is non-metal; metals on the last layer have 1 to 3 electrons.

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

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

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

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

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

Task 4

The table below shows some of the characteristics of covalent and ionic types of chemical bonding.

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

In calcium chloride _____________
In a hydrogen molecule _____________

Solution

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

Using it, it can be determined 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 non-polar, since this molecule consists of atoms of the same element is hydrogen.

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

Solution

The next task is to test the knowledge of the main classes of inorganic substances.

The table must fill in the empty cells. In the first two cases, the formulas of substances are given, it is necessary to attribute them to 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 for 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.

Lastly, write the formula for salt. Salts are complex substances consisting of metal atoms and an acidic 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 during the combustion of phosphorus in air and is a white powder. This substance is very active and reacts with water with the release of a large amount of heat, therefore it is used as a desiccant for gases and liquids, a water-removing agent in organic syntheses.

The reaction product 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), find the widest application. They are introduced into the composition of detergents and washing powders, used to reduce the hardness of water. At the same time, the ingress of excess amounts of phosphates with wastewater into water bodies contributes to the rapid development of algae (water bloom), which makes it necessary to carefully control the content of phosphates 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, it is necessary to formulate an equation for the reaction of phosphorus with oxygen and answer the question why the product of this reaction is used as a drying agent.

We write the reaction equation and arrange 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 reactions (compounds, decompositions, substitutions, exchanges) the interaction of phosphoric acid with sodium hydroxide belongs to.

Answer: ________

Solution

In the seventh task, it is necessary to draw up 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 PO 4 + 3 NaOH = Na 3 PO 4 + 3 H 2 O

Here we see that hydrogen and sodium in the reaction products are reversed.

Task 8

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

Answer: ________

2) Specify the sign of this reaction.

Answer: ________

Solution

Let us write the reaction equation in a reduced 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 in the form of a gas, or forms a low-dissociating substance (for example, water). The solubility table will help us with this.

Na 3 PO 4 + 3 AgNO 3 = Ag 3 PO 4 + 3 NaNO 3

The downward arrow next to silver phosphate 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 reaction equations. We write the full ionic equation for this reaction:

Now we cross out the ions that have moved from the left side of the equation to the right side without changing their charge:

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

Everything that is not crossed out, we write out in an abbreviated ionic equation:

PO 4 3– + 3Ag + = Ag3PO4

Task 9

The scheme of the redox reaction is given.

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

1. Make an electronic balance of this reaction.

Answer: ________

2. Specify the oxidizing agent and reducing agent.

Answer: ________

3. Arrange the coefficients in the reaction equation.

Answer: ________

Solution

The next task is 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 this compound. Do not forget that in total all the 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 –

Mn2+ (O 2– H +) 2 + K+Br 5+ O 3 2– → Mn4+ O 2 2– + K + Br – + H 2 + O 2 –

Mn 2+ –2e → Mn 4+ The process of electron donation is oxidation. At the same time, the oxidation state of the element increases during the reaction. This element is a reducing agent, it restores bromine.

Br 5+ +6e → Br - The process of accepting electrons - recovery. 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 at the same time (the oxidation state of an element decreases).

A reducing agent is a substance that donates electrons and is oxidized at the same time (the oxidation state of an element decreases). At school, it is written as follows.

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

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

If necessary, we arrange 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 factor of 3 in front of the water.

The transformation scheme is given:

Cu → CuCl 2 → Cu(OH) 2 → Cu(NO 3) 2

Write the molecular equations of the reactions by which these transformations can be carried out.

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 is in the series of voltages of metals 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, therefore, 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 in the table the selected numbers under the corresponding letters.

1. Methanol is an alcohol. The names of monohydric alcohols end in -ol, so A2.

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

3. Glucose is a carbohydrate, a monosaccharide. Therefore, we choose IN 1.

In the proposed schemes of 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 Reaction of reduction of acetaldehyde to ethyl alcohol.

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

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

Answer: ________

Task. We write down a brief condition of the problem:

m (CH 3 COOS 3 H 7) \u003d?

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

We substitute the numbers: n \u003d 300 g: 60 g / mol \u003d 5 mol. Thus, acetic acid reacted with propyl alcohol in an amount of 5 mol. Next, we determine how many moles of CH 3 COOS 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 ester is also formed, since there are no coefficients in the reaction equation. Therefore, if we take an acid in the amount of 5 mol, then the ether will also be 5 mol. Because they react in a 1:1 ratio.

Well, it remains to calculate the mass of 5 moles of ether using this triangle.

Substituting the numbers, we get: 5 mol 102 g / mol \u003d 510 g.

Answer: mass of ether = 510 g.

Acetylene is used as a fuel for 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 scheme below, write the equations for the reactions characteristic of acetylene. When writing reaction equations, use the structural formulas of organic substances.

Solution

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

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

Ringer's solution is widely used in medicine as a regulator of water-salt balance, a substitute for plasma 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 its brief condition:

m(H 2 O) \u003d 1000 g.

m(CaCl 2) \u003d 0.33 g.

m(KCl) = 0.3 g.

m(NaCl) = 8.6 g.

Since the density of water is unity, 1 liter of water will have a mass equal to 1000 grams. Next, to find the mass fraction in percent of the solution, we use the magic triangle,

m (in-va) - the mass of the substance;

m(r-ra) - mass of the solution;

ω is the mass fraction of a substance in percent in a given solution.

We derive a formula for finding ω% in solution. It will look like this:

ω% (p-ra NaCl)

In order to immediately proceed to finding the mass fraction in percent of the NaCl solution, we must know two other values, that is, the mass of the substance and the mass of the solution. The mass of the substance is known to us from the conditions of the problem, and the mass of the solution should 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 calculating is simple: m (in-va) \u003d 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:

Plugging in the numbers, we get:

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

VPR All-Russian Verification Work - Chemistry Grade 11

Explanations to the sample of the All-Russian verification work

When familiarizing yourself with the sample test work, it should be borne 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 the All-Russian test work in chemistry. The purpose of the test work sample 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.

Work instructions

The test work includes 15 tasks. To complete the work in chemistry, 1 hour 30 minutes (90 minutes) is allotted.
Prepare answers in the text of the work according to the instructions for the tasks. If you write down an incorrect answer, cross it out and write down a new one next to it.
When performing work, it is allowed to use the following additional materials:
– Periodic system of chemical elements D.I. Mendeleev;
- table of solubility of salts, acids and bases in water;
– electrochemical series of voltages of metals;
- non-programmable calculator.
When completing assignments, you can use a draft. Draft entries 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 the task that you can't complete right away and move on to the next one. If after completing all the work you have time left, you can return to the missed tasks.
The points you get 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 the course of chemistry, you know the following methods for separating mixtures: sedimentation, filtration, distillation (distillation), magnet action, evaporation, crystallization. Figures 1-3 show examples of some of these methods.

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

iron filings are attracted by a magnet

during distillation, after the 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, perform the following tasks:
1) determine the chemical element whose atom has such an electronic structure;
2) indicate the period number and group number in the Periodic system of chemical elements D.I. Mendeleev, in which this element is located;
3) determine whether a simple substance that forms this chemical element belongs to metals or non-metals.
Record your answers in a table.
Answer:

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

to determine the chemical element, you should calculate 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 that, we determine the group number (vertical column) (5) and the nature of this element (non-metal)

3. Periodic system of chemical elements D.I. Mendeleev- a rich repository of information about chemical elements, their properties and properties of their compounds, about the patterns of changes in these properties, about methods for obtaining substances, as well as about their presence in nature. So, for example, it is known that with an increase in the ordinal number of a chemical element in periods, the radii of atoms decrease, and in groups they increase.
Given 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 correct sequence.

Answer: ____________________________

N → C → Si → Al

4. The table below lists the characteristic properties of 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 scheme, for each of the four groups, enter the missing group names or chemical formulas of substances (one example of the formulas) belonging to this group.

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

CaO, bases, HCl, salts

Read the following text and do tasks 6-8.

In the food industry, the food additive E526 is used, which is calcium hydroxide Ca (OH) 2. It finds application in the production of: fruit juices, baby food, pickled cucumbers, table salt, confectionery and sweets.
Production of calcium hydroxide on an industrial scale is possible by mixing calcium oxide with water, this process is called quenching.
Calcium hydroxide has been 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 a calcium hydroxide solution is used to measure the amount of 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 raise the pH of water, since natural water contains substances (for example, 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 \u003d Ca (OH) 2
2) When calcium oxide interacts with water, a large amount of
the amount of heat, so the water boils and hisses, as if it hits hot coal, when the fire is extinguished with water (or “extinguishing this process is called because slaked lime is formed as a result”)

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 make it possible to use it to detect
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, clouding of the initial solution is observed, which makes it possible to judge the presence of carbon dioxide in the air (qualitative
reaction to CO 2)

1. Make an abbreviated ionic equation of 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 rise

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

9. The scheme of the redox reaction is given.

H 2 S + Fe 2 O 3 → FeS + S + H 2 O

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

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

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

1) Compiled electronic balance:

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 is composed:
3H 2 S + Fe 2 O 3 \u003d 2FeS + S + 3H 2 O

10. The scheme of transformations is given:

Fe → FeCl 2 → Fe(NO 3) 2 → Fe(OH) 2

Write the molecular reaction equations that can be used to carry out
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 \u003d Fe (NO 3) 2 + 2AgCl
3) Fe(NO 3) 2 + 2KOH = Fe(OH) 2 + 2KNO 3
(Other ones that do not contradict the condition of setting the equation are allowed
reactions.)

11. Establish a correspondence between the formula of organic matter 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 in the table the selected numbers under the corresponding letters.
Answer:

A	B	V

C3H8 - CnH2n + 2 - alkane
C3H6 - CnH2n- alkene
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 emissions of toxic substances 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 (N.O.) is formed during the complete combustion of 4.4 g of propane?
Write down a detailed solution to the problem.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

1) An equation for the reaction of propane combustion was compiled:
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
2) n (C 3 H 8) \u003d 4.4 / 44 \u003d 0.1 mol
n (CO 2) \u003d 3n (C 3 H 8) \u003d 0.3 mol
3) V (O 2) \u003d 0.3 22.4 \u003d 6.72 l

14. Isopropyl alcohol is used as a universal solvent: it is part of household chemicals, perfumes and cosmetics, windshield washer fluids for cars. In accordance with the scheme below, make up the equations for the reactions for obtaining this alcohol. When writing reaction equations, use the structural formulas of organic substances.

1) _______________________________________________________
2) _______________________________________________________
3) _______________________________________________________

The reaction equations are written corresponding to the scheme:

(Other ones are allowed that do not contradict the condition of setting the reaction equation.)

15. Physiological saline in medicine is called 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. Write down a detailed solution to the problem.
Answer:__________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________

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

m(r-ra) = 500g m(salt) = x

x/500 * 100%= 0.9%

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

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

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

VPR options in chemistry 11th grade 2017

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

To get acquainted with the exemplary options for work on the official website of FIPI, demo options with answers and descriptions are posted.

VPR samples in chemistry 11th grade 2017 (demo version)

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

When performing work, it is allowed to use the following additional materials:

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

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

– electrochemical series of voltages of metals;

- non-programmable calculator.

The structure and content of the All-Russian verification work of the VPR in chemistry

Each version of the VLOOKUP contains 15 tasks of various types and levels of complexity. The options include assignments of various formats.

These tasks have differences in the required form of recording the answer. So, for example, the answer can be: a sequence of numbers, symbols; the 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 difficult, since their implementation involves the complex application of the following skills:

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

Explain the conditionality of the properties and methods of obtaining substances by their composition and structure;

– to model a chemical experiment based on its description.

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

VLOOKUP. Chemistry. Grade 11. 10 options for typical tasks. Drozdov A.A.

M.: 20 1 7. - 9 6 p.

This manual fully complies with the federal state educational standard (second generation). The book contains 10 variants of typical tasks of the All-Russian Testing Work (VPR) in chemistry for 11th grade students. The collection is intended for 11th grade students, teachers and methodologists who use standard tasks to prepare for the All-Russian test work in chemistry.

Format: pdf

The size: 3.4 MB

Watch, download:drive.google

Work instructions, 4
Option 1 5
Option 2 12
Option 3 19
Option 4 26
Option 5 33
Option 6 40
Option 7 47
Option 8 54
Option 9 61
Option 10 68
Assessment system for test work 75
Answers 76
Applications 93

The test work includes 15 tasks. To complete the work in chemistry, 1 hour 30 minutes (90 minutes) is allotted.
Prepare answers in the text of the work according to the instructions for the tasks. If you write down an incorrect answer, cross it out and write down a new one next to it.
When performing work, it is allowed to use the following additional materials:
- Periodic system of chemical elements D.I. Mendeleev;
- table of solubility of salts, acids and bases in water;
- electrochemical series of voltages of metals;
- non-programmable calculator.
When completing assignments, you can use a draft. Draft entries 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 the task that you can't complete right away and move on to the next one. If after completing all the work you have time left, you can return to the missed tasks.
The points you get for completed tasks are summed up. Try to complete as many tasks as possible and score the most points.