Tasks 34 in chemistry ege. Ege in chemistry
Task 34
When a sample of calcium carbonate was heated, part of the substance decomposed. In this case, 4.48 l (n.o.) of carbon dioxide was released. The mass of the solid residue was 41.2 g. This residue was added to 465.5 g of a solution of hydrochloric acid, taken in excess. Determine the mass fraction of salt in the resulting solution.
In the response, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations (indicate the units of measurement of the desired quantities).
The handbook contains detailed theoretical material on all topics checked by the exam in chemistry. After each section, multilevel tasks in the form of the exam are given. For the final control of knowledge at the end of the directory are given training options corresponding to the exam. Students will not have to look for additional information on the Internet and buy other benefits. In this guide they will find everything you need for independent and effective preparation for the exam. The handbook is addressed to high school students to prepare for the exam in chemistry.
Answer: We write a brief condition for this problem.
After all the preparations are given, we proceed to the solution.
1) Determine the amount of CO 2 contained in 4.48 liters. his.
n(CO 2) \u003d V / Vm \u003d 4.48 L / 22.4 L / mol \u003d 0.2 mol
2) Determine the amount of calcium oxide formed.
According to the reaction equation, 1 mol of CO 2 and 1 mol of CaO are formed
Consequently: n(CO 2) \u003d n(CaO) and equals 0.2 mol
3) Determine the mass of 0.2 mol CaO
m(CaO) \u003d n(CaO) M(CaO) \u003d 0.2 mol 56 g / mol \u003d 11.2 g
Thus, a solid residue weighing 41.2 g consists of 11.2 g CaO and (41.2 g - 11.2 g) 30 g CaCO 3
4) Determine the amount of CaCO 3 contained in 30 g
n(CaCO 3) \u003d m(CaCO 3) / M(CaCO 3) \u003d 30 g / 100 g / mol \u003d 0.3 mol
The attention of schoolchildren and applicants for the first time offered a training manual for preparation for the exam in chemistry, which contains training tasks collected on topics. The book presents tasks of different types and levels of difficulty for all the tested topics of the chemistry course. Each of the sections of the manual includes at least 50 tasks. The tasks correspond to the modern educational standard and the regulation on the conduct of a unified state exam in chemistry for graduates of secondary educational institutions. Fulfillment of the proposed training tasks on the topics will allow you to qualitatively prepare for the exam in chemistry. The manual is addressed to high school students, applicants and teachers.
CaO + HCl \u003d CaCl 2 + H 2 O
CaCO 3 + HCl \u003d CaCl 2 + H 2 O + CO 2
5) Determine the amount of calcium chloride formed as a result of these reactions.
0.3 mol of CaCO 3 and 0.2 mol of CaO in total 0.5 mol reacted.
Accordingly, 0.5 mol of CaCl 2 is formed.
6) Calculate the mass of 0.5 mol of calcium chloride
M(CaCl 2) \u003d n(CaCl 2) M(CaCl 2) \u003d 0.5 mol · 111 g / mol \u003d 55.5 g.
7) Determine the mass of carbon dioxide. 0.3 mol of calcium carbonate was involved in the decomposition reaction, therefore:
n(CaCO 3) \u003d n(CO 2) \u003d 0.3 mol,
m(CO 2) \u003d n(CO 2) M(CO 2) \u003d 0.3 mol · 44g / mol \u003d 13.2g.
8) Find the mass of the solution. It consists of the mass of hydrochloric acid + the mass of the solid residue (CaCO 3 + CaO) minutes the mass of released CO 2. We write this in the form of a formula:
m(solution) \u003d m(CaCO 3 + CaO) + m(HCl) - m(CO 2) \u003d 465.5 g + 41.2 g - 13.2 g \u003d 493.5 g.
The new reference book contains all the theoretical material on the chemistry course needed to pass the exam. It includes all content elements that are checked by control and measuring materials, and helps to generalize and systematize knowledge and skills for a course of a secondary (full) school. The theoretical material is presented in a short, accessible form. Each section is accompanied by examples of training tasks that allow you to test your knowledge and degree of preparedness for the certification exam. Practical exercises correspond to the exam format. At the end of the manual answers are given to tasks that will help to objectively assess the level of your knowledge and the degree of preparedness for the certification exam. The manual is addressed to high school students, applicants and teachers.
9) And finally, we will answer the question of the problem. We find the mass fraction in% of salt in the solution using the following magic triangle:
ω% (CaCI 2) \u003d m(CaCI 2) / m(solution) \u003d 55.5 g / 493.5 g \u003d 0.112 or 11.2%
Answer: ω% (CaCI 2) \u003d 11.2%
Option No. 1380120
Missions 34 (C5). Sergei Shirokopoyas: Chemistry - preparation for the exam 2016
When performing tasks with a short answer, write in the field for the answer a number that corresponds to the number of the correct answer, or a number, word, sequence of letters (words) or numbers. The answer should be written without spaces or any additional characters. Separate the fractional part from the whole decimal point. Units do not need to be written. The answer to tasks 1-29 is a sequence of numbers or a number. For the full correct answer in tasks 7-10, 16-18, 22-25, 2 points are put; if one mistake is made - 1 point; for an incorrect answer (more than one error) or its absence - 0 points.
If the option is set by the teacher, you can enter or upload answers to tasks with a detailed answer into the system. The teacher will see the results of the tasks with a short answer and will be able to evaluate the downloaded answers to the tasks with a detailed answer. The points set by the teacher will be displayed in your statistics.
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Some organic matter A contains, by weight, 11.97% nitrogen, 51.28% carbon, 27.35% oxygen, and hydrogen. And formed by the interaction of substance B with propanol-2 in a molar ratio of 1: 1. It is known that substance B has a natural origin.
4) Write the equation of the reaction for obtaining substance A from substance B and propanol-2.
The combustion of 40.95 g of organic matter gave 39.2 liters of carbon dioxide (n.a.), 3.92 l of nitrogen (n.a.) and 34.65 g of water. When heated with hydrochloric acid, this substance undergoes hydrolysis, the products of which are compounds of the composition and secondary alcohol.
Based on the data of the problem condition:
2) establish its molecular formula;
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The primary amine salt reacted with silver nitrate, as a result of which a precipitate was formed and organic substance A was formed, containing 29.79% nitrogen, 51.06% oxygen and 12.77% carbon by weight.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of substance A;
2) establish its molecular formula;
3) compose the structural formula of this substance A, which reflects the order of atomic bonds in the molecule;
4) write the equation of the reaction for obtaining substance A from the salt of the primary amine and.
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When burning a dipeptide of natural origin weighing 2.64 g, 1.792 l of carbon dioxide (n.a.), 1.44 g of water and 448 ml of nitrogen (n.a.) were obtained. Upon hydrolysis of this substance in the presence of hydrochloric acid, only one salt was formed.
Based on the data of the problem condition:
2) establish its molecular formula;
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On the next page you will be asked to check them yourself.
Some organic substance A contains 13.58% nitrogen, 46.59% carbon and 31.03% oxygen by mass, and is formed by the interaction of substance B with ethanol in a 1: 1 molar ratio. It is known that substance B is of natural origin.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of substance A;
2) establish its molecular formula;
3) compose the structural formula of substance A, which reflects the order of atomic bonds in the molecule;
4) write the equation of the reaction for obtaining substance A from substance B and ethanol.
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Some organic substance A contains 10.68% nitrogen, 54.94% carbon and 24.39% oxygen by mass, and is formed by the interaction of substance B with propanol-1 in a 1: 1 molar ratio. It is known that substance B is a natural amino acid.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of substance A;
2) establish its molecular formula;
3) compose the structural formula of substance A, which reflects the order of atomic bonds in the molecule;
4) write the equation of the reaction for obtaining substance A from substance B and n-propanol.
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A certain substance, which is a salt of organic origin, contains 12.79% nitrogen, 43.84% carbon and 32.42% chlorine by mass and is formed by the reaction of a primary amine with chloroethane.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of the starting organic matter;
2) establish its molecular formula;
3) make up the structural formula of this substance, which reflects the order of atomic bonds in the molecule;
4) write the equation of the reaction for obtaining this substance from primary amine and chloroethane.
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When burning a dipeptide of natural origin weighing 3.2 g, 2.688 L of carbon dioxide (n.a.), 448 ml of nitrogen (n.a.) and 2.16 g of water are obtained. Upon hydrolysis of this substance in the presence of potassium hydroxide, only one salt was formed.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of the dipeptide;
2) establish its molecular formula;
3) compose the structural formula of the dipeptide, which reflects the order of the bonds of atoms in the molecule;
4) write the equation for the hydrolysis of this dipeptide in the presence of potassium hydroxide.
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When burning a naturally occurring dipeptide weighing 6.4 g, 5.376 l of carbon dioxide (n.a.), 896 ml of nitrogen (n.a.) and 4.32 g of water are obtained. Upon hydrolysis of this substance in the presence of hydrochloric acid, only one salt was formed.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of the dipeptide;
2) establish its molecular formula;
3) compose the structural formula of the dipeptide, which reflects the order of the bonds of atoms in the molecule;
4) write the equation for the hydrolysis of this dipeptide in the presence of hydrochloric acid.
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The combustion of some organic matter weighing 4.12 g yielded 3.584 l of carbon dioxide (n.a.), 448 ml of nitrogen (n.a.) and 3.24 g of water. When heated with hydrochloric acid, this substance undergoes hydrolysis, the products of which are compounds of the composition and alcohol.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of the starting organic matter;
2) establish its molecular formula;
3) make up the structural formula of this substance, which reflects the order of atomic bonds in the molecule;
4) write the equation of the hydrolysis reaction of this substance in the presence of hydrochloric acid.
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Upon the combustion of some organic matter weighing 4.68 g, 4.48 L of carbon dioxide (n.a.), 448 ml of nitrogen (n.a.) and 3.96 g of water were obtained. When heated with a solution of sodium hydroxide, this substance undergoes hydrolysis, the products of which are a salt of a natural amino acid and secondary alcohol.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of the starting organic matter;
2) establish its molecular formula;
3) make up the structural formula of this substance, which reflects the order of atomic bonds in the molecule;
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The combustion of some organic matter weighing 17.55 g gave 16.8 liters of carbon dioxide (n.a.), 1.68 l of nitrogen (n.a.) and 14.85 g of water. When heated with a solution of sodium hydroxide, this substance undergoes hydrolysis, the products of which are a salt of a natural amino acid and secondary alcohol.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of the starting organic matter;
2) establish its molecular formula;
3) make up the structural formula of this substance, which reflects the order of atomic bonds in the molecule;
4) write the equation of the hydrolysis reaction of this substance in the presence of sodium hydroxide.
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On the next page you will be asked to check them yourself.
The combustion of some organic matter weighing 35.1 g gave 33.6 l of carbon dioxide (n.a.), 3.36 l of nitrogen (n.a.) and 29.7 g of water. When heated with a solution of potassium hydroxide, this substance undergoes hydrolysis, the products of which are a salt of a natural amino acid and secondary alcohol.
Based on the data of the problem condition:
1) make the calculations necessary to find the formula of the starting organic matter;
2) establish its molecular formula;
In our last article, we talked about the basic tasks in the exam in chemistry in 2018. Now, we have to analyze in more detail the tasks of an increased (in the USE in chemistry codifier 2018, a high level of complexity) difficulty level, previously referred to as part C.
Only five (5) tasks are assigned to the tasks of an increased level of complexity - №30.311.32.22.33.34 and 35. Let us consider the topics of the tasks, how to prepare for them and how to solve complex tasks in the Unified State Examination in chemistry in 2018.
Example of task 30 in the exam in chemistry 2018
It is aimed at testing the student's knowledge about redox reactions (OVR). The task always gives an equation of a chemical reaction with omissions of substances on either side of the reaction (the left side is the reagents, the right side is the products). You can get a maximum of three (3) points for this assignment. The first point is given for the correct filling of gaps in the reaction and the correct equalization of the reaction (arrangement of coefficients). The second point can be obtained by correctly painting the balance of the OVR, and the last point is given for correctly determining who is the oxidizing agent in the reaction and who is the reducing agent. We will analyze the solution to task No. 30 from the demo version of the exam in chemistry in 2018:
Using the electronic balance method, make up the reaction equation
Na 2 SO 3 + ... + KOH à K 2 MnO 4 + ... + H 2 O
Identify the oxidizing agent and reducing agent.
The first thing to do is to arrange the charges of the atoms indicated in the equation, it turns out:
Na + 2 S +4 O 3 -2 + ... + K + O -2 H + à K + 2 Mn +6 O 4 -2 + ... + H + 2 O -2
Often after this action, we immediately see the first pair of elements, which changed the oxidation state (CO), that is, from different sides of the reaction, the same atom has a different oxidation state. In this particular task, we do not observe this. Therefore, it is necessary to use additional knowledge, namely, on the left side of the reaction, we see potassium hydroxide ( KOH), the presence of which tells us that the reaction proceeds in an alkaline environment. On the right side, we see potassium manganate, and we know that in an alkaline reaction medium, potassium manganate is obtained from potassium permanganate, therefore, a pass on the left side of the reaction is potassium permanganate ( KMnO 4 ). It turns out that on the left we had manganese in CO +7, and on the right in CO +6, so we can write the first part of the balance of the OVR:
Mn +7 +1 e — à Mn +6
Now, we can assume what else should happen in the reaction. If manganese receives electrons, then someone had to give them to him (we observe the law of conservation of mass). Consider all the elements on the left side of the reaction: hydrogen, sodium and potassium are already in CO +1, which is maximum for them, oxygen will not give its electrons to manganese, which means that sulfur remains in CO +4. We conclude that sulfur gives up electrons and goes into a state of sulfur with CO +6. Now we can write the second part of the balance:
S +4 -2 e — à S +6
Looking at the equation, we see that on the right side, there is no sulfur and sodium anywhere, which means they should be in the gap, and sodium sulfate is the logical connection to fill it ( NaSO 4 ).
Now the OVR balance is written (we get the first point) and the equation takes the form:
Na 2 SO 3 + KMnO 4 + KOHà K 2 MnO 4 + NaSO 4 + H 2 O
| Mn +7 +1 e — à Mn +6 | 1 | 2 |
| S +4 -2e -à S +6 | 2 | 1 |
It is important to immediately write in this place who is the oxidizing agent and who is the reducing agent, since students often focus on balancing the equation and simply forget to do this part of the task, thereby losing a point. By definition, an oxidizing agent is that particle that receives electrons (in our case, manganese), and a reducing agent is that particle that gives off electrons (in our case, sulfur), so we get:
Oxidizing agent: Mn +7 (KMnO 4 )
Reducing agent: S +4 (Na 2 SO 3 )
Here it must be remembered that we indicate the state of the particles in which they were when they began to exhibit the properties of an oxidizing agent or a reducing agent, and not the states to which they came as a result of the general reaction analysis.
Now, to get the last score, you need to correctly equate the equation (put the coefficients). Using balance, we see that in order for it to turn sulfur +4, it will go into state +6, two manganese +7, should become manganese +6, and therefore we put 2 in front of manganese:
Na 2 SO 3 + 2KMnO 4 + KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O
Now we see that we have 4 potassium on the right, and only three on the left, so we need to put 2 in front of potassium hydroxide:
Na 2 SO 3 + 2KMnO 4 + 2KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O
As a result, the correct answer to the task No. 30 is as follows:
Na 2 SO 3 + 2KMnO 4 + 2KOHà 2K 2 MnO 4 + NaSO 4 + H 2 O
| Mn +7 + 1e -à Mn +6 | 1 | 2 |
| S +4 -2e -à S +6 | 2 | 1 |
Oxidizing agent: Mn +7 (KMnO 4)
Reducing agent: S +4 (Na 2 SO 3 )
The solution of task 31 in the exam in chemistry
This is a chain of inorganic transformations. For the successful completion of this task, it is necessary to be well versed in the reactions characteristic of inorganic compounds. The task consists of four (4) reactions, for each of which you can get one (1) point, in total for the task you can get four (4) points. It is important to remember the rules for completing the assignment: all equations must be equalized, even if the student wrote the equation correctly, but did not equalize, he will not get a point; it’s not necessary to solve all the reactions, you can make one and get one (1) point, two reactions and get two (2) points, etc., while it is not necessary to execute the equations strictly in order, for example, the student can make reaction 1 and 3, then that’s what you need to do, and get two (2) points, the main thing is to indicate that these are reactions 1 and 3. We will analyze the solution to task No. 31 from the USE trial in chemistry in 2018:
Iron was dissolved in hot concentrated sulfuric acid. The resulting salt was treated with excess sodium hydroxide solution. The precipitated brown precipitate was filtered off and calcined. The resulting material was heated with iron.
Write the equations of the four reactions described.
For the convenience of the solution, in a draft, you can draw up the following diagram:
To complete the task, of course, you need to know all the proposed reactions. However, there are always hidden clues in the condition (concentrated sulfuric acid, excess sodium hydroxide, brown precipitate, calcined, heated with iron). For example, a student does not remember what happens to iron when interacting with conc. sulfuric acid, but he remembers that the brown precipitate of iron, after treatment with alkali, is most likely iron hydroxide 3 ( Y = Fe(OH) 3 ) Now we have the opportunity, substituting Y in the written scheme, to try to make equations 2 and 3. The subsequent actions are purely chemical, so we will not write them in such detail. The student must remember that heating iron hydroxide 3 leads to the formation of iron oxide 3 ( Z = Fe 2 O 3 ) and water, and heating iron oxide 3 with pure iron will lead them to the middle state - iron oxide 2 ( FeO) Substance X, which is a salt obtained after reaction with sulfuric acid, while yielding iron hydroxide 3 after treatment with alkali, will be iron 3 sulfate ( X = Fe 2 (SO 4 ) 3 ) It is important to remember to equalize equations. As a result, the correct answer to the task No. 31 is as follows:
| 1) 2Fe + 6H 2 SO 4 (k) à Fe 2 (SO 4) 3+ 3SO 2 + 6H 2 O |
| 2) Fe 2 (SO 4) 3+ 6NaOH (g) à 2 Fe (OH) 3+3Na 2 SO 4 |
| 3) 2Fe (OH) 3à Fe 2 O 3 + 3H 2 O |
| 4) Fe 2 O 3 + Fe à 3feo |
Task 32 exam in chemistry
Very similar to task number 31, only it gives a chain of organic transformations. The design requirements and decision logic are similar to task 31, the only difference is that task 32 gives five (5) equations, which means you can get five (5) points in total. Due to the similarity with task No. 31, we will not consider it in detail.
The solution of task 33 in chemistry 2018
The calculation task, for its implementation it is necessary to know the basic calculation formulas, be able to use a calculator and draw logical parallels. For task number 33 you can get four (4) points. Consider the part of the solution to task No. 33 from the demo exam in chemistry in 2018:
Determine the mass fractions (in%) of iron (II) sulfate and aluminum sulfide in the mixture if, when 25 g of this mixture was treated with water, gas was released that completely reacted with 960 g of a 5% solution of copper sulfate. In the response, write down the reaction equations that are indicated in the condition of the problem, and give all the necessary calculations (indicate the units of measurement of the sought physical quantities).
The first (1) point we get for writing the reactions that occur in the task. Obtaining this particular point depends on the knowledge of chemistry, the remaining three (3) points can be obtained only through calculations, therefore, if a student has problems with mathematics, he must receive at least one (1) point for completing task 33:
| Al 2 S 3 + 6H 2 Oà 2Al (OH) 3 + 3H 2 S |
| CuSO 4 + H 2 Sà CuS + H 2 SO 4 |
Since further actions are purely mathematical, we will not begin to analyze them here. You can watch the selection analysis on our YouTube channel (link to the video of the analysis of task No. 33).
Formulas that will be required to solve this task:
Task 34 in chemistry 2018
The calculated task, which differs from task No. 33 as follows:
- If in task No. 33 we know what substances interact between, then in task No. 34 we must find what reacted;
- In task number 34, organic compounds are given, while in task number 33, inorganic processes are most often given.
In fact, task number 34 is the opposite of task number 33, and therefore the logic of the task is the opposite. For task No. 34 you can get four (4) points, while, as in task No. 33, only one of them (in 90% of cases) is obtained for knowledge of chemistry, the remaining 3 (less than 2) points are obtained for mathematical calculations . For the successful completion of task number 34, you must:
Know the general formulas of all the main classes of organic compounds;
Know the basic reactions of organic compounds;
Be able to write an equation in a general way.
Once again, I want to note that the theoretical bases necessary for the successful passing of the exam in chemistry in 2018 have not changed, which means that all the knowledge that your child received at school will help him pass the chemistry exam in 2018. In our center for preparation for the Unified State Exam and Unified State Examination Godograph, your child will receive all necessary for the preparation of theoretical materials, and in the classroom will consolidate the knowledge gained for successful implementation all exam assignments. The best teachers who have passed a very large competition and complex entrance tests will work with him. Classes are held in small groups, which allows the teacher to take time for each child and form his individual strategy for exam work.
We have no problems with the lack of tests of a new format, our teachers write them themselves, based on all the recommendations of the codifier, specifier and demo of the exam in chemistry in 2018.
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Preparation for the exam in chemistry is covered by our experts in this section - analysis of problems, reference data and theoretical material. Preparing for the exam is now easy and free with our sections on each subject! We are sure that you will pass the unified state exam in 2019 for the maximum score!
General Exam Information
The exam in chemistry consists of two parts and 34 tasks .
First part contains 29 tasks with a short answer, including 20 tasks of the basic difficulty level: No. 1–9, 12–17, 20–21, 27–29. Nine tasks of an increased level of difficulty: No. 9–11.17–19, 22–26.
Second part contains 5 tasks of high complexity with a detailed answer: No. 30–34
Tasks of a basic level of complexity with a short answer test the assimilation of the content of the most important sections of the school chemistry course: theoretical foundations of chemistry, inorganic chemistry, organic chemistry, methods of knowledge in chemistry, chemistry and life.
Tasks advanced difficulty with a short answer, they are focused on checking the essential elements of the content of the basic educational programs in chemistry not only at the basic, but also at an advanced level. Compared with the tasks of the previous group, they provide for a wider variety of actions for applying knowledge in a changed, non-standard situation (for example, to analyze the nature of the studied types of reactions), as well as the ability to systematize and generalize the knowledge gained.
Tasks from detailed answer , unlike the tasks of the two previous types, provide a comprehensive verification of the mastery at an in-depth level of several content elements from various content blocks.