New options for OGE early chemistry. Tests by topic
Training test for preparation for the OGE - 2018 in chemistry in 9th grade
Instructions for performing the work
2 hours (120 minutes) are allotted to complete the work. The work consists of 2 parts, including 22 tasks. Part 1 contains 19 short-answer tasks, part 2 contains 3 long-answer tasks.
Answers to tasks 1-15 are written in the form of one number, which corresponds to the number of the correct answer.
Answers to tasks 16-19 are written as a sequence of numbers.
For tasks 20-22, you should give a complete, detailed answer, including the necessary reaction equations and the solution to the problem.
When doing work you can use Periodic table chemical elements D.I. Mendeleev, a table of solubility of salts, acids and bases in water, an electrochemical series of metal voltages and a non-programmable calculator.
Part 1
1.Chemical element The 2nd period of the VIA group corresponds to the electron distribution scheme
1) Fig. 1
2) Fig. 2
3) Fig. 3
4) Fig. 4
Answer:
2. The nonmetallic properties of simple substances increase in the series
1) phosphorus → silicon → aluminum
2) fluorine → chlorine → bromine
3) selenium → sulfur → oxygen
4) nitrogen → phosphorus → arsenic
Answer:
3. Covalent polar connection realized in matter
1) CuO
2) P 4
3)SO2
4) MgCl 2
Answer:
4 . In which compound does the oxidation state of chlorine equal to +7?
1)HCl
2) Cl 2 O
3) KClO 3
4) KClO 4
Answer:
5. Substances whose formulas are ZnO and Na 2 SO 4 , are respectively
1) basic oxide and acid
2) amphoteric hydroxide and salt
3) amphoteric oxide and salt
4) main oxide and base
Answer:
6. A reaction whose equation is
2NaOH + CuCl 2 = Cu(OH) 2 + 2NaCl
refers to reactions
1) decomposition
2) connections
3) substitution
4) exchange
Answer:
7. The smallest amount of positive ions is formed during the dissociation of 1 mol
1) nitric acid
2) sodium carbonate
3) aluminum sulfate
4) potassium phosphate
Answer:
8. The irreversible occurrence of the ion exchange reaction between solutions of barium hydroxide and potassium carbonate is due to the interaction of ions
1) K + and OH -
2) K + and CO 3 2―
3) Ba 2+ and CO 3 2―
4) Ba 2+ and OH -
Answer:
9. Copper reacts with solution
1) AgNO3
2) Al 2 (SO 4 ) 3
3) Fe SO 4
4) NaOH
Answer:
10 . Copper(II) oxide can react with each substance in the pair
1) HCl, O 2
2) Ag, SO 3
3) H 2, SO 4
4) Al, N 2
Answer:
11 . Determine the formula of the unknown substance in the reaction scheme:
KOH + …→ K 2 CO 3 +H2O
1) CO
2) CO2
3) CH 4
4) C
Answer:
12. You can convert CaNO3 to CaSO3 using
1) hydrogen sulfide
2) barium sulfite
3) sodium sulfite
4) sulfur dioxide
Answer:
13. Are the judgments about the methods of separating mixtures correct?
A. Evaporation is referred to as physical means separation of mixtures.
B. Separation of a mixture of water and ethanol is possible by filtration.
1) only A is correct
2) only B is correct
3) both judgments are correct
4) both judgments are incorrect
Answer:
14. In the reaction 3CuO + 2NH 3 = 3Cu+ N 2 + 3H 2 O
The change in the oxidation state of the oxidizing agent corresponds to the diagram
1) +2 → 0
2) −3 → 0
3) −2 → 0
4) 0 → +2
Answer:
15 . On which diagram is the distribution of mass fractions of elements
corresponds to NHNO 3
Part 2
16. When completing a task, choose two correct ones from the proposed list of answers and write down the numbers under which they are indicated.
In the series of chemical elements Be- Mg- Ca
1) atomic radius increases
2) increases highest degree oxidation
3) the value of electronegativity increases
4) the basic properties of the formed hydroxides increase
5) the number of electrons in the external level decreases
Answer:
18. Establish a correspondence between two substances and a reagent that can be used to distinguish between these substances.
SUBSTANCES | REAGENT |
|
A) NaNO 3 and Ca(NO 3) 2 B) FeCl 2 and FeCl 3 B) H 2 SO 4 and HNO 3 | 1) BaCl2 2) Na 2 CO 3 3) HCl 4) NaOH |
Write down the numbers in your answer, arranging them in the order corresponding to the letters:
19. Match the substance with the reagents with which it can react.
Answer:
20. Using the electronic balance method, arrange the coefficients in the reaction equation, the diagram of which
P + H 2 SO 4 →H 3 PO 4 + SO 2 + H 2 0
Identify the oxidizing and reducing agent
2, H 2 SO 4, CaCO 3
Assessment system for test work in chemistry
Correct completion of each task part 1 basic level Difficulty (1–15) is scored 1 point.
Correct completion of each task part 1 increased level of complexity (16–19) is assessed with a maximum of 2 points. Tasks 16 and 17 are considered correctly completed if two answer options are correctly selected in each of them. For an incomplete answer - one of two answers is correctly named or three answers are named, two of which are correct - 1 point is given. The remaining answer options are considered incorrect and are scored 0 points.
Tasks 18 and 19 are considered completed correctly if three correspondences are correctly established. An answer in which two out of three matches are established is considered partially correct; it is worth 1 point. The remaining options are considered an incorrect answer and are scored 0 points.
Part 1
Part 2
20. Using the electronic balance method, arrange the coefficients in the reaction equation, the diagram of which is:
HNO 3 + Zn = Zn(NO 3 ) 2 + NO + H 2 O
Specify the oxidizing agent and the reducing agent.
Response Elements |
|
1) Let's create an electronic balance: S +6 + 2ē = S +4 │2 │5 P 0 - 5ē = P +5 │5 │2 2) We point out that S +6 (H 2 SO 4 ) is an oxidizing agent, and P 0 (P)-reducing agent 3) Let’s arrange the coefficients in the reaction equations: 2P + 5H 2 SO 4 →2H 3 PO 4 + 5SO 2 + 2H 2 0 |
|
Evaluation criteria | Points |
There was an error in only one of the elements in the answer. | |
There were errors in two elements in the answer | |
Maximum score | |
21. When an excess solution of potassium carbonate reacted with a 10% solution of barium nitrate, 3.94 g of sediment precipitated. Determine the mass of the barium nitrate solution taken for the experiment.
Response Elements (other wording of the answer is allowed that does not distort its meaning) |
|
Explanation.
K 2 CO 3 + Ba(NO 3 ) 2 = ↓ + 2KNO 3 2) The amount of barium carbonate substance and the mass of barium nitrate are calculated: N(BaCO 3 ) = m(BaCO 3 ) / M(BaCO 3 ) = 3.94: 197 = 0.02 mol n (Ba(NO 3 ) 2 ) = n(BaCO 3 ) = 0.02 mol m (Ba(NO 3 ) 2 ) = n (Ba(NO 3 ) 2 ) M (Ba(NO 3 ) 2 ) = 0.02 261 = 5.22 g. 3) The mass of the barium nitrate solution was determined: M (solution) = m(Ba(NO 3) 2 / ω (Ba(NO 3) 2 = 5.22 / 0.1 = 52.2 g Answer: 52.2 g. |
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Evaluation criteria | Points |
The answer is correct and complete, includes all the named elements | |
2 elements from the above are written correctly | |
Correctly written 1 element from the above (1st or 2nd) | |
All elements of the answer are written incorrectly | |
Maximum score | |
22. Given substances: CuO, NaCl, KOH, MnO 2, H 2 SO 4, CaCO 3
Using water and the necessary substances only from this list, obtain copper(II) chloride in two stages. Describe the signs of the reactions being carried out. For the second reaction, write the abbreviated ionic equation for the reaction.
Response Elements (other wording of the answer is allowed that does not distort its meaning) |
|
Let's write 2 reaction equations: 2NaCl + H 2 SO 4 = 2HCl+ Na 2 SO 4 CuO +2HCl =CuCl 2 +H 2 O Let us indicate the signs of reactions. For the first reaction - gas evolution. For the dissolution reaction of CuO - a color change, the formation of a blue solution. Let's create a shortened ionic equation for the first reaction: CuO +2H + =Cu 2+ +H 2 O |
|
Evaluation criteria | Points |
The answer is correct and complete, includes all the named elements | |
Four elements of the answer are written correctly | |
Three elements of the answer are written correctly | |
Two elements of the answer are written correctly | |
One element of the answer is correctly written | |
All elements of the answer are written incorrectly | |
Maximum score | |
2018
The maximum number of points that an examinee can receive for completing all exam paper(without a real experiment), - 34 points.
Table 4
Conversion scale primary score for completing an exam paper with a mark on a five-point scale (work without a real experiment, demo version 1)
- 0-8 points - mark “2”
- 9-17 points - mark “3”
- 18-26 points - mark “4”
- 27-34 points - mark “5”
It is recommended to mark “5” if, out of the total amount of points sufficient to obtain this mark, the graduate scored 5 or more points for completing tasks in Part 3. The exam results can be used when admitting students to specialized classes high school. The guideline for selection into specialized classes can be an indicator whose lower limit corresponds to 23 points.
Option
control measuring materials for
holding in 2017 the main state
CHEMISTRY exam
Instructions for performing the work
The examination paper consists of two parts, including 22 tasks. Part 1 contains 19 short-answer tasks, part 2 contains 3 long-answer tasks.
You are given 2 hours (120 minutes) to complete the exam paper in chemistry.
Answers to tasks 1–15 are written as one number, which corresponds to the number of the correct answer. Write this figure in the answer field in the text of the work, and then transfer it to answer form No. 1.
Answers to tasks 16–19 are written as a sequence of numbers. Write this sequence of numbers in the answer field in the text of the work, and then transfer it to answer form No. 1.
For tasks 20–22, you should give a complete, detailed answer, including the necessary reaction equations and calculations. The tasks are completed on answer form No. 2.
When performing work, you can use the Periodic Table of Chemical Elements D.I. Mendeleev, a table of solubility of salts, acids and bases in water, an electrochemical series of metal voltages and a non-programmable calculator.
When completing assignments, you can use a draft. Entries in the draft are not taken into account when grading work.
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!
Part 1
The answer to tasks 1–15 is one number, which corresponds to the number of the correct answer. Write this number in the answer field in the text of the work, and then transfer it to ANSWER FORM No. 1 to the right of the number of the corresponding task, starting from the first cell.
A chemical element of the 3rd period of the VA group corresponds to a diagram of the distribution of electrons across layers:
An ionic bond is characteristic of each of two substances:
1) potassium chloride and hydrogen chloride
2) barium chloride and sodium oxide
3) sodium chloride and carbon monoxide (IV)
4) lithium oxide and chlorine
Answer:Each of the two substances is complex
1) water and chlorine
2) water and hydrogen
3) hydrogen and quartz
4) benzene and water
Answer:The largest number of positive ions is formed during the dissociation of 1 mol
1) sulfuric acid
2) sodium phosphate
3) iron(III) nitrate
4) potassium sulfide
Answer:Both substances of the pair react with water at room temperature
1) oxygen and carbon
2) magnesium and sulfur
3) aluminum and phosphorus
4) sodium and calcium
Answer:Among the substances: Zn, Al 2 O 3, Cu(OH) 2, BaCl 2 - reacts with a solution of sulfuric acid (s)
4) four
Answer:Are the judgments about the methods of separating mixtures correct?
A. Evaporation is a physical method for separating mixtures.
B. Separation of a mixture of water and ethanol is possible by filtration.
1) only A is correct
2) only B is correct
3) both judgments are correct
4) both judgments are incorrect
| The element carbon is an oxidizing agent in the reaction |
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The answer to tasks 16–19 is a sequence of numbers that should be written in ANSWER FORM No. 1 to the right of the number of the corresponding task, starting from the first cell. Write your answer without spaces, commas or other additional characters.
Write each number in a separate box in accordance with the samples given in the form.
When completing tasks 16, 17, from the proposed list of answers, select two correct ones and write down the numbers under which they are indicated in the table.
16. What magnesium and silicon have in common is
1) the presence of three electronic layers in their atoms
2) the existence of simple substances corresponding to them in the form of diatomic molecules
3) the fact that they belong to metals
4) the fact that their electronegativity value is less than that of phosphorus
5) their formation of higher oxides with the general formula EO 2
Answer:When completing tasks 18, 19, for each element of the first column, select the corresponding element from the second column. Write down the selected numbers in the table under the corresponding letters. The numbers in the answer may be repeated.
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| Match the name of the substance with the reagents with which this substance can interact. |
Do not forget to transfer all answers to answer form No. 1 in accordance with the instructions for completing the work.
Part 2
For answers to tasks 20–22, use ANSWER FORM No. 2.
First write down the number of the task (20, 21 or 22), and then the detailed answer to it. Write down your answers clearly and legibly.
20. Using the electron balance method, create an equation for the reaction
KMnO 4 + KOH → K 2 MnO 4 + O 2 + H 2 O
Identify the oxidizing agent and the reducing agent.
21. An excess of potassium hydroxide solution was added to 376 g of a solution with a mass fraction of copper (II) nitrate of 7.5%. Determine the mass of the deposited sediment.
22. Given substances: Zn, HCl (solution), H 3 PO 4, AgNO 3, NH 4 Cl, Ba(NO 3) 2. Using water and the necessary substances only from this list, obtain zinc nitrate in two stages. Describe the signs of the reactions being carried out. For an ion exchange reaction, write the abbreviated ionic equation for the reaction.
Tips for preparing for the Unified State Exam in chemistry on the website website
How to competently pass the Unified State Exam (and Unified State Exam) in chemistry? If you only have 2 months and you’re not ready yet? And don’t be friends with chemistry...
It offers tests with answers for each topic and task, by passing which you can study the basic principles, patterns and theory found in the Unified State Exam in chemistry. Our tests allow you to find answers to most questions encountered in the Unified State Exam in chemistry, and our tests allow you to consolidate the material, find weak spots, and work through the material.
All you need is the Internet, stationery, time and a website. It is best to have a separate notebook for formulas/solutions/notes and a dictionary of trivial names of compounds.
- From the very beginning, you need to assess your current level and the number of points you need, for this it is worth going through. If everything is very bad and you need excellent performance, congratulations, even now all is not lost. Train yourself to successful completion You can do it without the help of a tutor.
Decide on minimum quantity points you want to score, this will allow you to understand how many tasks you must solve accurately to get the score you need.
Naturally, take into account that everything may not go so smoothly and solve it as best you can. larger number tasks, or better yet, everything. The minimum that you have determined for yourself - you must decide ideally. - Let's move on to the practical part - training for the solution.
Most effective method- next. Select only the exam you are interested in and solve the corresponding test. About 20 solved tasks guarantee that you will meet all types of problems. As soon as you begin to feel that you know how to solve every task you see from beginning to end, proceed to the next task. If you don’t know how to solve a task, use the search on our website. There is almost always a solution on our website, otherwise just write to the tutor by clicking on the icon in the lower left corner - it’s free. - At the same time, we repeat the third point for everyone on our website, starting with.
- When the first part is given to you at least at an average level, you begin to decide. If one of the tasks is difficult, and you made a mistake in completing it, then return to the tests on this task or the corresponding topic with tests.
- Part 2. If you have a tutor, focus on studying this part with him. (provided that you are able to solve the rest at least 70%). If you started part 2, then you should score a passing grade without any problems 100% of the time. If this does not happen, it is better to stay on the first part for now. When you are ready for part 2, we recommend that you get a separate notebook where you will write down only the solutions to part 2. The key to success is solving as many tasks as possible, just like in part 1.
Part 1 contains 19 tasks with a short answer, including 15 tasks of a basic level of complexity (the serial numbers of these tasks: 1, 2, 3, 4, ...15) and 4 tasks of an increased level of complexity (the serial numbers of these tasks: 16, 17, 18, 19). Despite all their differences, the tasks in this part are similar in that the answer to each of them is written briefly in the form of one number or a sequence of numbers (two or three). The sequence of numbers is written on the answer form without spaces or other additional characters.
Part 2, depending on the CMM model, contains 3 or 4 tasks high level complexity, with a detailed answer. Difference exam models 1 and 2 consists of the content and approaches to completing the last tasks of the exam options:
Examination model 1 contains task 22, which involves performing a “thought experiment”;
Examination model 2 contains tasks 22 and 23, which involve completing laboratory work(real chemical experiment).
Scale for converting points to grades:
"2"– from 0 to 8
"3"– from 9 to 17
"4"– from 18 to 26
"5"– from 27 to 34
System for assessing the performance of individual tasks and the examination work as a whole
Correct completion of each of tasks 1–15 is scored 1 point. Correct completion of each of tasks 16–19 is assessed with a maximum of 2 points. Tasks 16 and 17 are considered completed correctly if two answer options are correctly selected in each of them. For an incomplete answer - one of two answers is correctly named or three answers are named, two of which are correct - 1 point is given. The remaining answer options are considered incorrect and are scored 0 points. Tasks 18 and 19 are considered completed correctly if three correspondences are correctly established. An answer in which two out of three matches are established is considered partially correct; it is worth 1 point. The remaining options are considered an incorrect answer and are scored 0 points.
The tasks of Part 2 (20–23) are checked by a subject commission. Maximum score for a correctly completed task: for tasks 20 and 21 - 3 points each; in model 1 for task 22 – 5 points; in model 2 for task 22 - 4 points, for task 23 - 5 points.
To complete the examination work in accordance with model 1, 120 minutes are allotted; according to model 2 – 140 minutes
In this section, I systematize the analysis of problems from the OGE in chemistry. Similar to the section, you will find detailed analyzes with instructions for the solution typical tasks in chemistry in the 9th grade OGE. Before analyzing each block of typical problems, I provide theoretical information, without which solving this task is impossible. There is only as much theory as is enough to know to successfully complete the task on the one hand. On the other hand, I tried to describe the theoretical material in an interesting and understandable language. I am sure that after completing the training using my materials, you will not only successfully pass the OGE in chemistry, but also fall in love with this subject.
General information about the exam
OGE in chemistry consists of three parts.
In the first part 15 tasks with one answer- this is the first level and the tasks in it are not difficult, provided, of course, basic knowledge in chemistry. These tasks do not require calculations, with the exception of task 15.
The second part consists of four questions- in the first two - 16 and 17, you need to choose two correct answers, and in 18 and 19, correlate the values or statements from the right column with the left.
The third part is problem solving. At 20 you need to equalize the reaction and determine the coefficients, and at 21 you need to solve the calculation problem.
Fourth part - practical, is not difficult, but you need to be careful and careful, as always when working with chemistry.
Total amount given for work 140 minutes.
Disassembled below standard options tasks accompanied by the theory necessary for solution. All tasks are thematic - opposite each task a topic is indicated for general understanding.