Preliminary exam in chemistry. Demonstration versions of the exam in chemistry (Grade 11)
However, it is often chosen by students who want to enter universities of the corresponding direction. This test is necessary for those who want to further study chemistry, chemical engineering and medicine, or will specialize in biotechnology. It is inconvenient that the date of the exam coincides with the exam in history and literature.
However, these subjects are rarely taken together - they are too different in focus for universities to require the results of the USE in such a set. This exam is quite difficult - the percentage of those who fail it ranges from 6 to 11%, and the average test score is about 57. All this does not contribute to the popularity of this subject - chemistry is only seventh in the ranking of popularity among graduates of past years.
The exam in chemistry is important for future doctors, chemists and biotechnologists
Demo version of the USE-2016
USE dates in chemistry
early period
- April 2, 2016 (Sat) - Main exam
- April 21, 2016 (Thu) - Reserve
main stage
- June 20, 2016 (Mon) - Main exam
- June 22, 2016 (Wed) - Reserve
Changes in the USE-2016
Unlike last year, some general innovations appeared in the examination in this discipline. In particular, the number of tests that will have to be solved at the basic level has been reduced (from 28 to 26), and the maximum number of primary points in chemistry is now 64. As for the specific features of the 2016 exam, some of the tasks have undergone changes in the answer format, which should give the student.
- In task No. 6, you need to demonstrate whether you know the classification of inorganic compounds and choose 3 answers out of 6 options offered in the test;
- Tests numbered 11 and 18 are designed to determine whether the student knows the genetic links between organic and inorganic compounds. The correct answer involves choosing 2 options from the 5 specified formulations;
- Tests No. 24, 25 and 26 require an answer in the form of a number that must be determined independently, while a year ago, schoolchildren had the opportunity to choose an answer from the proposed options;
- In numbers 34 and 35, students should not only choose answers, but establish a correspondence. These tasks are related to the topic "Chemical properties of hydrocarbons".
In 2016, the chemistry exam includes 40 tasks General information
The exam in chemistry will last 210 minutes (3.5 hours). The examination ticket includes 40 tasks, which are divided into three categories:
- A1–A26- relate to tasks that allow assessing the basic training of graduates. The correct answer to these tests makes it possible to score 1 primary point. You should spend 1-4 minutes to complete each task;
- B1–B9- these are tests with an increased level of complexity, they will require schoolchildren to briefly formulate the correct answer and in total they will make it possible to score 18 primary points. Each task is given 5-7 minutes;
- С1–С5- belong to the category of tasks of increased complexity. In this case, the student is required to formulate a detailed answer. In total, you can get another 20 primary points for them. Each task can be given up to 10 minutes.
The minimum score in this subject must be at least 14 primary points (36 test points).
How to prepare for the exam?
To pass the statewide chemistry exam, you can download and work through the demo version of the back exam in advance. The proposed materials give an idea of what you will have to face at the exam in 2016. Systematic work with tests will allow you to analyze gaps in knowledge. Practicing on the demo version allows students to quickly navigate through the real exam - you do not waste time trying to calm down, focus and understand the wording of the questions.
To solve problems of this type, it is necessary to know the general formulas for classes of organic substances and the general formulas for calculating the molar mass of substances of these classes:
Majority Decision Algorithm tasks to find the molecular formula includes the following steps:
- writing reaction equations in general form;
- finding the amount of substance n, for which the mass or volume is given, or the mass or volume of which can be calculated according to the condition of the problem;
- finding the molar mass of the substance M = m / n, the formula of which must be established;
- finding the number of carbon atoms in a molecule and compiling the molecular formula of a substance.
Examples of solving problem 35 of the Unified State Examination in chemistry to find the molecular formula of organic matter by combustion products with an explanation
The combustion of 11.6 g of organic matter produces 13.44 liters of carbon dioxide and 10.8 g of water. The vapor density of this substance in air is 2. It has been established that this substance interacts with an ammonia solution of silver oxide, is catalytically reduced by hydrogen to form a primary alcohol, and is capable of being oxidized by an acidified solution of potassium permanganate to a carboxylic acid. Based on these data:
1) establish the simplest formula of the starting substance,
2) make its structural formula,
3) give the reaction equation for its interaction with hydrogen.
Solution: the general formula of organic matter is CxHyOz.
Let's translate the volume of carbon dioxide and the mass of water into moles using the formulas:
n = m/M And n = V/ Vm,
Molar volume Vm = 22.4 l/mol
n (CO 2) \u003d 13.44 / 22.4 \u003d 0.6 mol, => the original substance contained n (C) \u003d 0.6 mol,
n (H 2 O) \u003d 10.8 / 18 \u003d 0.6 mol, => the original substance contained twice as much n (H) \u003d 1.2 mol,
This means that the desired compound contains oxygen in the amount:
n(O)= 3.2/16 = 0.2 mol
Let's look at the ratio of C, H and O atoms that make up the original organic matter:
n(C) : n(H) : n(O) = x: y: z = 0.6: 1.2: 0.2 = 3: 6: 1
We found the simplest formula: C 3 H 6 O
To find out the true formula, we find the molar mass of an organic compound using the formula:
M (CxHyOz) = Dair (CxHyOz) * M (air)
M ist (CxHyOz) \u003d 29 * 2 \u003d 58 g / mol
Let's check if the true molar mass corresponds to the molar mass of the simplest formula:
M (C 3 H 6 O) \u003d 12 * 3 + 6 + 16 \u003d 58 g / mol - corresponds, \u003d\u003e the true formula coincides with the simplest.
Molecular formula: C 3 H 6 O
From the data of the problem: "this substance interacts with an ammonia solution of silver oxide, is catalytically reduced by hydrogen to form a primary alcohol, and is capable of being oxidized by an acidified solution of potassium permanganate to a carboxylic acid" we conclude that it is an aldehyde.
2) When 18.5 g of saturated monobasic carboxylic acid reacted with an excess of sodium bicarbonate solution, 5.6 l (N.O.) of gas was released. Determine the molecular formula of the acid.
3) Some limiting carboxylic monobasic acid with a mass of 6 g requires the same mass of alcohol for complete esterification. This yields 10.2 g of ester. Set the molecular formula of the acid.
4) Determine the molecular formula of acetylenic hydrocarbon if the molar mass of the product of its reaction with an excess of hydrogen bromide is 4 times greater than the molar mass of the initial hydrocarbon
5) During the combustion of organic matter with a mass of 3.9 g, carbon monoxide (IV) with a mass of 13.2 g and water with a mass of 2.7 g were formed. Derive the formula of the substance, knowing that the hydrogen vapor density of this substance is 39.
6) During the combustion of organic matter weighing 15 g, carbon monoxide (IV) with a volume of 16.8 l and water with a mass of 18 g were formed. Derive the formula of the substance, knowing that the vapor density of this substance in terms of hydrogen fluoride is 3.
7) During the combustion of 0.45 g of gaseous organic matter, 0.448 l (n.o.) of carbon dioxide, 0.63 g of water and 0.112 l (n.o.) of nitrogen were released. The density of the initial gaseous substance in nitrogen is 1.607. Find the molecular formula of this substance.
8) The combustion of oxygen-free organic matter produced 4.48 l (n.o.) of carbon dioxide, 3.6 g of water and 3.65 g of hydrogen chloride. Determine the molecular formula of the burned compound.
9) During the combustion of organic matter weighing 9.2 g, carbon monoxide (IV) was formed with a volume of 6.72 l (n.o.) and water with a mass of 7.2 g. Set the molecular formula of the substance.
10) During the combustion of organic matter weighing 3 g, carbon monoxide (IV) was formed with a volume of 2.24 l (n.o.) and water with a mass of 1.8 g. It is known that this substance reacts with zinc.
Based on these conditions of the assignment:
1) make the calculations necessary to establish the molecular formula of an organic substance;
2) write down the molecular formula of the original organic matter;
3) make a structural formula of this substance, which unambiguously reflects the order of bonding of atoms in its molecule;
4) write the equation for the reaction of this substance with zinc.
Demonstration options for the exam in chemistry for grade 11 consist of two parts. The first part includes tasks to which you need to give a short answer. To the tasks from the second part it is necessary to give a detailed answer.
Everything demonstration versions of the exam in chemistry contain correct answers to all tasks and assessment criteria for tasks with a detailed answer.
There are no changes compared to.
Demonstration options for the exam in chemistry
Note that chemistry demos are presented in pdf format, and in order to view them you need to have installed, for example, the freely distributed Adobe Reader software package on your computer.
| Demonstration version of the exam in chemistry for 2002 |
| Demonstration version of the exam in chemistry for 2004 |
| Demonstration version of the exam in chemistry for 2005 |
| Demonstration version of the exam in chemistry for 2006 |
| Demonstration version of the exam in chemistry for 2008 |
| Demonstration version of the exam in chemistry for 2009 |
| Demonstration version of the exam in chemistry for 2010 |
| Demonstration version of the exam in chemistry for 2011 |
| Demonstration version of the exam in chemistry for 2012 |
| Demonstration version of the exam in chemistry for 2013 |
| Demonstration version of the exam in chemistry for 2014 |
| Demonstration version of the exam in chemistry for 2015 |
| Demonstration version of the exam in chemistry for 2016 |
| Demonstration version of the exam in chemistry for 2017 |
| Demonstration version of the exam in chemistry for 2018 |
| Demonstration version of the exam in chemistry for 2019 |
Changes in the demonstration versions of the exam in chemistry
Demonstration versions of the exam in chemistry for grade 11 for 2002 - 2014 consisted of three parts. The first part included tasks in which you need to choose one of the proposed answers. The tasks from the second part were required to give a short answer. To the tasks from the third part it was necessary to give a detailed answer.
In 2014 in demonstration version of the exam in chemistry the following changes:
- all calculation tasks, the performance of which was estimated at 1 point, were placed in part 1 of the work (A26-A28),
- topic "Redox Reactions" tested with assignments IN 2 And C1;
- topic "Hydrolysis of salts" checked only with the task AT 4;
- a new task has been included(in position AT 6) to check the topics "qualitative reactions to inorganic substances and ions", "qualitative reactions of organic compounds"
- total number of jobs in each variant was 42 (instead of 43 in the work of 2013).
In 2015, there were fundamental changes have been made:
- The grading system has been changed. tasks for finding the molecular formula of a substance. The maximum score for its implementation - 4 (instead of 3 points in 2014).
Option became be in two parts(part 1 - short answer questions, part 2 - open-ended questions).
Numbering assignments has become through throughout the variant without letter designations A, B, C.
Was the form of recording the answer in tasks with a choice of answers has been changed: the answer has become necessary to write down the number with the number of the correct answer (and not mark with a cross).
It was the number of tasks of the basic level of complexity has been reduced from 28 to 26 tasks.
Maximum score for completing all the tasks of the examination paper in 2015 became 64 (instead of 65 points in 2014).
IN 2016 year in demo in chemistrysignificant changes have been made compared to the previous year 2015 :
Part 1 changed the format of tasks 6, 11, 18, 24, 25 and 26 basic level of difficulty with a short answer.
Changed the format of tasks 34 and 35 increased level of complexity : these tasks now require you to match instead of selecting multiple correct answers from a suggested list.
The distribution of tasks by difficulty level and types of skills being tested has been changed.
In 2017 compared to demo version of 2016 in chemistrythere have been significant changes. The structure of the examination paper has been optimized:
Was changed the structure of the first part demo version: tasks with a choice of one answer were excluded from it; tasks were grouped into separate thematic blocks, each of which began to contain tasks of both basic and advanced levels of complexity.
It was reduced the total number of tasks up to 34.
Was grading scale changed(from 1 to 2 points) completing tasks of the basic level of complexity, which test the assimilation of knowledge about the genetic relationship of inorganic and organic substances (9 and 17).
Maximum score for the completion of all tasks of the examination paper was reduced to 60 points.
In 2018 in demo version of the exam in chemistry compared with demo version of 2017 in chemistry the following changes:
It was task 30 added high level of complexity with a detailed answer,
Maximum score for the completion of all tasks of the examination work remained without change by changing the scale for grading tasks in part 1.
IN demo version of the USE 2019 in chemistry compared with demo version of 2018 in chemistry there were no changes.
On our website you can also get acquainted with the training materials prepared by the teachers of our training center "Resolventa" for preparing for the exam in mathematics.
For students in grades 10 and 11 who want to prepare well and pass USE in mathematics or Russian language for a high score, the training center "Resolventa" conducts
We also have organized for schoolchildren
Specification
control measuring materials
for holding the unified state exam in 2016
in chemistry
1. Appointment of KIM USE
The Unified State Examination (hereinafter referred to as the Unified State Examination) is a form of objective assessment of the quality of training of persons who have mastered the educational programs of secondary general education, using tasks in a standardized form (control measuring materials).
The USE is conducted in accordance with Federal Law No. 273-FZ of December 29, 2012 “On Education in the Russian Federation”.
Control measuring materials allow to establish the level of development by graduates of the Federal component of the state standard of secondary (complete) general education in chemistry, basic and profile levels.
The results of the unified state examination in chemistry are recognized by educational institutions of secondary vocational education and educational institutions of higher professional education as the results of entrance examinations in chemistry.
2. Documents defining the content of KIM USE
3. Approaches to the selection of content, the development of the structure of the KIM USE
The basis of approaches to the development of KIM USE 2016 in chemistry was those general methodological guidelines that were determined during the formation of examination models of previous years. The essence of these settings is as follows.
- KIM are focused on testing the assimilation of the knowledge system, which is considered as an invariant core of the content of existing programs in chemistry for general education organizations. In the standard, this system of knowledge is presented in the form of requirements for the preparation of graduates. These requirements correspond to the level of presentation in the KIM of the content elements being checked.
- In order to ensure the possibility of a differentiated assessment of the educational achievements of graduates of the KIM USE, they check the development of basic educational programs in chemistry at three levels of complexity: basic, advanced and high. The educational material on the basis of which tasks are built is selected on the basis of its significance for the general educational preparation of secondary school graduates.
- The fulfillment of the tasks of the examination work involves the implementation of a certain set of actions. Among them, the most indicative are, for example, such as: to identify the classification features of substances and reactions; determine the degree of oxidation of chemical elements according to the formulas of their compounds; explain the essence of a particular process, the relationship of the composition, structure and properties of substances. The ability of the examinee to carry out various actions when performing work is considered as an indicator of the assimilation of the studied material with the necessary depth of understanding.
- The equivalence of all variants of the examination work is ensured by maintaining the same ratio of the number of tasks that test the assimilation of the main elements of the content of the key sections of the chemistry course.
4. The structure of KIM USE
Each version of the examination work is built according to a single plan: the work consists of two parts, including 40 tasks. Part 1 contains 35 tasks with a short answer, including 26 tasks of a basic level of complexity (the serial numbers of these tasks: 1, 2, 3, 4, ... 26) and 9 tasks of an increased level of complexity (the serial numbers of these tasks: 27, 28, 29, ...35).
Part 2 contains 5 tasks of a high level of complexity, with a detailed answer (the serial numbers of these tasks: 36, 37, 38, 39, 40).