Determine the atoms of which of the elements indicated in the series in the ground state contain one unpaired electron.
Write down the numbers of the selected elements in the answer field.
Answer:
Answer: 23
Explanation:
Let's write down electronic formula for each of the indicated chemical elements and depict the electron-graphic formula of the last electronic level:
1) S: 1s 2 2s 2 2p 6 3s 2 3p 4
2) Na: 1s 2 2s 2 2p 6 3s 1
3) Al: 1s 2 2s 2 2p 6 3s 2 3p 1
4) Si: 1s 2 2s 2 2p 6 3s 2 3p 2
5) Mg: 1s 2 2s 2 2p 6 3s 2
From the chemical elements indicated in the row, select three metal elements. Arrange the selected elements in ascending order of restorative properties.
Write in the answer field the numbers of the selected elements in the desired sequence.
Answer: 352
Explanation:
In the main subgroups of the periodic table, metals are located under the boron-astatine diagonal, as well as in side subgroups. Thus, the metals from this list include Na, Al, and Mg.
metallic and therefore restorative properties elements increase when moving to the left in the period and down in the subgroup.
In this way, metallic properties the metals listed above increase in the series Al, Mg, Na
From among the elements indicated in the row, select two elements that, in combination with oxygen, exhibit an oxidation state of +4.
Write down the numbers of the selected elements in the answer field.
Answer: 14
Explanation:
The main oxidation states of elements from the list presented in complex substances:
Sulfur - "-2", "+4" and "+6"
Sodium Na - "+1" (single)
Aluminum Al - "+3" (the only one)
Silicon Si - "-4", "+4"
Magnesium Mg - "+2" (single)
From the proposed list of substances, select two substances in which an ionic chemical bond is present.
Answer: 12
Explanation:
In the overwhelming majority of cases, it is possible to determine the presence of an ionic type of bond in a compound by the fact that its structural units simultaneously include atoms typical metal and nonmetal atoms.
Based on this criterion, ion type bonding takes place in the compounds KCl and KNO 3 .
In addition to the above feature, the presence of an ionic bond in a compound can be said if it contains structural unit contains the ammonium cation (NH 4 + ) or its organic analogs - alkylammonium cations RNH 3 + , dialkylammonium R 2NH2+ , trialkylammonium R 3NH+ and tetraalkylammonium R 4N+ , where R is some hydrocarbon radical. For example, an ionic type of bond occurs in the compound (CH 3 ) 4 NCl between cation (CH 3 ) 4 + and chloride ion Cl − .
Establish a correspondence between the formula of a 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.
Answer: 241
Explanation:
N 2 O 3 - non-metal oxide. All non-metal oxides except N 2 O, NO, SiO and CO are acidic.
Al 2 O 3 - metal oxide in the oxidation state +3. Metal oxides in the oxidation state +3, +4, as well as BeO, ZnO, SnO and PbO, are amphoteric.
HClO 4 is a typical representative of acids, because. during dissociation in an aqueous solution, only H + cations are formed from cations:
HClO 4 \u003d H + + ClO 4 -
From the proposed list of substances, select two substances, with each of which zinc interacts.
1) nitric acid (solution)
2) iron(II) hydroxide
3) magnesium sulfate (solution)
4) sodium hydroxide (solution)
5) aluminum chloride (solution)
Write down the numbers of the selected substances in the answer field.
Answer: 14
Explanation:
1) Nitric acid is a strong oxidizing agent and reacts with all metals except platinum and gold.
2) Iron hydroxide (ll) - insoluble base. Metals do not react with insoluble hydroxides at all, and only three metals react with soluble (alkalis) - Be, Zn, Al.
3) Magnesium sulfate - more salt active metal than zinc, and therefore the reaction does not proceed.
4) Sodium hydroxide - alkali (soluble metal hydroxide). Only Be, Zn, Al work with metal alkalis.
5) AlCl 3 - a salt of a more active metal than zinc, i.e. reaction is not possible.
From the proposed list of substances, select two oxides that react with water.
Write down the numbers of the selected substances in the answer field.
Answer: 14
Explanation:
Of the oxides, only oxides of alkali and alkaline earth metals react with water, as well as all acid oxides except for SiO 2 .
Thus, answer options 1 and 4 are suitable:
BaO + H 2 O \u003d Ba (OH) 2
SO 3 + H 2 O \u003d H 2 SO 4
1) hydrogen bromide
3) sodium nitrate
4) sulfur oxide (IV)
5) aluminum chloride
Write in the table the selected numbers under the corresponding letters.
Answer: 52
Explanation:
Salts among these substances are only sodium nitrate and aluminum chloride. All nitrates, like sodium salts, are soluble, and therefore sodium nitrate cannot precipitate in principle with any of the reagents. Therefore, salt X can only be aluminum chloride.
A common mistake among those who pass the exam in chemistry is a misunderstanding that in an aqueous solution ammonia forms a weak base - ammonium hydroxide due to the reaction:
NH 3 + H 2 O<=>NH4OH
In this regard, an aqueous solution of ammonia gives a precipitate when mixed with solutions of metal salts that form insoluble hydroxides:
3NH 3 + 3H 2 O + AlCl 3 \u003d Al (OH) 3 + 3NH 4 Cl
In a given transformation scheme
Cu X > CuCl 2 Y > CuI
substances X and Y are:
Answer: 35
Explanation:
Copper is a metal located in the activity series to the right of hydrogen, i.e. does not react with acids (except H 2 SO 4 (conc.) and HNO 3). Thus, the formation of copper (ll) chloride is possible in our case only by reaction with chlorine:
Cu + Cl 2 = CuCl 2
Iodide ions (I -) cannot coexist in the same solution with divalent copper ions, because are oxidized:
Cu 2+ + 3I - \u003d CuI + I 2
Establish a correspondence between the reaction equation and the oxidizing substance in this reaction: for each position indicated by a letter, select the corresponding position indicated by a number.
|
REACTION EQUATION A) H 2 + 2Li \u003d 2LiH B) N 2 H 4 + H 2 \u003d 2NH 3 C) N 2 O + H 2 \u003d N 2 + H 2 O D) N 2 H 4 + 2N 2 O \u003d 3N 2 + 2H 2 O |
OXIDIZER |
Write in the table the selected numbers under the corresponding letters.
Answer: 1433
Explanation:
An oxidizing agent in a reaction is a substance that contains an element that lowers its oxidation state.
Establish a correspondence between the formula of a substance and the reagents, with each of which this substance can interact: for each position indicated by a letter, select the corresponding position indicated by a number.
| SUBSTANCE FORMULA | REAGENTS |
| A) Cu (NO 3) 2 | 1) NaOH, Mg, Ba (OH) 2 2) HCl, LiOH, H 2 SO 4 (solution) 3) BaCl 2 , Pb(NO 3) 2 , S 4) CH 3 COOH, KOH, FeS 5) O 2, Br 2, HNO 3 |
Write in the table the selected numbers under the corresponding letters.
Answer: 1215
Explanation:
A) Cu(NO 3) 2 + NaOH and Cu(NO 3) 2 + Ba(OH) 2 - similar interactions. Salt with metal hydroxide reacts if the starting materials are soluble, and the products contain a precipitate, a gas, or a low-dissociating substance. Both for the first and for the second reaction, both requirements are met:
Cu(NO 3) 2 + 2NaOH = 2NaNO 3 + Cu(OH) 2 ↓
Cu(NO 3) 2 + Ba(OH) 2 = Na(NO 3) 2 + Cu(OH) 2 ↓
Cu (NO 3) 2 + Mg - the salt reacts with the metal if the free metal is more active than what is included in the salt. Magnesium in the activity series is located to the left of copper, which indicates its greater activity, therefore, the reaction proceeds:
Cu(NO 3) 2 + Mg = Mg(NO 3) 2 + Cu
B) Al (OH) 3 - metal hydroxide in the oxidation state +3. Metal hydroxides in the oxidation state +3, +4, and also, as exceptions, the hydroxides Be (OH) 2 and Zn (OH) 2, are amphoteric.
By definition, amphoteric hydroxides call those that react with alkalis and almost all soluble acids. For this reason, we can immediately conclude that answer 2 is appropriate:
Al(OH) 3 + 3HCl = AlCl 3 + 3H 2 O
Al (OH) 3 + LiOH (solution) \u003d Li or Al (OH) 3 + LiOH (solid) \u003d to \u003d\u003e LiAlO 2 + 2H 2 O
2Al(OH) 3 + 3H 2 SO 4 = Al 2 (SO 4) 3 + 6H 2 O
C) ZnCl 2 + NaOH and ZnCl 2 + Ba (OH) 2 - interaction of the "salt + metal hydroxide" type. The explanation is given in p.A.
ZnCl 2 + 2NaOH = Zn(OH) 2 + 2NaCl
ZnCl 2 + Ba(OH) 2 = Zn(OH) 2 + BaCl 2
It should be noted that with an excess of NaOH and Ba (OH) 2:
ZnCl 2 + 4NaOH \u003d Na 2 + 2NaCl
ZnCl 2 + 2Ba(OH) 2 = Ba + BaCl 2
D) Br 2, O 2 are strong oxidizing agents. Of the metals, they do not react only with silver, platinum, gold:
Cu + Br2 t° > CuBr2
2Cu + O2 t° > 2CuO
HNO 3 is an acid with strong oxidizing properties, because oxidizes not with hydrogen cations, but with an acid-forming element - nitrogen N +5. Reacts with all metals except platinum and gold:
4HNO 3 (conc.) + Cu \u003d Cu (NO 3) 2 + 2NO 2 + 2H 2 O
8HNO 3 (razb.) + 3Cu \u003d 3Cu (NO 3) 2 + 2NO + 4H 2 O
Match between general formula homologous series and the name of the substance belonging to this series: 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: 231
Explanation:
From the proposed list of substances, select two substances that are isomers of cyclopentane.
1) 2-methylbutane
2) 1,2-dimethylcyclopropane
3) pentene-2
4) hexene-2
5) cyclopentene
Write down the numbers of the selected substances in the answer field.
Answer: 23
Explanation:
Cyclopentane has the molecular formula C 5 H 10 . Let's write the structural and molecular formulas of the substances listed in the condition
| Substance name | Structural formula | Molecular formula |
| cyclopentane | C 5 H 10 | |
| 2-methylbutane | C 5 H 12 | |
| 1,2-dimethylcyclopropane | C 5 H 10 | |
| pentene-2 | C 5 H 10 | |
| hexene-2 | C 6 H 12 | |
| cyclopentene | C 5 H 8 |
From the proposed list of substances, select two substances, each of which reacts with a solution of potassium permanganate.
1) methylbenzene
2) cyclohexane
3) methyl propane
Write down the numbers of the selected substances in the answer field.
Answer: 15
Explanation:
Of the hydrocarbons with an aqueous solution of potassium permanganate, those that contain in their structural formula C=C or C≡C bonds, as well as benzene homologues (except for benzene itself).
Thus methylbenzene and styrene are suitable.
From the proposed list of substances, select two substances with which phenol interacts.
1) hydrochloric acid
2) sodium hydroxide
4) nitric acid
5) sodium sulfate
Write down the numbers of the selected substances in the answer field.
Answer: 24
Explanation:
Phenol has weak acid properties more pronounced than in alcohols. For this reason, phenols, unlike alcohols, react with alkalis:
C 6 H 5 OH + NaOH = C 6 H 5 ONa + H 2 O
Phenol contains in its molecule hydroxyl group directly attached to benzene ring. The hydroxy group is an orientant of the first kind, that is, it facilitates substitution reactions in the ortho and para positions:
From the proposed list of substances, select two substances that undergo hydrolysis.
1) glucose
2) sucrose
3) fructose
5) starch
Write down the numbers of the selected substances in the answer field.
Answer: 25
Explanation:
All listed substances are carbohydrates. Monosaccharides do not undergo hydrolysis from carbohydrates. Glucose, fructose, and ribose are monosaccharides, sucrose is a disaccharide, and starch is a polysaccharide. Consequently, sucrose and starch from the specified list are subjected to hydrolysis.
The following scheme of transformations of substances is given:
1,2-dibromoethane → X → bromoethane → Y → ethyl formate
Determine which of the following substances are substances X and Y.
2) ethanal
4) chloroethane
5) acetylene
Write in the table the numbers of the selected substances under the corresponding letters.
Answer: 31
Explanation:
Match the title starting material and a product that is mainly formed during the interaction of this substance with bromine: 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: 2134
Explanation:
Substitution at the secondary carbon atom proceeds in more than with the primary. Thus, the main product of propane bromination is 2-bromopropane and not 1-bromopropane:
Cyclohexane is a cycloalkane with a ring size of more than 4 carbon atoms. Cycloalkanes with a ring size of more than 4 carbon atoms, when interacting with halogens, enter into a substitution reaction with preservation of the cycle:
Cyclopropane and cyclobutane - cycloalkanes with a minimum ring size mainly enter into addition reactions, accompanied by ring break:
The substitution of hydrogen atoms at the tertiary carbon atom occurs to a greater extent than at the secondary and primary. Thus, the bromination of isobutane proceeds mainly as follows:
Establish a correspondence between the reaction scheme and the organic substance that is the product of this reaction: 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: 6134
Explanation:
Heating aldehydes with freshly precipitated copper hydroxide results in the oxidation of the aldehyde group to a carboxyl group:
Aldehydes and ketones are reduced by hydrogen in the presence of nickel, platinum or palladium to alcohols:
Primary and secondary alcohols are oxidized by hot CuO to aldehydes and ketones, respectively:
Under the action of concentrated sulfuric acid on ethanol during heating, two different products are possible. When heated to temperatures below 140°C, intermolecular dehydration predominantly occurs with the formation of diethyl ether, and when heated above 140°C, intramolecular dehydration occurs, resulting in the formation of ethylene:
From the proposed list of substances, select two substances, the reaction thermal decomposition which is redox.
1) aluminum nitrate
2) potassium bicarbonate
3) aluminum hydroxide
4) ammonium carbonate
5) ammonium nitrate
Write down the numbers of the selected substances in the answer field.
Answer: 15
Explanation:
Redox reactions are such reactions as a result of which the chemical one or more chemical elements change their oxidation state.
Decomposition reactions of absolutely all nitrates are redox reactions. Metal nitrates from Mg to Cu inclusive decompose to metal oxide, nitrogen dioxide and molecular oxygen:
All metal bicarbonates decompose already with slight heating (60 ° C) to metal carbonate, carbon dioxide and water. In this case, there is no change in oxidation states:
Insoluble oxides decompose when heated. The reaction in this case is not a redox reaction, because not a single chemical element changes its oxidation state as a result of it:
Ammonium carbonate decomposes when heated into carbon dioxide, water and ammonia. The reaction is not redox:
Ammonium nitrate decomposes into nitric oxide (I) and water. The reaction refers to OVR:
From the proposed list, select two external influences that lead to an increase in the rate of the reaction of nitrogen with hydrogen.
1) lowering the temperature
2) pressure increase in the system
5) use of an inhibitor
Write in the answer field the numbers of the selected external influences.
Answer: 24
Explanation:
1) lowering the temperature:
The rate of any reaction decreases with decreasing temperature.
2) pressure increase in the system:
An increase in pressure increases the rate of any reaction in which at least one gaseous substance takes part.
3) decrease in hydrogen concentration
Decreasing the concentration always slows down the rate of the reaction.
4) increase in nitrogen concentration
Increasing the concentration of reactants always increases the rate of the reaction
5) use of an inhibitor
Inhibitors are substances that slow down the rate of a reaction.
Establish a correspondence between the formula of a substance and the products of electrolysis aqueous solution of this substance on inert electrodes: 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: 5251
Explanation:
A) NaBr → Na + + Br -
Na + cations and water molecules compete for the cathode.
2H 2 O + 2e - → H 2 + 2OH -
2Cl - -2e → Cl 2
B) Mg (NO 3) 2 → Mg 2+ + 2NO 3 -
Mg 2+ cations and water molecules compete for the cathode.
Cations alkali metals, as well as magnesium and aluminum, are not able to recover in an aqueous solution due to their high activity. For this reason, instead of them, water molecules are restored in accordance with the equation:
2H 2 O + 2e - → H 2 + 2OH -
NO 3 anions and water molecules compete for the anode.
2H 2 O - 4e - → O 2 + 4H +
So the answer is 2 (hydrogen and oxygen).
C) AlCl 3 → Al 3+ + 3Cl -
Alkali metal cations, as well as magnesium and aluminum, are not able to recover in an aqueous solution due to their high activity. For this reason, instead of them, water molecules are restored in accordance with the equation:
2H 2 O + 2e - → H 2 + 2OH -
Cl anions and water molecules compete for the anode.
Anions consisting of one chemical element(except F -) win competition from water molecules for oxidation at the anode:
2Cl - -2e → Cl 2
Thus answer 5 (hydrogen and halogen) is appropriate.
D) CuSO 4 → Cu 2+ + SO 4 2-
Metal cations to the right of hydrogen in the activity series are easily reduced in an aqueous solution:
Cu 2+ + 2e → Cu 0
Acid residues containing acid-forming element in the highest degree oxidation, lose competition to water molecules for oxidation at the anode:
2H 2 O - 4e - → O 2 + 4H +
Thus, answer 1 (oxygen and metal) is appropriate.
Establish a correspondence between the name of the salt and the medium of the aqueous solution of this salt: 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: 3312
Explanation:
A) iron (III) sulfate - Fe 2 (SO 4) 3
formed by a weak "base" Fe(OH) 3 and strong acid H2SO4. Conclusion - acidic environment
B) chromium (III) chloride - CrCl 3
formed by a weak "base" Cr(OH) 3 and a strong acid HCl. Conclusion - acidic environment
C) sodium sulfate - Na 2 SO 4
Educated strong base NaOH and strong acid H 2 SO 4 . Conclusion - the medium is neutral
D) sodium sulfide - Na 2 S
Formed by the strong base NaOH and the weak acid H2S. Conclusion - the environment is alkaline.
Establish a correspondence between the method of influencing an equilibrium system
CO (g) + Cl 2 (g) COCl 2 (g) + Q
and shift direction chemical equilibrium as a result of this impact: 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: 3113
Explanation:
Equilibrium shift under external impact on the system occurs in such a way as to minimize the effect of this external impact (Le Chatelier's principle).
A) An increase in the concentration of CO leads to a shift in the equilibrium towards the direct reaction, since as a result of it the amount of CO decreases.
B) An increase in temperature will shift the equilibrium in the direction endothermic reaction. Since the forward reaction is exothermic (+Q), the equilibrium will shift towards the reverse reaction.
C) A decrease in pressure will shift the equilibrium in the direction of the reaction as a result of which an increase in the amount of gases occurs. As a result of the reverse reaction, more gases are formed than as a result of the forward reaction. Thus, the equilibrium will shift in the direction of the reverse reaction.
D) An increase in the concentration of chlorine leads to a shift in the equilibrium towards a direct reaction, since as a result of it the amount of chlorine decreases.
Establish a correspondence between two substances and a reagent with which these substances can be distinguished: for each position indicated by a letter, select the corresponding position indicated by a number.
|
SUBSTANCES A) FeSO 4 and FeCl 2 B) Na 3 PO 4 and Na 2 SO 4 C) KOH and Ca (OH) 2 D) KOH and KCl |
REAGENT |
Write in the table the selected numbers under the corresponding letters.
Answer: 3454
Explanation:
It is possible to distinguish two substances with the help of a third only if these two substances interact with it in different ways, and, most importantly, these differences are outwardly distinguishable.
A) Solutions of FeSO 4 and FeCl 2 can be distinguished using a solution of barium nitrate. In the case of FeSO 4, a white precipitate of barium sulfate is formed:
FeSO 4 + BaCl 2 = BaSO 4 ↓ + FeCl 2
In the case of FeCl 2, no visible signs there is no interaction because the reaction does not proceed.
B) Solutions Na 3 PO 4 and Na 2 SO 4 can be distinguished using a solution of MgCl 2. A solution of Na 2 SO 4 does not enter into the reaction, and in the case of Na 3 PO 4 a white precipitate of magnesium phosphate precipitates:
2Na 3 PO 4 + 3MgCl 2 = Mg 3 (PO 4) 2 ↓ + 6NaCl
AT) KOH solutions and Ca(OH) 2 can be distinguished with a solution of Na 2 CO 3 . KOH does not react with Na 2 CO 3, but Ca (OH) 2 gives a white precipitate of calcium carbonate with Na 2 CO 3:
Ca(OH) 2 + Na 2 CO 3 = CaCO 3 ↓ + 2NaOH
D) KOH and KCl solutions can be distinguished using a MgCl 2 solution. KCl does not react with MgCl 2, and mixing solutions of KOH and MgCl 2 leads to the formation of a white precipitate of magnesium hydroxide:
MgCl 2 + 2KOH \u003d Mg (OH) 2 ↓ + 2KCl
Establish a correspondence between the substance and its scope: 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: 2331
Explanation:
Ammonia is used in the production of nitrogenous fertilizers. In particular, ammonia is a raw material for the production nitric acid, from which, in turn, fertilizers are obtained - sodium, potassium and ammonium nitrate (NaNO 3, KNO 3, NH 4 NO 3).
Carbon tetrachloride and acetone are used as solvents.
Ethylene is used to make macromolecular compounds(polymers), namely polyethylene.
| The answer to tasks 27-29 is a number. Write this number in the answer field in the text of the work, while observing the specified degree of accuracy. Then transfer this number to the ANSWER FORM No. 1 to the right of the number of the corresponding task, starting from the first cell. Write each character in a separate box in accordance with the samples given in the form. Units physical quantities no need to write. In reaction thermochemical equation which MgO (tv.) + CO 2 (g) → MgCO 3 (tv.) + 102 kJ, entered 88 g of carbon dioxide. How much heat will be released in this case? (Write down the number to the nearest integer.) Answer: ___________________________ kJ. Answer: 204 Explanation: Calculate the amount of carbon dioxide substance: n (CO 2) \u003d n (CO 2) / M (CO 2) \u003d 88/44 \u003d 2 mol, According to the reaction equation, the interaction of 1 mol of CO 2 with magnesium oxide releases 102 kJ. In our case, the amount of carbon dioxide is 2 mol. Denoting the amount of heat released in this case as x kJ, we can write the following proportion: 1 mol CO 2 - 102 kJ 2 mol CO 2 - x kJ Therefore, the following equation is valid: 1 ∙ x = 2 ∙ 102 Thus, the amount of heat that will be released when 88 g of carbon dioxide participates in the reaction with magnesium oxide is 204 kJ. Determine the mass of zinc that reacts with hydrochloric acid to produce 2.24 liters (N.O.) of hydrogen. (Write down the number to tenths.) Answer: ___________________________ Answer: 6.5 Explanation: Let's write the reaction equation: Zn + 2HCl \u003d ZnCl 2 + H 2 Calculate the amount of hydrogen substance: n (H 2) \u003d V (H 2) / V m \u003d 2.24 / 22.4 \u003d 0.1 mol. Since in the reaction equation zinc and hydrogen are preceded by equal odds, this means that the amounts of substances of zinc that entered into the reaction and hydrogen formed as a result of it are also equal, i.e. n (Zn) \u003d n (H 2) \u003d 0.1 mol, therefore: m(Zn) = n(Zn) ∙ M(Zn) = 0.1 ∙ 65 = 6.5 g.
|
The exams in history and chemistry, which completed the USE-2017 on the main dates, were held in the normal mode without failures and serious violations. Approximately 135,000 people applied to take part in the exam in history. For conducting the exam more than 2.8 thousand examination points (EPS) were involved in this subject. During the main period, 83.5 thousand people applied for participation in the Unified State Examination in Chemistry, the exam was held in more than 2.6 thousand PES. The course of exams in the regions was controlled by public observers and employees of Rosobrnadzor.
"The main wave of the exam ended today, it passed in a normal operating mode, calmly and without serious failures. We still have reserve days, we continue to monitor the course of the exams and the level of organization that has been achieved, we should in no case reduce- said at a press conference at the Situational Information Center of Rosobrnadzor the head of the department, Sergei Kravtsov.
The established minimum score for the exam in history is 32 points, for the exam in chemistry - 36 points. Participants of the exams will know their results no later than July 4, 2017.
Sergey Kravtsov also noted that during the USE-2017, the technological component of the exams was improved. More than half of the PES works using new technologies for printing KIM and scanning the works of participants in the Unified State Exam in the PES, a third of the works of participants, thanks to the spread of scanning technology, are cross-checked in other regions. It was possible to significantly reduce the time for checking the work and processing the results. The head of Rosobrnadzor said that the results of the Unified State Examination in the most massive compulsory subject, the Russian language, will be issued five days ahead of schedule and graduates will be able to receive school leaving certificates before graduation balls.
Basic USE period-2017 continue reserve days for the delivery of all USE subjects, which will be held from June 20 to July 1. Graduates of previous years, as well as graduates of current year who received an unsatisfactory result in one of the compulsory subjects(Russian language and mathematics), or missed the exam for a good reason.
/ Monday, June 19, 2017 /
Russian schoolchildren have finished taking the exam in history and chemistry. These exams were the last in the main period of the Unified State Examination of this year and were held as usual, the press service of Rosobrnadzor reported.
According to the department, 135 thousand people signed up for the exam in history, 83.5 thousand in chemistry. . . . . .
As recalled, in turn, the deputy head Federal Commission control developers measuring materials Unified State Examination of Chemistry Dmitry Dobrotin, there are no more tasks with the choice of one answer in the control and measuring materials on the subject. As a result, students are required to be more independent in the formulation of answers, a greater variety of skills that they must demonstrate when completing tasks.
. . . . .
More than 218 thousand people will write the Unified State Exam (USE) in history and chemistry on June 19, 2017, during the main period of the exam. This was reported in the press service Federal Service for Supervision in Education and Science (Rosobrnadzor).
"According to the schedule of the unified state exam June 19, 2017, during the main period of the Unified State Examination, exams in history and chemistry are held. . . . . . More than 2,800 examination points (PET) will be used to conduct the Unified State Examination in this subject ", - said the press service.
They explained that examination paper on history consists of two parts, including 25 tasks. It takes 3 hours 55 minutes (235 minutes) to complete it. . . . . . 600 PPE ", added the press release.
They specified that the examination paper consists of two parts, including 34 tasks. It takes 3.5 hours (210 minutes) to complete it. USE participants in Chemistry can use a non-programmable calculator in the exam. . . . . . From June 20 to July 1, reserve days are provided for the delivery of all items. . . . . .
The unified state exam is final form certification of school students. The Unified State Examination in Chemistry is an elective exam, so it is taken by students who require this subject when entering universities.
The exam includes 34 tasks divided into two blocks. In the first part contains 29 tasks different levels of difficulty, where a short answer is required. For other tasks, the solution is written in text form. Be sure to take into account the design of the work, which affects the final grade.
During the main period chemistry test scheduled for June 19. The results are checked on various levels and takes a certain amount of time. The final date is set when the results of the exam in chemistry in 2017 will be known.
Stages of verification activities in the case of the chemistry exam are as follows:
- analysis of testing results by regions - 23.06 ;
- re-processing by federal responsible services - 30.06 ;
- sending results to the regions - 01.07 ;
- official approval - 03.07 ;
- date of publication of the results - 04.07.
Where to find the results of the exam in chemistry
After July 4, the results are published in free access for students and their representatives. USE results in chemistry 2017 provided by any of the established methods:
- Hotlines of Rosobrnadzor.
The department has opened a telephone line through which questions and suggestions regarding the exam, the procedure for conducting, the features and results of the exam are received. The number +7 495 984-89-19 receives calls from exam participants, their parents, teachers and organizers.
In many subjects of the Russian Federation, their own lines are opened, through which participants and other interested parties are informed, including on the results of testing. You can find out the required numbers on the USE website or in local educational authorities. - Official port of the Unified State Examination.
The official USE portal - ege.edu.ru provides an opportunity for participants to find out their results. For this, the appropriate section is selected or the address check.ege.edu.ru is typed in the browser.
After clicking on the link, a form opens where information about the participant is entered (name, registration code or passport number). Additionally, a region is selected and a verification digital combination is entered. - Exam points.
Test results are available at the points where the procedure took place. Graduates can also view the results in their schools. Educational institutions publish the results on information boards. All interested persons can get acquainted with them.
- Public Services Portal.
If they register on the website of public services, students can check the number of points scored in the subjects they have passed. To do this, select the section on education in the catalog of services. For getting necessary information an application is filled in, where the full name, registration code and region are indicated.
Information is provided for all subjects, and detailed information is given for each of them. - Local sites of educational committees.
Test results are published on regional portals educational departments. Their list is given on official portal USE in the "Contacts" section.
Where can I see the text of the work?
Possibility see your work students are not provided. The exception is when an appeal is filed.
In case of disagreement with the amount of points received, the participant has the right to appeal the decision. For this within two business days after the official publication of the results, you need to apply for an appeal. Applications are accepted by the institution that sent the participant for testing.
The complaint is being considered by a special commission. The participant and his representatives are also present at the appeal. Based on the results of an additional check, the amount of points may increase, decrease or remain unchanged.
During the appeal, the participant is given the opportunity to review the work. For this, copies of the completed forms are provided. If necessary, explanations of the tasks performed are given.
Test results are the sum of points that the student scores for correctly completed tasks.
For grading the following system is used:
- 2 - less than 36 points;
- 3 – 36 – 45;
- 4 – 45 – 50;
- 5 – 50 – 60.
If the score is 36, the exam is considered passed. For admission to higher educational institutions you need to score at least 37 points.
The results of the early exam in chemistry
Participants who have completed school curriculum and received satisfactory marks in all subjects. Participants could be tested under valid circumstances (medical indications, relocation, participation in competitions).
Early Unified State Examination in Chemistry 2017 took place March 29. Testing was carried out according to deadlines. Final date of publication of the results - April 11.
AT early period an extra day was provided for passing the exam- April 10th. In this case, the results were available from 25 April.
Applications for participation in the filing exam more than 4.8 thousand people. Testing took place in 140 points of the Unified State Examination.
When can I retake the exam in chemistry
Upon receipt of an unsatisfactory grade, participants are given the right to retake. Since chemistry is an elective subject, a retake is possible only next year.
Resubmit work also allowed to improve their results. If a student has been removed from an exam for using teaching materials or phone, then he will also have to take the test again next year.
News and statistics
The chemistry exam ended without failures and violations. An application for surrender was submitted by 83.5 thousand schoolchildren. For them, more than 2.6 thousand points for the USE were organized. Minimum score amounted to 36. Generalized statistics will be known after verification.
