The electron configuration corresponds to the excited state of an atom
1) 1s 2 2s 2 2p 6 3s 1
2) 1s 2 2s 2 2p 6 3s 2 3p 6
3) 1s 2 2s 2 2p 6 3s 1 3p 2
Answer: 3
Explanation:
The energy of the 3s-sublevel is lower than the energy of the 3p-sublevel, but the 3s-sublevel, which should contain 2 electrons, is not completely filled. Consequently, such an electronic configuration corresponds to an excited state of an atom (aluminum).
The fourth option is not an answer due to the fact that although the 3d level is not filled, its energy is higher than the 4s sublevel, i.e. in this case, it is filled in last.
In which row are the chemical elements in the order of decreasing their atomic radius?
1) Rb → K → Na
2) Mg → Ca → Sr
3) Si → Al → Mg
Answer: 1
Explanation:
The atomic radius of the elements decreases with a decrease in the number of electron shells (the number of electron shells corresponds to the number of the period of the periodic table chemical elements) and in the transition to non-metals (i.e., with an increase in the number of electrons at the outer level). Consequently, in the table of chemical elements, the atomic radius of the elements decreases from bottom to top and from left to right.
A chemical bond is formed between atoms with the same relative electronegativity
2) covalent polar
3) covalent non-polar
Answer: 3
Explanation:
A covalent non-polar bond is formed between atoms with the same relative electronegativity, since there is no shift in the electron density.
The oxidation states of sulfur and nitrogen in (NH 4) 2 SO 3 are respectively equal
1) +4 and -3 2) -2 and +5 3) +6 and +3 4) -2 and +4
Answer: 1
Explanation:
(NH 4) 2 SO 3 (ammonium sulfite) is a salt formed by sulfurous acid and ammonia, therefore, the oxidation states of sulfur and nitrogen are +4 and -3, respectively (the oxidation state of sulfur in sulfurous acid is +4, the oxidation state of nitrogen in ammonia is 3).
Atomic crystal lattice It has
1) white phosphorus
3) silicon
Answer: 3
Explanation:
White phosphorus has a molecular crystal lattice, the formula of the white phosphorus molecule is P 4.
Both allotropic sulfur modifications (rhombic and monoclinic) have molecular crystal lattices with cyclic crown-shaped S 8 molecules at their sites.
Lead is a metal and has a metallic crystal lattice.
Silicon has a diamond-type crystal lattice, however, due to the longer length of the Si-Si bond in compared C-C inferior to diamond in hardness.
From the listed substances, select three substances that belong to amphoteric hydroxides.
Answer: 245
Explanation:
Amphoteric metals include Be, Zn, Al (you can remember "BeZnAl"), as well as Fe III and Cr III. Therefore, of the proposed answer options, amphoteric hydroxides include Be (OH) 2, Zn (OH) 2, Fe (OH) 3.
The compound Al (OH) 2 Br is a basic salt.
Are the following judgments about the properties of nitrogen correct?
A. Under normal conditions, nitrogen reacts with silver.
B. Nitrogen under normal conditions in the absence of a catalyst does not react with hydrogen.
1) only A is true
2) only B is true
3) both statements are true
Answer: 2
Explanation:
Nitrogen is a very inert gas and does not react with metals other than lithium under normal conditions.
The interaction of nitrogen with hydrogen refers to industrial production ammonia. The process is exothermic and reversible and only takes place in the presence of catalysts.
Carbon monoxide (IV) reacts with each of two substances:
1) oxygen and water
2) water and calcium oxide
3) potassium sulfate and sodium hydroxide
4) silicon oxide (IV) and hydrogen
Answer: 2
Explanation:
Carbon monoxide (IV) (carbon dioxide) is an acidic oxide, therefore, interacts with water to form unstable carbonic acid, alkalis and oxides of alkali and alkaline earth metals to form salts:
CO 2 + H 2 O ↔ H 2 CO 3
CO 2 + CaO → CaCO 3
Each of two reacts with sodium hydroxide solution
3) H 2 O and P 2 O 5
Answer: 4
Explanation:
NaOH is an alkali (has basic properties), therefore, it can interact with acidic oxide - SO 2 and amphoteric metal hydroxide - Al (OH) 3:
2NaOH + SO 2 → Na 2 SO 3 + H 2 O or NaOH + SO 2 → NaHSO 3
NaOH + Al (OH) 3 → Na
Calcium carbonate interacts with the solution
1) sodium hydroxide
2) hydrogen chloride
3) barium chloride
Answer: 2
Explanation:
Calcium carbonate is a water-insoluble salt and therefore does not interact with salts and bases. Calcium carbonate dissolves in strong acids with the formation of salts and the release of carbon dioxide:
CaCO 3 + 2HCl → CaCl 2 + CO 2 + H 2 O
In the transformation scheme
1) iron (II) oxide
2) iron (III) hydroxide
3) iron (II) hydroxide
4) iron (II) chloride
Answer: X-5; Y-2
Explanation:
Chlorine is a strong oxidizing agent (the oxidizing ability of halogens increases from I 2 to F 2), it oxidizes iron to Fe +3:
2Fe + 3Cl 2 → 2FeCl 3
Iron (III) chloride is a soluble salt and enters into exchange reactions with alkalis with the formation of a precipitate - iron (III) hydroxide:
FeCl 3 + 3NaOH → Fe (OH) 3 ↓ + NaCl
Homologues are
1) glycerin and ethylene glycol
2) methanol and butanol-1
3) propyne and ethylene
Answer: 2
Explanation:
Homologues - substances belonging to the same class organic compounds and differing by one or more CH 2 -groups.
Glycerin and ethylene glycol are trihydric and dihydric alcohols, respectively, differing in the number of oxygen atoms, therefore, they are neither isomers nor homologues.
Methanol and butanol-1 are primary alcohols with an unbranched skeleton, differ by two CH 2 -groups, therefore, they are homologous.
Propyne and ethylene belong to the classes of alkynes and alkenes, respectively, contain different numbers of carbon and hydrogen atoms, therefore, they are neither homologues nor isomers.
Propanone and propanal belong to different classes organic compounds, but contain 3 carbon atoms, 6 hydrogen atoms and 1 oxygen atom, therefore, are isomers in terms of the functional group.
For butene-2 impossible reaction
1) dehydration
2) polymerization
3) halogenation
Answer: 1
Explanation:
Butene-2 \u200b\u200bbelongs to the class of alkenes, enters into addition reactions with halogens, hydrogen halides, water and hydrogen. In addition, unsaturated hydrocarbons polymerize.
The dehydration reaction is a reaction that occurs with the elimination of a water molecule. Since butene-2 \u200b\u200bis a hydrocarbon, i.e. does not contain heteroatoms, water elimination is impossible.
Phenol does not interact with
1) nitric acid
2) sodium hydroxide
3) bromine water
Answer: 4
Explanation:
With phenol, nitric acid and bromic water enter into the reaction of electrophilic substitution on the benzene ring, resulting in the formation of nitrophenol and bromophenol, respectively.
Phenol, which has weak acidic properties, reacts with alkalis to form phenolates. In this case, sodium phenolate is formed.
Alkanes do not react with phenol.
Acetic acid methyl ester reacts with
1) NaCl 2) Br 2 (solution) 3) Cu (OH) 2 4) NaOH (solution)
Answer: 4
Explanation:
Acetic acid methyl ester (methyl acetate) belongs to the class of esters and undergoes acid and alkaline hydrolysis. Under conditions of acid hydrolysis, methyl acetate is converted into acetic acid and methanol, under conditions of alkaline hydrolysis with sodium hydroxide - sodium acetate and methanol.
Butene-2 \u200b\u200bcan be obtained by dehydration
1) butanone 2) butanol-1 3) butanol-2 4) butanal
Answer: 3
Explanation:
One of the ways to obtain alkenes is the reaction of intramolecular dehydration of primary and secondary alcohols, which takes place in the presence of anhydrous sulfuric acid and at temperatures above 140 o C. The elimination of a water molecule from an alcohol molecule proceeds according to Zaitsev's rule: a hydrogen atom and a hydroxyl group are split off from adjacent carbon atoms, moreover, hydrogen is split off from the carbon atom at which there is the smallest number of hydrogen atoms. Thus, intramolecular dehydration primary alcohol - butanol-1 leads to the formation of butene-1, intramolecular dehydration of the secondary alcohol, butanol-2, to the formation of butene-2.
Methylamine can react with (c)
1) alkalis and alcohols
2) alkalis and acids
3) oxygen and alkalis
4) acids and oxygen
Answer: 4
Explanation:
Methylamine belongs to the class of amines and has basic properties due to the presence of a lone electron pair on the nitrogen atom. In addition, the main properties of methylamine are more pronounced than that of ammonia, due to the presence of a methyl group, which has a positive inductive effect. Thus, possessing basic properties, methylamine interacts with acids to form salts. In an oxygen atmosphere, methylamine is burned to carbon dioxide, nitrogen and water.
In a given transformation scheme
substances X and Y, respectively, are
1) ethanediol-1,2
3) acetylene
4) diethyl ether
Answer: X-2; Y-5
Explanation:
Bromoethane in an aqueous alkali solution enters into a nucleophilic substitution reaction with the formation of ethanol:
CH 3 -CH 2 -Br + NaOH (aq.) → CH 3 -CH 2 -OH + NaBr
Under conditions of concentrated sulfuric acid at temperatures above 140 ° C, intramolecular dehydration occurs with the formation of ethylene and water:
All alkenes easily react with bromine:
CH 2 \u003d CH 2 + Br 2 → CH 2 Br-CH 2 Br
Substitution reactions include the interaction
1) acetylene and hydrogen bromide
2) propane and chlorine
3) ethene and chlorine
4) ethylene and hydrogen chloride
Answer: 2
Explanation:
Addition reactions include the interaction of unsaturated hydrocarbons (alkenes, alkynes, alkadienes) with halogens, hydrogen halides, hydrogen and water. Acetylene (ethyne) and ethylene belong to the classes of alkynes and alkenes, respectively; therefore, they enter into addition reactions with hydrogen bromide, hydrogen chloride and chlorine.
Alkanes enter into substitution reactions with halogens in the light or at elevated temperatures. The reaction proceeds by a chain mechanism with the participation of free radicals - particles with one unpaired electron:
At speed chemical reaction
HCOOCH 3 (l) + H 2 O (l) → HCOOH (l) + CH 3 OH (l)
does not render influence
1) pressure increase
2) temperature rise
3) change in the concentration of HCOOCH 3
4) using a catalyst
Answer: 1
Explanation:
The reaction rate is influenced by changes in temperature and concentrations of the starting reagents, as well as the use of a catalyst. According to Van't Hoff's rule of thumb, for every 10 degrees, the rate constant of a homogeneous reaction increases by 2-4 times.
The use of a catalyst also speeds up the reactions without having a catalyst in the products.
The starting materials and reaction products are in the liquid phase; therefore, a change in pressure does not affect the rate of this reaction.
Abbreviated ionic equation
Fe +3 + 3OH - \u003d Fe (OH) 3 ↓
corresponds to the molecular reaction equation
1) FeCl 3 + 3NaOH \u003d Fe (OH) 3 ↓ + 3NaCl
2) 4Fe (OH) 2 + O 2 + 2H 2 O \u003d 4Fe (OH) 3 ↓
3) FeCl 3 + 3NaHCO 3 \u003d Fe (OH) 3 ↓ + 3CO 2 + 3NaCl
Answer: 1
Explanation:
In an aqueous solution, soluble salts, alkalis and strong acids dissociate into ions, insoluble bases, insoluble salts, weak acids, gases, and simple substances are recorded in molecular form.
The condition of solubility of salts and bases corresponds to the first equation, in which the salt enters into an exchange reaction with alkali to form an insoluble base and another soluble salt.
The complete ionic equation is written as follows:
Fe +3 + 3Cl - + 3Na + + 3OH - \u003d Fe (OH) 3 ↓ + 3Cl - + 3Na +
Which of the following gases is toxic and has a strong odor?
1) hydrogen
2) carbon monoxide (II)
Answer: 3
Explanation:
Hydrogen and carbon dioxide are odorless, non-toxic gases. Carbon monoxide and chlorine are toxic, but unlike CO, chlorine has a pungent odor.
The polymerization reaction enters
1) phenol 2) benzene 3) toluene 4) styrene
Answer: 4
Explanation:
All substances from the proposed options are aromatic hydrocarbons, but polymerization reactions are not typical for aromatic systems. The styrene molecule contains a vinyl radical, which is a fragment of the ethylene molecule, which is characterized by polymerization reactions. Thus, styrene polymerizes to form polystyrene.
To 240 g of a solution with a mass fraction of 10% salt was added 160 ml of water. Determine the mass fraction of salt in the resulting solution. (Write the number down to whole integers.)
Answer: 6%Explanation:
The mass fraction of salt in the solution is calculated by the formula:
Based on this formula, we calculate the mass of salt in the original solution:
m (in-va) \u003d ω (in-va in the outgoing solution). m (original solution) / 100% \u003d 10%. 240 g / 100% \u003d 24 g
When water is added to the solution, the mass of the resulting solution will be 160 g + 240 g \u003d 400 g (water density 1 g / ml).
The mass fraction of salt in the resulting solution will be:
Calculate how much nitrogen (n.o.) is formed when 67.2 l (n.o.) of ammonia is completely burned. (Write the number down to tenths.)
Answer: 33.6 L
Explanation:
The complete combustion of ammonia in oxygen is described by the equation:
4NH 3 + 3O 2 → 2N 2 + 6H 2 O
A consequence of Avogadro's law is that the volumes of gases under the same conditions relate to each other in the same way as the number of moles of these gases. Thus, according to the reaction equation
ν (N 2) \u003d 1 / 2ν (NH 3),
therefore, the volumes of ammonia and nitrogen are related to each other in the same way:
V (N 2) \u003d 1 / 2V (NH 3)
V (N 2) \u003d 1 / 2V (NH 3) \u003d 67.2 l / 2 \u003d 33.6 l
What is the volume (in liters at standard level) of oxygen formed during the decomposition of 4 mol of hydrogen peroxide? (Write the number down to tenths).
Answer: 44.8 L
Explanation:
In the presence of a catalyst - manganese dioxide, peroxide decomposes to form oxygen and water:
2H 2 O 2 → 2H 2 O + O 2
According to the reaction equation, the amount of formed oxygen is half the amount of hydrogen peroxide:
ν (O 2) \u003d 1/2 ν (H 2 O 2), therefore ν (O 2) \u003d 4 mol / 2 \u003d 2 mol.
The volume of gases is calculated by the formula:
V \u003d V m ν , where V m is the molar volume of gases at normal conditions, equal to 22.4 l / mol
The volume of oxygen generated during the decomposition of peroxide is:
V (O 2) \u003d V m · ν (O 2) \u003d 22.4 L / mol · 2 mol \u003d 44.8 L
Establish a correspondence between the classes of compounds and the trivial name of the substance, which is its representative.
Answer: A-3; B-2; IN 1; G-5
Explanation:
Alcohols are organic substances containing one or more hydroxyl groups (-OH) directly bonded to a saturated carbon atom. Ethylene glycol is a dihydric alcohol containing two hydroxyl groups: CH 2 (OH) -CH 2 OH.
Carbohydrates are organic substances containing carbonyl and several hydroxyl groups, the general formula of carbohydrates is written as C n (H 2 O) m (where m, n\u003e 3). Of the proposed options, carbohydrates include starch - a polysaccharide, a high molecular weight carbohydrate consisting of a large number of monosaccharide residues, the formula of which is written as (C 6 H 10 O 5) n.
Hydrocarbons are organic substances that contain only two elements - carbon and hydrogen. The hydrocarbons from the proposed options include toluene - an aromatic compound consisting only of carbon and hydrogen atoms and does not contain functional groups with heteroatoms.
Carboxylic acids are organic substances, the molecules of which contain a carboxyl group, consisting of interconnected carbonyl and hydroxyl groups. The class of carboxylic acids includes butyric (butanoic) acid - C 3 H 7 COOH.
Establish a correspondence between the reaction equation and the change in the oxidation state of the oxidizing agent in it.
| EQUATION OF REACTION
A) 4NH 3 + 5O 2 \u003d 4NO + 6H 2 O B) 2Cu (NO 3) 2 \u003d 2CuO + 4NO 2 + O 2 B) 4Zn + 10HNO 3 \u003d NH 4 NO 3 + 4Zn (NO 3) 2 + 3H 2 O D) 3NO 2 + H 2 O \u003d 2HNO 3 + NO |
CHANGING THE DEGREE OF OXIDATION OF THE OXIDANT |
Answer: A-1; B-4; AT 6; G-3
Explanation:
Oxidizing agent is a substance that contains atoms that can attach electrons during a chemical reaction and thus reduce the oxidation state.
Reducing agent is a substance that contains atoms that can donate electrons during a chemical reaction and thus increase the oxidation state.
A) Oxidation of ammonia with oxygen in the presence of a catalyst leads to the formation of nitrogen monoxide and water. The oxidizing agent is molecular oxygen, initially having an oxidation state of 0, which, by attaching electrons, is reduced to an oxidation state of -2 in NO and H 2 O compounds.
B) Copper nitrate Cu (NO 3) 2 is a salt containing an acid residue with nitric acid. The oxidation states of nitrogen and oxygen in the nitrate anion are +5 and -2, respectively. During the reaction, the nitrate anion is converted to nitrogen dioxide NO 2 (with the oxidation state of nitrogen +4) and oxygen O 2 (with the oxidation state 0). Therefore, nitrogen is an oxidizing agent, since it lowers the oxidation state from +5 in nitrate ion to +4 in nitrogen dioxide.
C) In this redox reaction, the oxidizing agent is nitric acid, which, converting into ammonium nitrate, reduces the oxidation state of nitrogen from +5 (in nitric acid) to -3 (in the ammonium cation). The oxidation state of nitrogen in acid residues of ammonium nitrate and zinc nitrate remains unchanged, i.e. same as nitrogen in HNO 3.
D) In \u200b\u200bthis reaction, the nitrogen in the dioxide disproportionates, i.e. both increases (from N +4 in NO 2 to N +5 in HNO 3) and decreases (from N +4 in NO 2 to N +2 in NO) its oxidation state.
Establish a correspondence between the formula of a substance and the products of electrolysis of its aqueous solution, which were released on inert electrodes.
Answer: A-4; B-3; IN 2; G-5
Explanation:
Electrolysis is a redox process that occurs on the electrodes during the passage of a constant electric current through a solution or molten electrolyte. At the cathode, the reduction of those cations that have the highest oxidative activity occurs predominantly. At the anode, first of all, those anions that have the highest reducing ability are oxidized.
Electrolysis of aqueous solution
1) Electrolysis process aqueous solutions at the cathode does not depend on the material of the cathode, but depends on the position of the metal cation in the electrochemical series of voltages.
For cations in a row
Li + - Al 3+ reduction process:
2H 2 O + 2e → H 2 + 2OH - (H 2 is evolved at the cathode)
Zn 2+ - Pb 2+ recovery process:
Me n + + ne → Me 0 and 2H 2 O + 2e → H 2 + 2OH - (H 2 and Me are released at the cathode)
Cu 2+ - Au 3+ reduction process Me n + + ne → Me 0 (Me is released at the cathode)
2) The process of electrolysis of aqueous solutions at the anode depends on the material of the anode and on the nature of the anion. If the anode is insoluble, i.e. is inert (platinum, gold, coal, graphite), the process will depend only on the nature of the anions.
For anions F -, SO 4 2-, NO 3 -, PO 4 3-, OH - the oxidation process:
4OH - - 4e → O 2 + 2H 2 O or 2H 2 O - 4e → O 2 + 4H + (oxygen is evolved at the anode)
halide ions (except F -) oxidation process 2Hal - - 2e → Hal 2 (free halogens are released)
organic acids oxidation process:
2RCOO - - 2e → R-R + 2CO 2
Total electrolysis equation:
A) Na 2 CO 3 solution:
2H 2 O → 2H 2 (at the cathode) + O 2 (at the anode)
B) Cu (NO 3) 2 solution:
2Cu (NO 3) 2 + 2H 2 O → 2Cu (at the cathode) + 4HNO 3 + O 2 (at the anode)
B) AuCl 3 solution:
2AuCl 3 → 2Au (at the cathode) + 3Cl 2 (at the anode)
D) BaCl 2 solution:
BaCl 2 + 2H 2 O → H 2 (at the cathode) + Ba (OH) 2 + Cl 2 (at the anode)
Establish a correspondence between the name of the salt and the ratio of this salt to hydrolysis.
Answer: A-2; B-3; IN 2; G-1
Explanation:
Hydrolysis of salts - the interaction of salts with water, leading to the addition of the hydrogen cation H + of the water molecule to the anion of the acid residue and (or) the hydroxyl group OH - of the water molecule to the metal cation. Salts formed by cations corresponding to weak bases and anions corresponding to weak acids undergo hydrolysis.
A) Sodium stearate is a salt formed by stearic acid (a weak monobasic carboxylic acid of the aliphatic series) and sodium hydroxide (an alkali - a strong base), therefore, undergoes hydrolysis at the anion.
C 17 H 35 COONa → Na + + C 17 H 35 COO -
C 17 H 35 COO - + H 2 O ↔ C 17 H 35 COOH + OH - (formation of weakly dissociating carboxylic acid)
Solution medium alkaline (pH\u003e 7):
C 17 H 35 COONa + H 2 O ↔ C 17 H 35 COOH + NaOH
B) Ammonium phosphate is a salt formed by weak orthophosphoric acid and ammonia (weak base), therefore, it undergoes hydrolysis both by cation and anion.
(NH 4) 3 PO 4 → 3NH 4 + + PO 4 3-
PO 4 3- + H 2 O ↔ HPO 4 2- + OH - (formation of weakly dissociating hydrogen phosphate ion)
NH 4 + + H 2 O ↔ NH 3 · H 2 O + H + (formation of ammonia dissolved in water)
The solution medium is close to neutral (pH ~ 7).
C) Sodium sulfide is a salt formed by weak hydrosulfuric acid and sodium hydroxide (alkali is a strong base), therefore, it undergoes hydrolysis at the anion.
Na 2 S → 2Na + + S 2-
S 2- + H 2 O ↔ HS - + OH - (formation of weakly dissociating hydrosulfide ion)
Solution medium alkaline (pH\u003e 7):
Na 2 S + H 2 O ↔ NaHS + NaOH
D) Beryllium sulfate is a salt formed by strong sulfuric acid and beryllium hydroxide (weak base), therefore, undergoes hydrolysis at the cation.
BeSO 4 → Be 2+ + SO 4 2-
Be 2+ + H 2 O ↔ Be (OH) + + H + (formation of a weakly dissociating cation Be (OH) +)
The solution medium is acidic (pH< 7):
2BeSO 4 + 2H 2 O ↔ (BeOH) 2 SO 4 + H 2 SO 4
Establish a correspondence between the way of influencing the equilibrium system
MgO (solid) + CO 2 (g) ↔ MgCO 3 (solid) + Q
and a shift in chemical equilibrium as a result of this effect
Answer: A-1; B-2; IN 2; G-3Explanation:
This reaction is in chemical equilibrium, i.e. in a state where the speed of the forward reaction is equal to the speed of the reverse. A shift in equilibrium in the desired direction is achieved by changing the reaction conditions.
Le Chatelier's principle: if an equilibrium system is influenced from the outside, changing any of the factors that determine the equilibrium position, then the direction of the process that weakens this effect will increase in the system.
Equilibrium factors:
— pressure: an increase in pressure shifts the equilibrium towards a reaction leading to a decrease in volume (conversely, a decrease in pressure shifts the equilibrium towards a reaction leading to an increase in volume)
— temperature: an increase in temperature shifts the equilibrium towards an endothermic reaction (conversely, a decrease in temperature shifts the equilibrium towards an exothermic reaction)
— concentration of starting substances and reaction products: an increase in the concentration of starting substances and the removal of products from the reaction sphere shift the equilibrium towards the direct reaction (on the contrary, a decrease in the concentration of starting substances and an increase in reaction products shift the equilibrium towards the opposite reaction)
— catalysts do not affect the displacement of equilibrium, but only accelerate its achievement.
Thus,
A) since the reaction for obtaining magnesium carbonate is exothermic, a decrease in temperature will shift the equilibrium towards the direct reaction;
B) carbon dioxide is the initial substance in the production of magnesium carbonate, therefore, a decrease in its concentration will lead to a shift in equilibrium towards the initial substances, because in the direction of the reverse reaction;
B) Magnesium oxide and magnesium carbonate are solids, only CO 2 is a gas, so its concentration will affect the pressure in the system. With a decrease in the concentration of carbon dioxide, the pressure decreases, therefore, the equilibrium of the reaction shifts towards the starting substances (reverse reaction).
D) the introduction of the catalyst does not affect the displacement of the equilibrium.
Establish a correspondence between the formula of a substance and reagents, with each of which this substance can interact.
| FORMULA OF SUBSTANCE | REAGENTS
1) H 2 O, NaOH, HCl 2) Fe, HCl, NaOH 3) HCl, HCHO, H 2 SO 4 4) O 2, NaOH, HNO 3 5) H 2 O, CO 2, HCl |
Answer: A-4; B-4; IN 2; G-3
Explanation:
A) Sulfur is a simple substance that can burn in oxygen to form sulfur dioxide:
S + O 2 → SO 2
Sulfur (like halogens) in alkaline solutions disproportionates, resulting in the formation of sulfides and sulfites:
3S + 6NaOH → 2Na 2 S + Na 2 SO 3 + 3H 2 O
Concentrated nitric acid oxidizes sulfur to S +6, reducing to nitrogen dioxide:
S + 6HNO 3 (conc.) → H 2 SO 4 + 6NO 2 + 2H 2 O
B) Forfor (III) oxide is an acidic oxide, therefore, interacts with alkalis to form phosphites:
P 2 O 3 + 4NaOH → 2Na 2 HPO 3 + H 2 O
In addition, phosphorus (III) oxide is oxidized by atmospheric oxygen and nitric acid:
P 2 O 3 + O 2 → P 2 O 5
3P 2 O 3 + 4HNO 3 + 7H 2 O → 6H 3 PO 4 + 4NO
C) Iron (III) oxide - amphoteric oxide, because exhibits both acidic and basic properties (reacts with acids and alkalis):
Fe 2 O 3 + 6HCl → 2FeCl 3 + 3H 2 O
Fe 2 O 3 + 2NaOH → 2NaFeO 2 + H 2 O (fusion)
Fe 2 O 3 + 2NaOH + 3H 2 O → 2Na 2 (dissolution)
Fe 2 O 3 enters into a proportional reaction with iron with the formation of iron oxide (II):
Fe 2 O 3 + Fe → 3FeO
D) Cu (OH) 2 - a base insoluble in water, dissolves with strong acids, turning into the corresponding salts:
Cu (OH) 2 + 2HCl → CuCl 2 + 2H 2 O
Cu (OH) 2 + H 2 SO 4 → CuSO 4 + 2H 2 O
Cu (OH) 2 oxidizes aldehydes to carboxylic acids (similar to the "silver mirror" reaction):
HCHO + 4Cu (OH) 2 → CO 2 + 2Cu 2 O ↓ + 5H 2 O
Establish a correspondence between substances and a reagent with which they can be distinguished from each other.
Answer: A-3; B-1; IN 3; G-5
Explanation:
A) The two soluble salts CaCl 2 and KCl can be distinguished using a potassium carbonate solution. Calcium chloride enters into an exchange reaction with it, as a result of which calcium carbonate precipitates:
CaCl 2 + K 2 CO 3 → CaCO 3 ↓ + 2KCl
B) Solutions of sulfite and sodium sulfate can be distinguished by the indicator - phenolphthalein.
Sodium sulfite is a salt formed by a weak unstable sulfurous acid and sodium hydroxide (alkali is a strong base), therefore, it undergoes hydrolysis at the anion.
Na 2 SO 3 → 2Na + + SO 3 2-
SO 3 2- + H 2 O ↔ HSO 3 - + OH - (formation of low-dissociating hydrosulfite ion)
The solution medium is alkaline (pH\u003e 7), the color of phenolphthalein indicator in alkaline medium is raspberry.
Sodium sulfate is a salt formed by strong sulfuric acid and sodium hydroxide (alkali is a strong base), does not hydrolyze. The solution medium is neutral (pH \u003d 7), the color of the phenolphthalein indicator in a neutral medium is pale pink.
B) The salts of Na 2 SO 4 and ZnSO 4 can also be distinguished with a potassium carbonate solution. Zinc sulfate enters into an exchange reaction with potassium carbonate, as a result of which zinc carbonate precipitates:
ZnSO 4 + K 2 CO 3 → ZnCO 3 ↓ + K 2 SO 4
D) The salts FeCl 2 and Zn (NO 3) 2 can be distinguished by a solution of lead nitrate. When it interacts with iron chloride, a poorly soluble substance PbCl 2 is formed:
FeCl 2 + Pb (NO 3) 2 → PbCl 2 ↓ + Fe (NO 3) 2
Establish a correspondence between reactants and carbon-containing products of their interaction.
| REACTIVE SUBSTANCES
A) CH 3 -C≡CH + H 2 (Pt) → B) CH 3 -C≡CH + H 2 O (Hg 2+) → B) CH 3 -C≡CH + KMnO 4 (H +) → D) CH 3 -C≡CH + Ag 2 O (NH 3) → |
INTERACTION PRODUCT
1) CH 3 -CH 2 -CHO 2) CH 3 -CO-CH 3 3) CH 3 -CH 2 -CH 3 4) CH 3 -COOH and CO 2 5) CH 3 -CH 2 -COOAg 6) CH 3 -C≡CAg |
Answer: A-3; B-2; AT 4; G-6
Explanation:
A) Propyne adds hydrogen, converting in its excess into propane:
CH 3 -C≡CH + 2H 2 → CH 3 -CH 2 -CH 3
B) The addition of water (hydration) to alkynes in the presence of divalent mercury salts, resulting in the formation of carbonyl compounds, is a reaction of M.G. Kucherov. Hydration of propyne leads to the formation of acetone:
CH 3 -C≡CH + H 2 O → CH 3 -CO-CH 3
C) The oxidation of propyne with potassium permanganate in an acidic medium leads to the breaking of the triple bond in the alkyn, resulting in the formation of acetic acid and carbon dioxide:
5CH 3 -C≡CH + 8KMnO 4 + 12H 2 SO 4 → 5CH 3 -COOH + 5CO 2 + 8MnSO 4 + 4K 2 SO 4 + 12H 2 O
D) Silver propynide is formed and precipitated when propyne is passed through an ammonia solution of silver oxide. This reaction serves to detect alkynes with a triple bond at the end of the chain.
2CH 3 -C≡CH + Ag 2 O → 2CH 3 -C≡CAg ↓ + H 2 O
Establish a correspondence between the reactants and the organic matter that is the reaction product.
| INTERACTION PRODUCT
5) (CH 3 COO) 2 Cu |
Answer: A-4; B-6; IN 1; G-6
Explanation:
A) During oxidation ethyl alcohol with copper (II) oxide, acetaldehyde is formed, while the oxide is reduced to metal:
B) When concentrated sulfuric acid acts on alcohol at temperatures above 140 0 C, the reaction of intramolecular dehydration occurs - the elimination of a water molecule, which leads to the formation of ethylene:
C) Alcohols react violently with alkali and alkaline earth metals. The active metal replaces hydrogen in the hydroxyl group of the alcohol:
2CH 3 CH 2 OH + 2K → 2CH 3 CH 2 OK + H 2
D) In \u200b\u200ban alcoholic alkali solution, alcohols undergo an elimination (cleavage) reaction. In the case of ethanol, ethylene is formed:
CH 3 CH 2 Cl + KOH (alcohol) → CH 2 \u003d CH 2 + KCl + H 2 O
Using the electronic balance method, write the reaction equation:
In this reaction, chloric acid is an oxidizing agent, since the chlorine contained in it lowers the oxidation state from +5 to -1 in HCl. Therefore, the reducing agent is acidic phosphorus (III) oxide, where phosphorus increases the oxidation state from +3 to a maximum +5, converting into orthophosphoric acid.
Let's compose half-reactions of oxidation and reduction:
Cl +5 + 6e → Cl −1 | 2
2P +3 - 4e → 2P +5 | 3
We write the equation of the redox reaction in the form:
3P 2 O 3 + 2HClO 3 + 9H 2 O → 2HCl + 6H 3 PO 4
The copper was dissolved in concentrated nitric acid. The evolved gas was passed over the heated zinc powder. The resulting solid added to the sodium hydroxide solution. An excess of carbon dioxide was passed through the resulting solution, while the formation of a precipitate was observed.
Write the equations for the four reactions described.
1) When copper is dissolved in concentrated nitric acid, copper is oxidized to Cu +2, while a brown gas is released:
Cu + 4HNO 3 (conc.) → Cu (NO 3) 2 + 2NO 2 + 2H 2 O
2) When brown gas is passed over a heated zinc powder, zinc is oxidized, and nitrogen dioxide is reduced to molecular nitrogen (it is assumed by many, with reference to Wikipedia, that zinc nitrate does not form when heated, since it is thermally unstable):
4Zn + 2NO 2 → 4ZnO + N 2
3) ZnO - amphoteric oxide, dissolves in an alkali solution, turning into tetrahydroxozincate:
ZnO + 2NaOH + H 2 O → Na 2
4) When an excess of carbon dioxide is passed through a solution of sodium tetrahydroxozincate, an acidic salt is formed - sodium bicarbonate, zinc hydroxide precipitates:
Na 2 + 2CO 2 → Zn (OH) 2 ↓ + 2NaHCO 3
Write the reaction equations with which you can carry out the following transformations:
Use structural formulas when writing reaction equations organic matter.
1) The most characteristic reactions for alkanes are free radical substitution reactions, during which a hydrogen atom is replaced by a halogen atom. In the reaction of butane with bromine, the substitution of the hydrogen atom at the secondary carbon atom predominantly occurs, as a result of which 2-bromobutane is formed. This is due to the fact that the radical with an unpaired electron at the secondary carbon atom is more stable than the free radical with an unpaired electron at the primary carbon atom:
2) When 2-bromobutane interacts with alkali in an alcoholic solution, a double bond is formed as a result of the elimination of a hydrogen bromide molecule (Zaitsev's rule: when a hydrogen halide is cleaved from secondary and tertiary haloalkanes, a hydrogen atom is split off from the least hydrogenated carbon atom):
3) The interaction of butene-2 \u200b\u200bwith bromine water or a solution of bromine in an organic solvent leads to rapid discoloration of these solutions as a result of the addition of a bromine molecule to butene-2 \u200b\u200band the formation of 2,3-dibromobutane:
CH 3 -CH \u003d CH-CH 3 + Br 2 → CH 3 -CHBr-CHBr-CH 3
4) When interacting with a dibromo derivative, in which the halogen atoms are at neighboring carbon atoms (or at the same atom), an alcoholic alkali solution, two molecules of hydrogen halide are split off (dehydrohalogenation) and a triple bond is formed:
5) In the presence of divalent mercury salts, alkynes add water (hydration) to form carbonyl compounds:
A mixture of iron and zinc powders reacts with 153 ml of a 10% hydrochloric acid solution (ρ \u003d 1.05 g / ml). Interaction with the same mass of the mixture requires 40 ml of a 20% sodium hydroxide solution (ρ \u003d 1.10 g / ml). Determine the mass fraction of iron in the mixture.
In the answer, write down the reaction equations that are indicated in the condition of the problem, and provide all the necessary calculations.
Answer: 46.28%
When 2.65 g of organic matter was burned, 4.48 l of carbon dioxide (NU) and 2.25 g of water were obtained.
It is known that when this substance is oxidized with a potassium permanganate sulfate solution, a monobasic acid is formed and carbon dioxide is released.
Based on the given conditions of the assignment:
1) make the calculations necessary to establish the molecular formula of organic matter;
2) write down the molecular formula of the original organic matter;
3) make up the structural formula of this substance, which unambiguously reflects the order of the bonds of atoms in its molecule;
4) write the equation of the oxidation reaction of this substance with potassium permanganate sulfate solution.
Answer:
1) C x H y; x \u003d 8, y \u003d 10
2) C 8 H 10
3) C 6 H 5 -CH 2 -CH 3 - ethylbenzene
4) 5C 6 H 5 -CH 2 -CH 3 + 12KMnO 4 + 18H 2 SO 4 → 5C 6 H 5 -COOH + 5CO 2 + 12MnSO 4 + 6K 2 SO 4 + 28H 2 O
The Unified State Exam in Chemistry is an exam that is taken by graduates who plan to enter a university for certain specialties related to this discipline. Chemistry is not included in the list of compulsory subjects, according to statistics, 1 out of 10 graduates takes chemistry.
- The graduate receives 3 hours of time for testing and completing all tasks - planning and allocating time to work with all tasks is an important task for the test subject.
- Usually the exam includes 35-40 tasks, which are divided into 2 logical blocks.
- Like the rest of the USE, the chemistry test is divided into 2 logical blocks: testing (choosing the correct option or options from the proposed ones) and questions that require detailed answers. It is the second block that usually takes longer, so the subject needs to rationally allocate time.
- The main thing is to have reliable, deep theoretical knowledge that will help you successfully complete various tasks of the first and second blocks.
- You need to start preparing in advance in order to systematically work through all the topics - six months may not be enough. The best option is to start training in 10th grade.
- Identify the topics that make up for you biggest problemsso that you know what to ask when you ask a teacher or tutor.
- Learning to perform tasks typical for the exam in chemistry is not enough to know the theory, it is necessary to bring the skills of performing tasks and various tasks to automatism.
- Not always self-preparation effective, so it is worth finding a specialist to whom you can turn for help. The best option is a professional tutor. Also, don't be afraid to ask your school teacher questions. Do not neglect school education, carefully follow the assignments in the classroom!
- The exam has hints! The main thing is to learn how to use these sources of information. The student has a periodic table, tables of metal stress and solubility - this is about 70% of the data that will help to understand various tasks.
- Chemistry requires a thorough knowledge of mathematics - without this it will be difficult to solve problems. Be sure to repeat the work with percentages and proportions.
- Learn the formulas you need to solve chemistry problems.
- Study the theory: textbooks, reference books, collections of problems will be useful.
- The best way to reinforce theoretical assignments is to actively solve chemistry assignments. In online mode, you can solve in any quantity, improve the skills of solving problems of different types and difficulty levels.
- It is recommended to disassemble and analyze controversial points in assignments and mistakes with the help of a teacher or tutor.
Demo options USE in chemistry for grade 11 consist of two parts. The first part includes tasks to which you need to give a short answer. The tasks from the second part must be answered in detail.
All demonstration options for the exam in chemistry contain correct answers to all assignments and assessment criteria for items with a detailed answer.
There are no changes in comparison with.
Demonstration options for the exam in chemistry
Note that demonstration options for chemistry are presented in pdf format, and to view them it is necessary that, for example, the free software package Adobe Reader be installed on your computer.
| Demonstration version of the USE in chemistry for 2002 |
| Demonstration version of the exam in chemistry for 2004 |
| Demonstration version of the USE in chemistry for 2005 |
| Demonstration version of the USE in chemistry for 2006 |
| Demo version of the USE in chemistry for 2008 |
| Demonstration version of the USE 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 USE in chemistry for 2013 |
| Demonstration version of the exam in chemistry for 2014 |
| Demonstration version of the USE in chemistry for 2015 |
| Demonstration version of the exam in chemistry for 2016 |
| Demonstration version of the USE in chemistry for 2017 |
| Demo version of the exam in chemistry for 2018 |
| Demonstration version of the exam in chemistry for 2019 |
Changes in the demo 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 required a short answer. The tasks from the third part had to be given a detailed answer.
In 2014 in demo version of the exam in chemistrythe 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 using assignments IN 2 and C1;
- topic "Hydrolysis of salts" checked only with the task AT 4;
- a new task was included (in position AT 6) to check the topics “qualitative reactions to inorganic substances and ions "," qualitative reactions of organic compounds "
- total number of tasks in each version became 42 (instead of 43 in 2013 work).
In 2015, there were fundamental changes were 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).
The variant became be in two parts (part 1 - short answer assignments, part 2 - tasks with a detailed answer).
Numbering assignments became through throughout the entire version without letter designations A, B, C.
Was the form of recording the answer in tasks with a choice of answer has been changed:it became necessary to write the answer in a number with the number of the correct answer (and not mark with a cross).
It was reduced the number of tasks basic level difficulty from 28 to 26 tasks.
Maximum score for completing all tasks examination work 2015 became 64 (instead of 65 points in 2014).
IN 2016 year in demo version in chemistrysignificant changes were made compared to previous 2015 :
In 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 35advanced level of complexity : these tasks now require matching instead of choosing multiple correct answers from the suggested list.
Changed the distribution of tasks by difficulty level and types of tested skills.
In 2017 compared to demo version 2016 in chemistrythere have been significant changes. The structure of the examination work has been optimized:
Was the structure of the first part has been changed 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 difficulty.
It was reduced the total number of tasks up to 34.
Was grading scale changed (from 1 to 2 points) performing 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 completing all the tasks of the examination work was reduced to 60 points.
In 2018 in demo version of the exam in chemistry compared with demo 2017 in chemistrythe following changes:
It was added task 30 high level difficulties with a detailed answer,
Maximum score for completing all the tasks of the examination work remained without change by changing the grading scale for tasks in Part 1.
IN demo version of the 2019 exam in chemistry compared with demo 2018 in chemistrythere were no changes.
On our website, you can also familiarize yourself with the training materials prepared by the teachers of our training center "Resolvent" to prepare for the exam in mathematics.
For schoolchildren in grades 10 and 11 who want to prepare well and pass Unified State Exam in Mathematics or Russian for a high score, the Resolvent training center conducts
We also have organized
The result of the USE in chemistry not lower than the minimum established number of points gives the right to enter universities for specialties where the subject of chemistry is included in the list of entrance examinations.
Universities do not have the right to set a minimum threshold for chemistry below 36 points. Prestigious universitiestend to set their minimum threshold much higher. Because first-year students must have very good knowledge to study there.
On the official website of the FIPI, every year the versions of the Unified State Exam in Chemistry are published: a demonstration, an early period. It is these options that give an idea of \u200b\u200bthe structure of the future exam and the level of complexity of tasks and are sources of reliable information in preparation for the exam.
Early version of the exam in chemistry 2017
| Year | Download early version |
| 2017 | variant po himii |
| 2016 | download |
Demonstration version of the exam in chemistry 2017 from the FIPI
| Option of tasks + answers | Download demo version |
| Specification | demo variant himiya ege |
| Codifier | kodifikator |
There are changes in the versions of the USE in chemistry in 2017 compared to the CMM of the last 2016, therefore it is advisable to conduct training according to the current version, and use the options of previous years for the diverse development of graduates.
Additional materials and equipment
The following materials are attached to each variant of the exam paper in chemistry:
- periodic system of chemical elements D.I. Mendeleev;
- table of solubility of salts, acids and bases in water;
- electrochemical series of metal voltages.
It is allowed to use a non-programmable calculator during the examination work. The list of additional devices and materials, the use of which is permitted for the Unified State Exam, is approved by the order of the Ministry of Education and Science of Russia.
For those who want to continue their education at a university, the choice of subjects should depend on the list of entrance examinations in the chosen specialty
(direction of training).
The list of entrance examinations in universities for all specialties (areas of training) is determined by the order of the Ministry of Education and Science of Russia. Each university chooses from this list certain subjects that it indicates in its admission rules. You need to familiarize yourself with this information on the websites of the selected universities before applying for participation in the USE with a list of selected subjects.
Training options for the exam in chemistry
We have developed training tests in chemistry for the USE 2019 with answers and solutions.
In preparation, study 10 training options, compiled on the basis of the new one.
Features of tasks in tests of the exam in chemistry
Consider the typology and structure of some tasks in the first part:
- - the condition gives a number of chemical elements and questions about each of them, pay attention to the number of cells for the answer - there are two of them, therefore, there are two solutions;
- - correspondence between two sets: there will be two columns, in one formulas of substances, and in the second - a group of substances, it will be necessary to find correspondences.
- In the first part, there will also be tasks that require the behavior of a "mental chemical experiment" in which the student chooses formulas that allow him to find the correct answer to the exam question.
- The tasks of the second block are higher in complexity and require mastery of several content elements and several skills and abilities.
hint: when solving a problem, it is important to determine the class, group of substances and properties.
Tasks with detailed answers are aimed at checking knowledge of the main courses:
- Atomic structure;
- Periodic laws;
- Inorganic chemistry;
- Organic chemistry;
- Calculations by formulas;
- The use of chemistry in life.
Preparation for the exam in chemistry - quickly and efficiently
Fast - means, not less than six months:
- Pull up on math.
- Review the whole theory.
- Solve online trial options in chemistry, watch video tutorials.
Our site has provided for such an opportunity - come and train and get high scores on exams.