Among the listed substances, acid salts are.
Task No. 1
Among the listed substances, an acidic salt is
1) magnesium hydride
2) sodium bicarbonate
3) calcium hydroxide
4) copper hydroxycarbonate
Answer: 2
Explanation:
Acid salts are salts containing two types of cations: a metal (or ammonium) cation and a hydrogen cation, and a multiply charged anion of an acidic residue. The hydrogen cation gives the name of the salt the prefix “hydro”, for example, sodium hydrogen sulfate. Consequently, of all the proposed options, sodium bicarbonate NaHCO 3 belongs to the acidic salt.
In magnesium hydride MgH 2, hydrogen is an anion, i.e. has an oxidation state of -1.
Calcium hydroxide Ca(OH) 2 (slaked lime) is the basic hydroxide or base. Basic hydroxides are complex substances that consist of metal or ammonium cations and hydroxide ion (OH −) and in aqueous solution dissociate to form OH − anions and cations.
Copper hydroxycarbonate (CuOH) 2 CO 3 is the main salt of the metal copper and carbonic acid. Basic salts (hydroxo salts) are products of incomplete substitution of hydroxide groups in the molecules of polyacid bases with acidic residues.
Task No. 2
- 1. CaCO 3
- 2.CaCl2
- 3. CaO
- 4. Ca(OH) 2
Answer: 1
Explanation:
Calcium carbonate CaCO 3 is a water-insoluble salt and interacts with strong acids that can displace weak carbonic acid H 2 CO 3. In turn, carbonic acid is unstable and breaks down into carbon dioxide and water:
CaCO 3 + 2HCl = CaCl 2 + CO 2 + H 2 O
CaCl 2 + Na 2 CO 3 = CaCO 3 ↓ + 2NaCl
Therefore, substance X 2 is calcium carbonate CaCO 3.
Task No. 3
In the transformation scheme, the substance “X 2” is
- 1.FeO
- 2. Fe(OH) 3
- 3.FeCl2
- 4. FeCl 3
Answer: 2
Explanation:
Iron (II) chloride FeCl 2 is a soluble average salt of divalent iron and hydrochloric acid, reacts with alkalis to form a water-insoluble precipitate. white:
FeCl 2 + 2NaOH = Fe(OH) 2 ↓ + 2NaCl
The resulting iron (II) hydroxide Fe(OH) 2 darkens in humid air over time, oxidizing to ferric hydroxide Fe(OH) 3:/p>
4Fe(OH) 2 + O 2 + 2H 2 O = 4Fe(OH) 3
Task No. 4
Acid salts do not include a substance whose formula is
- 1.NH4Cl
- 2. NaHS
- 3. Ca(HCO 3) 2
- 4. NaH2PO4
Answer: 1
Explanation:
Acid salts are salts containing two types of cations: a metal (or ammonium) cation and a hydrogen cation, and an acid residue anion. Monobasic acids, for example, HNO 3, HNO 2, HCl, do not form acid salts.
Acid salts are sodium hydrosulfide NaHS (hydrogen sulfide salt H 2 S), calcium bicarbonate Ca(HCO 3) 2 (carbonic acid salt H 2 CO 3) and sodium dihydrogen phosphate NaH 2 PO 4 (phosphoric acid salt H 3 PO 4). Ammonium chloride NH 4 Cl is a middle salt (salt hydrochloric acid HCl).
Task No. 5
The interaction of sodium carbonate with a solution of calcium chloride is classified as a reaction
1) decomposition
3) substitution
4) connections
Answer: 2
Explanation:
The reaction between two soluble salts, sodium carbonate and a solution of calcium chloride, proceeds according to the following scheme:
Na 2 CO 3 + CaCl 2 = CaCO 3 ↓ + 2NaCl
This reaction belongs to exchange reactions (an exchange reaction is a reaction between complex substances in which they exchange their components). As a result of this reaction, a precipitate of calcium carbonate CaCO 3 is formed.
Compound reactions include reactions between two simple substances or two complex ones, resulting in the formation of one complex or one more compound.
Substitution reactions include reactions between complex and simple substances, in which atoms of a simple substance replace one of the atoms of a complex substance.
Task No. 6
Aluminum chloride in solution reacts with
- 1. K2SO4
- 2. MgSO 4
- 3.HNO3
- 4. Ca(OH) 2
Answer: 4
Explanation:
Aluminum chloride AlCl 3 is a soluble salt, therefore it can interact with alkalis and soluble salts in exchange reactions to form a precipitate.
AlCl 3 + K 2 SO 4 → the reaction does not occur: Al 3+, Cl −, K +, SO 4 2- ions do not form precipitation;
AlCl 3 + MgSO 4 → the reaction does not occur: Al 3+, Cl −, Mg 2+, SO 4 2- ions do not form precipitation;
AlCl 3 + HNO 3 → the reaction does not occur: Al 3+, Cl −, H +, NO 3 − ions do not form precipitation.
The exchange reaction between aluminum chloride AlCl 3 and alkali Ca(OH) 2 occurs due to the formation of insoluble aluminum hydroxide Al(OH) 3:
2AlCl 3 + 3Ca(OH) 2 = 2Al(OH) 3 ↓ + 3CaCl 2
Task No. 7
A solution of iron(II) chloride reacts with
1) silver phosphate
2) zinc nitrate
3) potassium sulfate
4) sodium sulfide
Answer: 4
Explanation:
Iron (II) chloride (FeCl 2) is a soluble salt, so it enters into exchange reactions with other soluble salts and alkalis if a precipitate is formed as a result.
Of all the proposed options, a precipitate is formed only when interacting with sodium sulfide Na 2 S due to the formation of a FeS precipitate.
Classification is not organic matter. Nomenclature inorganic substances(trivial and international). Classification of organic substances. Nomenclature of organic substances (trivial and international).
1. Among the listed substances, select three substances related to the main oxides
1) FeO
2) CaO
3) A1 2 Oz
4) K 2 O
5) CO 2
6)NO
2. Among the listed elements can form acid oxide
1) strontium
2) manganese
3) calcium
4) magnesium
5) sulfur
6) chrome
Write down the numbers under which they are indicated
3. Among the listed oxides
1) CO 2
2) Mn 2 O 7
3)SO2
4) Na 2 O
5) Cr 2 Oz
6) CrO
Write down the numbers under which they are indicated
4. Among the listed oxidesbelong to acid oxides
1) Na 2 O
2) MgO
3) Al 2 O 3
4) SO2
5) CO 2,
6) SiO 2
Write down the numbers under which they are indicated
5 . Among the listed substancesacids include
1) H 2 C 2 O 4,
2) HCN,
3) H2S
4) K 2 SO 3,
5) NaHSO 4
6) KNO 3
Write down the numbers under which they are indicated
6 . Among the listed oxides are salt-forming
1)SeO3
2) SiO 2,
3) Cl 2 O 7
4) CO
5) NO
6) N 2 O
Write down the numbers under which they are indicated
7 . Among the listed substances acid salts are
1) NaHCO 3
2) HCOOK
3) (NH 4 ) 2 SO 4
4) KHSO 3
5) Na 2 HPO 4
6) Na 3 PO 4
Write down the numbers under which they are indicated
8. Among the listed substancesnon-metal hydroxides are
1) H 2 SO 4,
2) HC1
3) HNO3
4) H 3 PO 4,
5) H2S
6) HCI
Write down the numbers under which they are indicated
9 . Among the listed substancesBasic oxides include
1) A1 2 O 3,
2) MgO,
3) Na 2 O
4) N 2 O
5) CuO
6) ZnO
Write down the numbers under which they are indicated
10. Among the listed substancesacid oxides include
1) СгО 3
2) CaO
3) A1 2 O 3
4)NO
5) SiO 2
6) Mn 2 O 7
Write down the numbers under which they are indicated
11. Among the listed substances, dibasic acids include
1) H 2 CO3
2) HC1 O 4
3) H 2 SO 3
4) HMnO4
5) H 2 ZnO 2
6) H 2 C rO 4
Write down the numbers under which they are indicated
12. Among the listed substances, acids include
1) H N O 3
2) KNSOz
3) HNO2
4) H 2 S
5) Na 2 SO 3
6) CH 3 COOH
Write down the numbers under which they are indicated
13. Among the listed substances, non-salt-forming oxides include
1) N 2 O 5
2) NO 2
3) N 2 O 3
4)NO
5) CO
6) N 2 O
Write down the numbers under which they are indicated
14. Among the listed oxides, amphoteric oxides include
1) sulfur oxide (IV)
2) aluminum oxide
3) lithium oxide
4) phosphorus oxide (V)
5) zinc oxide
6) iron (III) oxide
Write down the numbers under which they are indicated
15. Among the listed salts, medium salts include
1) Mn(NO 3 ) 2
2) Mg(H 2 P0 4 ) 2
3) A1 2 (SO 4 ) 3
4) (NH 4 ) 2 HP0 4
5) NaHSO3
6) (NH 4 ) 2 S
Write down the numbers under which they are indicated
16. Among the listed substances, alkanes include
1) C 3 H 6
2) C 2 H 4
3) CH 4
4) C 6 H 6
5) C 3 H 8
6) C 2 H 6
Write down the numbers under which they are indicated
17. Among the listed saltsacid salts include
1) Ag 2 CO 3
2) NaHS
3) Cu(NO 3 ) 2
4) Fe 2 (SO 4 ) 3
5) Ca(HCO 3 ) 2
6) KH 2 PO 4
Write down the numbers under which they are indicated
18. Among the listed substances, amino acids include
1) aniline
2) styrene
3) glycine
4) alanine
5) phenylalanine
6) phenol
Write down the numbers under which they are indicated
19. Among the listed substances, amphoteric compounds include
1) glucose
2) butanol-1
3) glycine
4) formic acid
5) zinc hydroxide
6) chromium hydroxide ( III)
Write down the numbers under which they are indicated
20. Among the listed substances, amphoteric hydroxides include
1) With a(OH) 2
2) Fe(OH) 3
3) Be(OH) 2
4) CON
5) Zn(OH) 2
6) Ba(OH) 2
Write down the numbers under which they are indicated
21. Among the listed substances, alcohols are
1) ethanol
2) toluene
3) ethylene glycol
4) glycerin
5) o-xylene
6) methanal
Write down the numbers under which they are indicated
Answers: 1-124, 2-256, 3-123, 4-456, 5-123, 6-123, 7-145, 8-134, 9-235, 10-156, 11-136, 12-346, 13-456, 14-256, 15-136, 16-356, 17-256, 18-345, 19-156, 20-235, 21-134.
Test
on the topic “Main classes of inorganic compounds”
Option 1
1. Acids include each of the 2 substances:
A) H 2 S, Na 2 CO 3 b) K 2 SO 4, Na 2 SO 4 c) H 3 PO 4, HNO 3 d) KOH, HCl
2. Copper (II) hydroxide corresponds to the formula:
A)
3. Sodium sulfate formula:
a) magnesium hydride b) sodium bicarbonate
c) calcium hydroxide d) copper hydroxychloride
5. Which element forms an acidic oxide?
a) strontium b) sulfur c) calcium d) magnesium
A) ZnO b) SiO 2 c) BaO d) Al 2 O 3
7. Carbon monoxide (IV) reacts with each of two substances:
a) water and calcium oxide
b) oxygen and sulfur oxide (IV)
d) phosphoric acid and hydrogen
Substance formulas
Interaction Products
A) Mg + HCl →
1) MgCl 2
b) Mg(OH) 2 + CO 2 →
2) MgCl 2 + H 2
V) Mg(OH) 2 + HCl →
3) MgCl 2 + H 2 O
4) MgCO 3 + H 2
5) MgCO 3 + H 2 O
A ) Fe→ Fe 2 O 3 → FeCl 3 → Fe(OH) 3 → Fe 2 O 3
b ) S→SO 2 →SO 3 → H 2 SO 4 → ZnSO 4
10. What mass of potassium sulfate is formed when 49 g of sulfuric acid reacts with potassium hydroxide?
Option 2
1. The bases include each of the 2 substances:
A) H 2 O, Na 2 O b) KOH, NaOH c) HPO 3, HNO 3 d) KOH, Na Cl
2. Copper oxide (II) corresponds to the formula:
A) Cu 2 O b) Cu (OH) 2 c) CuOH d) CuO
3. Sulfitasodium Formula:
A ) Na 2 SO 4 b) Na 2 S c) Na 2 SO 3 d) Na 2 SiO 3
4. Among the listed substances, an acidic salt is
a) barium hydroxide b) potassium hydroxycarbonate
c) copper bicarbonate d) calcium hydride;
5. Which element can form an amphoteric oxide?
a) sodium b) sulfur c) phosphorus d) aluminum
6. Basic oxides include
A) MgO b) SO 2 c) B 2 O 3 d) Al 2 O 3
7. Sodium oxide reacts with each of two substances:
a) water and calcium oxide
b) oxygen and hydrogen
c) potassium sulfate and sodium hydroxide
d) phosphoric acid and sulfur oxide (IV)
8. Establish a correspondence between the formula of the starting substances and the reaction products
Substance formulas
Interaction Products
A) Fe + HCl →
1) FeCl2
b) Fe(OH) 2 + CO 2 →
2) FeCl 2 + H 2
V) Fe(OH) 2 + HCl →
3) FeCl 2 + H 2 O
4) FeCO 3 + H 2
5) FeCO 3 + H 2 O
9. Carry out the chain of the following transformations:
A) Mg→ MgO→ MgCl 2 → Mg(OH) 2 → MgO
b) C→ CO 2 → Na 2 CO 3 → Na 2 SO 4 → BaSO 4
10. What mass of barium sulfate is formed when 30.6 g of barium oxide reacts with sufficient quantity sulfuric acid?
Answer table.
1 option
Option 2
1 in
1b
2 b
2 g
3 a
3 in
4 b
4 in
5 b
5 g
6 in
6 a
7 a
7 g
8 - 253
8 - 253
10- 87 g
10 – 46.6 g
Problem 1. One test tube contains a solution of magnesium chloride, the other contains aluminum chloride. Using what single reagent can you determine which test tubes contain these salts?
Solution. Aluminum differs from magnesium in that its hydroxide, Al(HE) 3 , amphoteric and soluble in alkalis. Therefore, when adding an excess of alkali solution to solution AlWITHl 3 a clear solution is formed:
Magnesium hydroxide is insoluble in alkalis, therefore, when an alkali solution is added to a solution of magnesium chloride, a precipitate forms:
MgWITHl2 + 2KON = Mg(OH)2↓ + 2KSl.
Problem 2. Write complete equations for the following reactions:
1) AlWITHl 3 + CON (izb.) →
2) AlWITHl 3 + NN 3(izb.) + N 2 ABOUT→
3) Al( NABOUT 3 ) 3 + NA 2 S+ N 2 ABOUT→
4) Na[Al(HE) 4 ] + CO 2 →
Solution. 1) When an alkali acts on aluminum salts, a precipitate of aluminum hydroxide precipitates, which dissolves in excess alkali to form aluminate:
A l C l 3 + 4KON = K[A l (OH)4] + ZKS l.
2) When a solution of ammonia and alkali reacts with aluminum salts, a precipitate of aluminum hydroxide precipitates.
AlWITHl3 + 3 NH3 + 3H2O = Al(OH)3↓ + 3 NH4Sl.
Unlike alkalis, ammonia solution does not dissolve aluminum hydroxide. That is why ammonia is used to completely precipitate aluminum from aqueous solutions of its salts.
3) Sodium sulfide enhances the hydrolysis of aluminum chloride and brings it to the end, to Al(HE) 3 . In turn, aluminum chloride enhances the hydrolysis of sodium sulfide and brings it to the end, to H 2 S:
2Al( NO3)3 + ZNa2S+ 6H2O = 2Al(OH)3↓ + ZN2S + 6 NаСl.
4) Sodium aluminate is formed by a very weak acid - aluminum hydroxide, so it is easily destroyed in an aqueous solution even under the influence of weak acids, for example, carbonic:
NA[Al(OH)4] + CO2 = Al(OH)3↓ + NaHCO3.
Problem 3. 25 g of 8% sodium hydroxide solution were added to 25 g of an 8% aluminum chloride solution. The resulting precipitate was filtered and calcined. Determine its mass and composition.
Solution. When alkalis act on solutions of aluminum salts, a precipitate of aluminum hydroxide is formed:
AlWITHl3 + ZNaOH = Al(OH)3↓ + ZNаСl.
Let's carry out the calculation using this equation. v(AlCl 3) = 25. 0.08 / 133.5 = 0.015, v(NaOH) = 25. 0.08 / 40 = 0.05. Al Cl 3 is in short supply. As a result of this reaction, 0.015 is consumed. 3 = 0.045 mol N aOH, and 0.015 mol Al (OH) 3 is formed. An excess of N aOH in the amount of 0.05-0.045 = 0.005 mol dissolves 0.005 mol Al (OH) 3 according to the equation:
A l (OH)3 + Na OH = Na [A l (OH)4].
Thus, 0.015-0.005 = 0.01 mol Al (OH) 3 remains in the precipitate. When this precipitate is calcined as a result of the reaction
2Al(OH)3Al2O3 + 3H2O
0.01/2 = 0.005 mol Al 2 O 3 is formed, weighing 0.005-102 = 0.51 g.
Answer. 0.51 g Al 2 ABOUT 3 .
Problem 4. What mass of alum KAl( SABOUT 4 ) 2 . 12H 2 O must be added to 500 g of 6% potassium sulfate solution to mass fraction the last one doubled? Find the volume of gas (at normal conditions) that will be released when the resulting solution is exposed to excess potassium sulfide.
Solution. The mass of the initial solution is 500 g, it contains 30 g of K 2 SABOUT 4 ( M= 174). Add x mol of alum KA to the solutionl( SABOUT 4 ) 2 . 12H 2 O (M = 474) (they contain x/2 mol K 2 SABOUT 4 ): m(KAl( SABOUT 4 ) 2 . 12H 2 O) = 474x, m(TO 2 SABOUT 4 ) = 30 +174 . x/2 = 30+87 x, m(r-ra) = 500+474x. According to the condition, mass fraction K 2 SABOUT 4 in the final solution is 12%, i.e.
(30+87x) / (500+474x) = 0.12,
whence x =1.00. The mass of added alum is equal to m (KAl (S O 4) 2. 12H 2 O) = 474. 1.00 = 474 g.
The resulting solution contains x/2 = 0.500 mol Al 2 (S O 4) 3, which reacts with excess K 2 S according to the equation:
Al2( SО4)3 + ЗК2S+ 6H2O = 2Al(OH)3↓ + ZN2S+ ZK2SO4.
According to this equation, v (H 2 S) = 3. v (Al 2 (S O 4) 3) = 3. 0.500 = 1.500 mol. V(H 2 S) = 1.500. 22.4=33.6 l.
Answer. 474 g KAl( SABOUT 4 ) 2 . 12H 2 ABOUT; 33.6 l H 2 S.
Examples of problem solving
Problem 1
NаСl→ NA→ NaN→ Naon→ NaNSO3.
Solution. Sodium is formed by electrolysis of molten sodium chloride:
2 NаСl = 2Na + Cl2 .
Sodium reacts with hydrogen:
2 N a + H2 = 2 N aH.
Sodium hydride is completely hydrolyzed by water:
NаН + Н2О =NaOH + H2 .
When excess sulfur dioxide is passed through a solution of sodium hydroxide, sodium hydrosulfite is formed:
N aOH + S O2 = N aH S O3.
Problem 2. When exposed to excess carbon dioxide on 32.9 g of an unknown metal compound with oxygen, solid substance “A” was formed, and gas “B” was released. Substance “A” was dissolved in water and an excess of barium nitrate solution was added, and 27.58 g of precipitate precipitated. Gas “B” was passed through a tube with hot copper, and the mass of the tube increased by 6.72 g. Establish the formula for the original compound.
Solution. From the problem statement it is clear that after passing CO 2 above the oxygen compound of the metal, a metal carbonate was formed, and an alkaline one at that (since only alkali metal carbonates are quite soluble in water), and oxygen was released. Let the formula of the original compound be Me x ABOUT y . Reaction equations:
2MehOy + xCO2 = xMe2CO3 + (y-0.5x)O2,
М2СО3 + ВаСl2 = 2MeSl+ BaCO3↓ ,
2C u + O2 = 2C u O.
The increase in the mass of the tube with heated copper is equal to the mass of the oxygen that reacted in the last reaction, therefore: v (O 2) = 6.72/32 = 0.21 mol.
According to the second reaction, v (BaCO 3) = 27.58 / 197 = 0.14 mol = v (Me 2 CO 3), therefore, v (Me) = 2 v (Me 2 CO 3) = 0.28 mol. The ratio of the coefficients in the reaction equation is equal to the ratio of the amounts of substances (in moles), therefore from the first equation it follows that x/(y-0.5x) = 0.14/0.21, from which we obtain that x:y = 1:2 . Therefore, we can conclude that the simplest formula of an oxygen compound is MeO 2.
Since v (MeO 2) = v (Me) = 0.28 mol, the molar mass of the oxygen compound is: M (MeO 2) = 39.2 / 0.28 = 117.5 g/mol, and the atomic mass of the metal: M(Me) = 117.5-32 = 85.5 g/mol. This metal is rubidium, Rb. The required formula is Rb O 2.
Answer.RbABOUT 2 .
Problem 3. When 6.0 g of metal interacted with water, 3.36 liters of hydrogen (n.s.) were released. Identify this metal if it is divalent in its compounds.
Solution. Since the metal is divalent, its reaction with water
described by the equation:
Me + 2H2O = Me(OH)2 + H2.
According to the equation, v (Me) = v (H 2) = 3.36/22.4 = 0.15 mol. Hence the atomic mass of the metal is A (Me) = m/v = 6.0/0.15 = 40 g/mol. This metal is calcium.
Answer. Calcium.
Problem 4. Write the reaction equations that allow the following transformations to occur:
Mg→ MgSO4→ Mg( NO3)2→ MgABOUT→ (CH3COO)2Mg.
Solution. Magnesium dissolves in dilute sulfuric acid:
Mg+ H2SO4 = MgSO4 + H2 .
Magnesium sulfate enters into an exchange reaction in an aqueous solution with barium nitrate:
MgSO4 + Ba(NO3)2 = BaSO4↓ + Mg( NO3)2.
When strongly heated, magnesium nitrate decomposes:
2Mg( NO3)2 = 2MgO+4NO2 + O2 .
Magnesium oxide is a typical basic oxide. It dissolves in acetic acid:
M g O + 2CH3COOH = (CH3COO)2M g + H2O.
Problem 5. There is a mixture of calcium, calcium oxide and calcium carbide with a molar ratio of components of 1:3:4 (in the order listed). What volume of water can enter chemical reaction with 35 g of this mixture?
Solution. Let the initial mixture containxmole Sa, thenv(CaO) = 3x, v(SaS 2 ) = 4 x. The total mass of the mixture is:
40. x + 56.3 x + 64.4 x = 35,
whence x = 0.0754 mol. When this mixture interacts with water, the following reactions occur:
Ca + 2H2O = Ca(OH)2 + H2,
CaO + H2 O = Ca(OH)2,
CaC2 + 2H2O = Ca(OH)2 + C2H2.
The first reaction involves 2x mol of H 2 O, the second - 3x, and the third - 2. 4x = 8x mol H 2 O, total - 13x = 13. 0.0754 = 0.980 mol. The mass of reacted water is 0.980. 18 = 17.6 g, volume of water is 17.6 g / 1 g/ml = 17.6 ml.
Answer. 17.6 ml N 2 ABOUT.