Black oxide Black oxide - Black oxide.
A black oxide on the surface of a metal obtained by immersion in a melt or hot salt solution.
(Source: “Metals and alloys. Directory.” Edited by Yu.P. Solntsev; NPO “Professional”, NPO “Peace and Family”; St. Petersburg, 2003)
See what “Black oxide” is in other dictionaries:
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DEFINITION
Oxides– a class of inorganic compounds, they are compounds of a chemical element with oxygen, in which oxygen exhibits an oxidation state of “-2”.
The exception is oxygen difluoride (OF 2), since the electronegativity of fluorine is higher than that of oxygen and fluorine always exhibits an oxidation state of "-1".
Oxides, depending on the chemical properties they exhibit, are divided into two classes - salt-forming and non-salt-forming oxides. Salt-forming oxides have an internal classification. Among them, acidic, basic and amphoteric oxides are distinguished.
Chemical properties of non-salt-forming oxides
Non-salt-forming oxides exhibit neither acidic, basic, nor amphoteric properties and do not form salts. Non-salt-forming oxides include oxides of nitrogen (I) and (II) (N 2 O, NO), carbon monoxide (II) (CO), silicon oxide (II) SiO, etc.
Despite the fact that non-salt-forming oxides are not capable of forming salts, when carbon monoxide (II) reacts with sodium hydroxide, an organic salt is formed - sodium formate (formic acid salt):
CO + NaOH = HCOONa.
When non-salt-forming oxides interact with oxygen, higher oxides of elements are obtained:
2CO + O 2 = 2CO 2 ;
2NO + O 2 = 2NO 2.
Chemical properties of salt-forming oxides
Among salt-forming oxides, basic, acidic and amphoteric oxides are distinguished, the first of which, when interacting with water, form bases (hydroxides), the second - acids, and the third - exhibit the properties of both acidic and basic oxides.
Basic oxides react with water to form bases:
CaO + 2H 2 O = Ca(OH) 2 + H 2 ;
Li 2 O + H 2 O = 2LiOH.
When basic oxides react with acidic or amphoteric oxides, salts are obtained:
CaO + SiO 2 = CaSiO 3;
CaO + Mn 2 O 7 = Ca(MnO 4) 2;
CaO + Al 2 O 3 = Ca(AlO 2) 2.
Basic oxides react with acids to form salts and water:
CaO + H 2 SO 4 = CaSO 4 + H 2 O;
CuO + H 2 SO 4 = CuSO 4 + H 2 O.
When basic oxides formed by metals in the activity series after aluminum interact with hydrogen, the metals included in the oxide are reduced:
CuO + H 2 = Cu + H 2 O.
Acidic oxides react with water to form acids:
P 2 O 5 + H 2 O = HPO 3 (metaphosphoric acid);
HPO 3 + H 2 O = H 3 PO 4 (orthophosphoric acid);
SO 3 + H 2 O = H 2 SO 4.
Some acidic oxides, for example, silicon (IV) oxide (SiO 2), do not react with water, therefore, the acids corresponding to these oxides are obtained indirectly.
When acidic oxides react with basic or amphoteric oxides, salts are obtained:
P 2 O 5 + 3CaO = Ca 3 (PO 4) 2;
CO 2 + CaO = CaCO 3 ;
P 2 O 5 +Al 2 O 3 = 2AlPO 4.
Acidic oxides react with bases to form salts and water:
P 2 O 5 + 6NaOH = 3Na 3 PO 4 + 3H 2 O;
Ca(OH) 2 + CO 2 = CaCO 3 ↓ + H 2 O.
Amphoteric oxides interact with acidic and basic oxides (see above), as well as with acids and bases:
Al 2 O 3 + 6HCl = 2AlCl 3 + 3H 2 O;
Al 2 O 3 + NaOH + 3H 2 O = 2Na;
ZnO + 2HCl = ZnCl 2 + H 2 O;
ZnO + 2KOH + H 2 O = K 2 4
ZnO + 2KOH = K 2 ZnO 2 .
Physical properties of oxides
Most oxides are solids at room temperature (CuO is a black powder, CaO is a white crystalline substance, Cr 2 O 3 is a green powder, etc.). Some oxides are liquids (water - hydrogen oxide - colorless liquid, Cl 2 O 7 - colorless liquid) or gases (CO 2 - colorless gas, NO 2 - brown gas). The structure of oxides is also different, most often molecular or ionic.
Obtaining oxides
Almost all oxides can be obtained by the reaction of a specific element with oxygen, for example:
2Cu + O 2 = 2CuO.
The formation of oxides also results from the thermal decomposition of salts, bases and acids:
CaCO 3 = CaO + CO 2;
2Al(OH) 3 = Al 2 O 3 + 3H 2 O;
4HNO 3 = 4NO 2 + O 2 + 2H 2 O.
Other methods for producing oxides include roasting binary compounds, for example, sulfides, oxidation of higher oxides to lower ones, reduction of lower oxides to higher ones, interaction of metals with water at high temperatures, etc.
Examples of problem solving
EXAMPLE 1
| Exercise | During the electrolysis of 40 mol of water, 620 g of oxygen were released. Determine the oxygen yield. |
| Solution | The yield of the reaction product is determined by the formula: η = m pr / m theor × 100%. The practical mass of oxygen is the mass indicated in the problem statement – 620 g. The theoretical mass of the reaction product is the mass calculated from the reaction equation. Let us write down the equation for the reaction of water decomposition under the influence of electric current: 2H 2 O = 2H 2 + O 2. According to the reaction equation n(H 2 O):n(O 2) = 2:1, therefore n(O 2) = 1/2×n(H 2 O) = 20 mol. Then, the theoretical mass of oxygen will be equal to: |
Oxides- these are binary oxygen compounds, that is, complex substances consisting of two elements, one of which is oxygen.
E 2 +n O n -2- general formula of oxides, where
n - oxidation state of the element
2 - oxidation state of oxygen
The names of oxides are made up of the word “oxide” and the name of the element forming the oxide in the genitive case (CaO - calcium oxide).
Oxides classification scheme
Oxide classification table with examples
|
Oxides classification |
Definition |
Examples of reactions |
Typical interactions |
|
Normal |
Oxides in which there are only bonds between oxygen and some element |
MgO, SO 3, SiO 2 |
See properties of acidic and basic oxides |
|
Peroxides |
Those in which there are bonds between two oxygen atoms |
Na 2 O 2 , H 2 O 2 |
See the table of properties of hydrogen peroxide |
|
Mixed oxides |
Those that are a mixture of two oxides of the same element in different oxidation states |
Pb 3 O 4 = 2РbО PbO 2 Fe 3 O 4 = FeO Fe 2 O 3 |
They have the same properties as their constituent oxides. |
|
Acidic or anhydrides |
Oxides that react with water to form acids; with bases and basic oxides - form salts |
SO 3, SO 2, Mn 2 O 7 |
SO 2 + H 2 O → H 2 SO 3 With bases and basic oxides: Mn 2 O 7 + 2KOH → 2KMnO 4 + H 2 O |
|
Basic oxides |
Those that react with water to form bases; form salts with acids and acid oxides |
CaO + H 2 O → Ca(OH) 2 With acids and acid oxides: Na 2 O + CO 2 → Na 2 CO 3 |
|
|
Amphoteric oxides |
Those that, depending on conditions, exhibit the properties of both acidic and basic oxides |
With acids: ZnO + 2HCl → ZnCl 2 + H 2 O With alkalis: ZnO + 2NaOH + H 2 O → Na 2 |
|
|
Indifferent (non-salt-forming) |
Oxides that do not react with either acids or bases. No salts are formed |
NO + H 2 O -/-> N 2 O + NaOH |
Methods for producing oxides table
Almost all chemical elements form oxides. At the moment, oxides of helium, neon and argon have not been obtained.
|
Methods for producing oxides |
Note |
|
|
Interaction of simple substances with oxygen |
S + O 2 → SO 2 4Al + 3O 2 → 2Al 2 0 3 |
This is how non-metal oxides are mainly obtained. |
|
Thermal decomposition of bases, salts, acids |
CaCO 3 t → CaO + CO 2 2H 3 BO 3 t → Bg 2 O 3 + H 2 O Mg(OH) 2 t → MgO + H 2 0 |
This is how metal oxides are mainly obtained |
|
Interaction of simple substances and salts with oxidizing acids |
C + 4HNO 3 (p-p) → CO 2 + 4N0 2 + H 2 O Сu + 4HNO 3 (cond.) → Cu(NO 3) 2 + 2NO 2 + 2H 2 O Na 2 SO 3 + 2H 2 SO 4 → 2NaHS0 4 + SO 2 + H 2 O |
Method for producing predominantly non-metal oxides |
Chemical properties of oxides table
|
Oxides classification |
Chemical properties of oxides |
Reaction examples |
|
Basic oxides |
1. Basic oxide* + water -> alkali |
K 2 O + H 2 O → 2KOH, BaO + H 2 O → Ba(OH) 2 |
|
2. Basic oxide + acid -> salt + water |
CuO + H 2 SO 4 → CuSO 4 + H 2 O |
|
|
3. Basic oxide + acidic oxide -> salt |
MgO + CO 2 → MgCO 3, ZCaO + P 2 O 5 → Ca 3 (PO 4) 2 |
|
|
Acidic oxides |
1. Acidic oxide + water -> acid |
SO 3 + H 2 O → H 2 SO 4 Cl 2 O 7 + H 2 O → 2HClO 4 SiO 2 + H 2 O -/-> no reaction (exception) |
|
2. Acid oxide + alkali -> salt + water |
SO 3 + 2NaOH → Na 2 SO 4 + H 2 O |
|
|
3. Acidic oxide + basic oxide -> salt |
SiO 2 + CaO t → CaSiO 3, R 2 O 4 + ZK 2 O → 2K 3 RO 4 |
|
|
Amphoteric oxides |
1. They react with acids as basic oxides |
ZnO + H 2 SO 4 → ZnSO 4 + H 2 O |
|
2. They react with bases (alkalis) as acid oxides |
ZnO + 2NaOH → Na 2 ZnO 2 + H 2 O |
_______________
A source of information: Nasonova A.E. Chemistry, school curriculum in tables and formulas, 1998
CuCl 2 + 4NH 3 = Cl 2
Na 2 + 4HCl = 2NaCl + CuCl 2 + 4H 2 O
2Cl + K 2 S = Cu 2 S + 2KCl + 4NH 3
When mixing solutions, hydrolysis occurs at both the weak base cation and the weak acid anion:
2CuSO 4 + Na 2 SO 3 + 2H 2 O = Cu 2 O + Na 2 SO 4 + 2H 2 SO 4
2CuSO 4 + 2Na 2 CO 3 + H 2 O = (CuOH) 2 CO 3 ↓ + 2Na 2 SO 4 + CO 2
Copper and copper compounds.
1) A direct electric current was passed through a solution of copper (II) chloride using graphite electrodes. The electrolysis product released at the cathode was dissolved in concentrated nitric acid. The resulting gas was collected and passed through a sodium hydroxide solution. The gaseous electrolysis product released at the anode was passed through a hot sodium hydroxide solution. Write the equations for the reactions described.
2) The substance obtained at the cathode during the electrolysis of molten copper (II) chloride reacts with sulfur. The resulting product was treated with concentrated nitric acid, and the liberated gas was passed through a solution of barium hydroxide. Write the equations for the reactions described.
3) The unknown salt is colorless and turns the flame yellow. When this salt is slightly heated with concentrated sulfuric acid, the liquid in which the copper dissolves is distilled off; the latter transformation is accompanied by the release of brown gas and the formation of a copper salt. During the thermal decomposition of both salts, one of the decomposition products is oxygen. Write the equations for the reactions described.
4) When a solution of salt A interacted with an alkali, a gelatinous, water-insoluble substance of blue color was obtained, which was dissolved in colorless liquid B to form a blue solution. The solid product remaining after careful evaporation of the solution was calcined; in this case, two gases were released, one of which is brown in color, and the second is part of the atmospheric air, and a black solid substance remains, which dissolves in liquid B to form substance A. Write the equations for the described reactions.
5) Copper turnings were dissolved in dilute nitric acid, and the solution was neutralized with potassium hydroxide. The released blue substance was separated, calcined (the color of the substance changed to black), mixed with coke and calcined again. Write the equations for the reactions described.
6) Copper shavings were added to the solution of mercury (II) nitrate. After the reaction was completed, the solution was filtered, and the filtrate was added dropwise to a solution containing sodium hydroxide and ammonium hydroxide. In this case, a short-term formation of a precipitate was observed, which dissolved to form a bright blue solution. When an excess of sulfuric acid solution was added to the resulting solution, a color change occurred. Write the equations for the reactions described.
7) Copper (I) oxide was treated with concentrated nitric acid, the solution was carefully evaporated and the solid residue was calcined. The gaseous reaction products were passed through a large amount of water and magnesium shavings were added to the resulting solution, resulting in the release of a gas used in medicine. Write the equations for the reactions described.
8) The solid formed when malachite is heated was heated in a hydrogen atmosphere. The reaction product was treated with concentrated sulfuric acid and added to a solution of sodium chloride containing copper filings, resulting in the formation of a precipitate. Write the equations for the reactions described.
9) The salt obtained by dissolving copper in dilute nitric acid was subjected to electrolysis using graphite electrodes. The substance released at the anode was reacted with sodium, and the resulting reaction product was placed in a vessel with carbon dioxide. Write the equations for the reactions described.
10) The solid product of the thermal decomposition of malachite was dissolved by heating in concentrated nitric acid. The solution was carefully evaporated and the solid residue was calcined to give a black substance, which was heated in excess of ammonia (gas). Write the equations for the reactions described.
11) A solution of dilute sulfuric acid was added to the black powdery substance and heated. A solution of caustic soda was added to the resulting blue solution until the precipitation stopped. The precipitate was filtered and heated. The reaction product was heated in a hydrogen atmosphere, resulting in a red substance. Write the equations for the reactions described.
12) An unknown red substance was heated in chlorine, and the reaction product was dissolved in water. Alkali was added to the resulting solution, the resulting blue precipitate was filtered and calcined. When the calcination product, which is black in color, was heated with coke, a red starting material was obtained. Write the equations for the reactions described.
13) The solution obtained by reacting copper with concentrated nitric acid was evaporated and the precipitate was calcined. The gaseous products are completely absorbed by water, and hydrogen is passed over the solid residue. Write the equations for the reactions described.
14) Black powder, which was formed by burning a red metal in excess air, was dissolved in 10% sulfuric acid. Alkali was added to the resulting solution, and the resulting blue precipitate was separated and dissolved in an excess of ammonia solution. Write the equations for the reactions described.
15) A black substance was obtained by calcining the precipitate that is formed by the interaction of sodium hydroxide and copper (II) sulfate. When this substance is heated with coal, a red metal is obtained, which dissolves in concentrated sulfuric acid. Write the equations for the reactions described.
16) Copper metal was treated with iodine by heating. The resulting product was dissolved in concentrated sulfuric acid while heating. The resulting solution was treated with potassium hydroxide solution. The precipitate that formed was calcined. Write the equations for the reactions described.
17) Excess soda solution was added to copper (II) chloride solution. The precipitate that formed was calcined, and the resulting product was heated in a hydrogen atmosphere. The resulting powder was dissolved in dilute nitric acid. Write the equations for the reactions described.
18) Copper was dissolved in dilute nitric acid. An excess of ammonia solution was added to the resulting solution, observing first the formation of a precipitate, and then its complete dissolution with the formation of a dark blue solution. The resulting solution was treated with sulfuric acid until the characteristic blue color of copper salts appeared. Write the equations for the reactions described.
19) Copper was dissolved in concentrated nitric acid. An excess of ammonia solution was added to the resulting solution, observing first the formation of a precipitate, and then its complete dissolution with the formation of a dark blue solution. The resulting solution was treated with excess hydrochloric acid. Write the equations for the reactions described.
20) The gas obtained by reacting iron filings with a solution of hydrochloric acid was passed over heated copper (II) oxide until the metal was completely reduced. the resulting metal was dissolved in concentrated nitric acid. The resulting solution was subjected to electrolysis with inert electrodes. Write the equations for the reactions described.
21) Iodine was placed in a test tube with concentrated hot nitric acid. The released gas was passed through water in the presence of oxygen. Copper(II) hydroxide was added to the resulting solution. The resulting solution was evaporated and the dry solid residue was calcined. Write the equations for the reactions described.
22) Orange copper oxide was placed in concentrated sulfuric acid and heated. An excess of potassium hydroxide solution was added to the resulting blue solution. the resulting blue precipitate was filtered, dried and calcined. The resulting solid black substance was placed in a glass tube, heated, and ammonia was passed over it. Write the equations for the reactions described.
23) Copper (II) oxide was treated with a solution of sulfuric acid. During electrolysis of the resulting solution on an inert anode, gas is released. The gas was mixed with nitric oxide (IV) and absorbed with water. Magnesium was added to a dilute solution of the resulting acid, as a result of which two salts were formed in the solution, but no gaseous product was released. Write the equations for the reactions described.
24) Copper (II) oxide was heated in a stream of carbon monoxide. The resulting substance was burned in a chlorine atmosphere. The reaction product was dissolved in water. The resulting solution was divided into two parts. A solution of potassium iodide was added to one part, and a solution of silver nitrate was added to the second. In both cases, the formation of a precipitate was observed. Write the equations for the reactions described.
25) Copper(II) nitrate was calcined and the resulting solid was dissolved in dilute sulfuric acid. The solution of the resulting salt was subjected to electrolysis. The substance released at the cathode was dissolved in concentrated nitric acid. Dissolution proceeds with the release of brown gas. Write the equations for the reactions described.
26) Oxalic acid was heated with a small amount of concentrated sulfuric acid. The released gas was passed through a solution of calcium hydroxide. In which a precipitate fell. Some of the gas was not absorbed; it was passed over a black solid obtained by calcination of copper (II) nitrate. The result was a dark red solid. Write the equations for the reactions described.
27) Concentrated sulfuric acid reacted with copper. The gas released during the process was completely absorbed by an excess of potassium hydroxide solution. The copper oxidation product was mixed with the calculated amount of sodium hydroxide until precipitation stopped. The latter was dissolved in excess hydrochloric acid. Write the equations for the reactions described.
Copper. Copper compounds.
1. CuCl 2 Cu + Cl 2
at the cathode at the anode
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
6NaOH (hor.) + 3Cl 2 = NaClO 3 + 5NaCl + 3H 2 O
2. CuCl 2 Cu + Cl 2
at the cathode at the anode
CuS + 8HNO 3 (conc. horizon) = CuSO 4 + 8NO 2 + 4H 2 O
or CuS + 10HNO 3 (conc.) = Cu(NO 3) 2 + H 2 SO 4 + 8NO 2 + 4H 2 O
4NO 2 + 2Ba(OH) 2 = Ba(NO 3) 2 + Ba(NO 2) 2 + 2H 2 O
3. NaNO 3 (tv.) + H 2 SO 4 (conc.) = HNO 3 + NaHSO 4
Cu + 4HNO 3(conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
2NaNO 3 2NaNO 2 + O 2
4. Cu(NO 3) 2 + 2NaOH = Cu(OH) 2 ↓ + 2NaNO 3
Cu(OH) 2 + 2HNO 3 = Cu(NO 3) 2 + 2H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + 2HNO 3 = Cu(NO 3) 2 + H 2 O
5. 3Cu + 8HNO 3(diluted) = 3Cu(NO 3) 2 + 2NO + 4H 2 O
Cu(NO 3) 2 + 2KOH = Cu(OH) 2 ↓ + 2KNO 3
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + C Cu + CO
6. Hg(NO 3) 2 + Cu = Cu(NO 3) 2 + Hg
Cu(NO 3) 2 + 2NaOH = Cu(OH) 2 ↓ + 2NaNO 3
(OH) 2 + 5H 2 SO 4 = CuSO 4 + 4NH 4 HSO 4 + 2H 2 O
7. Cu 2 O + 6HNO 3 (conc.) = 2Cu(NO 3) 2 + 2NO 2 + 3H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
4NO 2 + O 2 + 2H 2 O = 4HNO 3
10HNO3 + 4Mg = 4Mg(NO3)2 + N2O + 5H2O
8. (CuOH) 2 CO 3 2CuO + CO 2 + H 2 O
CuO + H2Cu + H2O
CuSO 4 + Cu + 2NaCl = 2CuCl↓ + Na 2 SO 4
9. 3Cu + 8HNO 3(diluted) = 3Cu(NO 3) 2 + 2NO + 4H 2 O
at the cathode at the anode
2Na + O 2 = Na 2 O 2
2Na 2 O 2 + CO 2 = 2Na 2 CO 3 + O 2
10. (CuOH) 2 CO 3 2CuO + CO 2 + H 2 O
CuO + 2HNO 3 Cu(NO 3) 2 + H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
11. CuO + H 2 SO 4 CuSO 4 + H 2 O
CuSO 4 + 2NaOH = Cu(OH) 2 + Na 2 SO 4
Cu(OH) 2 CuO + H 2 O
CuO + H2Cu + H2O
12. Cu + Cl 2 CuCl 2
CuCl 2 + 2NaOH = Cu(OH) 2 ↓ + 2NaCl
Cu(OH) 2 CuO + H 2 O
CuO + C Cu + CO
13. Cu + 4HNO 3 (conc.) = Cu (NO 3) 2 + 2NO 2 + 2H 2 O
4NO 2 + O 2 + 2H 2 O = 4HNO 3
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + H2Cu + H2O
14. 2Cu + O 2 = 2CuO
CuSO 4 + NaOH = Cu(OH) 2 ↓ + Na 2 SO 4
Сu(OH) 2 + 4(NH 3 H 2 O) = (OH) 2 + 4H 2 O
15. CuSO 4 + 2NaOH = Cu(OH) 2 + Na 2 SO 4
Cu(OH) 2 CuO + H 2 O
CuO + C Cu + CO
Cu + 2H 2 SO 4 (conc.) = CuSO 4 + SO 2 + 2H 2 O
16) 2Cu + I 2 = 2CuI
2CuI + 4H 2 SO 4 2CuSO 4 + I 2 + 2SO 2 + 4H 2 O
Cu(OH) 2 CuO + H 2 O
17) 2CuCl 2 + 2Na 2 CO 3 + H 2 O = (CuOH) 2 CO 3 + CO 2 + 4NaCl
(CuOH) 2 CO 3 2CuO + CO 2 + H 2 O
CuO + H2Cu + H2O
3Cu + 8HNO 3(dil.) = 3Cu(NO 3) 2 + 2NO + 4H 2 O
18) 3Cu + 8HNO 3(diluted) = 3Cu(NO 3) 2 + 2NO + 4H 2 O
(OH) 2 + 3H 2 SO 4 = CuSO 4 + 2(NH 4) 2 SO 4 + 2H 2 O
19) Cu + 4HNO 3(conc.) = Cu(NO 3) 2 + 2NO + 2H 2 O
Cu(NO 3) 2 + 2NH 3 H 2 O = Cu(OH) 2 ↓ + 2NH 4 NO 3
Cu(OH) 2 + 4NH 3 H 2 O = (OH) 2 + 4H 2 O
(OH) 2 + 6HCl = CuCl 2 + 4NH 4 Cl + 2H 2 O
20) Fe + 2HCl = FeCl 2 + H 2
CuO + H 2 = Cu + H 2 O
Cu + 4HNO 3(conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
2Cu(NO 3) 2 + 2H 2 O 2Cu + O 2 + 4HNO 3
21) I 2 + 10HNO 3 = 2HIO 3 + 10NO 2 + 4H 2 O
4NO 2 + 2H 2 O + O 2 = 4HNO 3
Cu(OH) 2 + 2HNO 3 Cu(NO 3) 2 + 2H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
22) Cu 2 O + 3H 2 SO 4 = 2CuSO 4 + SO 2 + 3H 2 O
СuSO 4 + 2KOH = Cu(OH) 2 + K 2 SO 4
Cu(OH) 2 CuO + H 2 O
3CuO + 2NH 3 3Cu + N 2 + 3H 2 O
23) CuO + H 2 SO 4 = CuSO 4 + H 2 O
4NO 2 + O 2 + 2H 2 O = 4HNO 3
10HNO3 + 4Mg = 4Mg(NO3)2 + NH4NO3 + 3H2O
24) CuO + CO Cu + CO 2
Cu + Cl 2 = CuCl 2
2CuCl 2 + 2KI = 2CuCl↓ + I 2 + 2KCl
CuCl 2 + 2AgNO 3 = 2AgCl↓ + Cu(NO 3) 2
25) 2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + H 2 SO 4 = CuSO 4 + H 2 O
2CuSO 4 + 2H 2 O 2Cu + O 2 + 2H 2 SO 4
Cu + 4HNO 3(conc.) = Cu(NO 3) 2 + 2NO 2 + 2H 2 O
26) H 2 C 2 O 4 CO + CO 2 + H 2 O
CO 2 + Ca(OH) 2 = CaCO 3 + H 2 O
2Cu(NO 3) 2 2CuO + 4NO 2 + O 2
CuO + CO Cu + CO 2
27) Cu + 2H 2 SO 4 (conc.) = CuSO 4 + SO 2 + 2H 2 O
SO 2 + 2KOH = K 2 SO 3 + H 2 O
СuSO 4 + 2NaOH = Cu(OH) 2 + Na 2 SO 4
Cu(OH) 2 + 2HCl CuCl 2 + 2H 2 O
Manganese. Manganese compounds.
I. Manganese.
In air, manganese is covered with an oxide film, which protects it from further oxidation even when heated, but in a finely crushed state (powder) it oxidizes quite easily. Manganese interacts with sulfur, halogens, nitrogen, phosphorus, carbon, silicon, boron, forming compounds with degree +2:
3Mn + 2P = Mn 3 P 2
3Mn + N 2 = Mn 3 N 2
Mn + Cl 2 = MnCl 2
2Mn + Si = Mn 2 Si
When reacting with oxygen, manganese forms manganese (IV) oxide:
Mn + O 2 = MnO 2
4Mn + 3O 2 = 2Mn 2 O 3
2Mn + O 2 = 2MnO
When heated, manganese reacts with water:
Mn+ 2H 2 O (steam) Mn(OH) 2 + H 2
In the electrochemical voltage series, manganese is located before hydrogen, therefore it easily dissolves in acids, forming manganese (II) salts:
Mn + H 2 SO 4 = MnSO 4 + H 2
Mn + 2HCl = MnCl 2 + H 2
Manganese reacts with concentrated sulfuric acid when heated:
Mn + 2H 2 SO 4 (conc.) MnSO 4 + SO 2 + 2H 2 O
With nitric acid under normal conditions:
Mn + 4HNO 3 (conc.) = Mn(NO 3) 2 + 2NO 2 + 2H 2 O
3Mn + 8HNO 3 (dil..) = 3Mn(NO 3) 2 + 2NO + 4H 2 O
Alkali solutions have practically no effect on manganese, but it reacts with alkaline melts of oxidizing agents, forming manganates (VI)
Mn + KClO 3 + 2KOH K 2 MnO 4 + KCl + H 2 O
Manganese can reduce the oxides of many metals.
3Mn + Fe 2 O 3 = 3MnO + 2Fe
5Mn + Nb 2 O 5 = 5MnO + 2Nb
II. Manganese compounds (II, IV, VII)
1) Oxides.
Manganese forms a number of oxides, the acid-base properties of which depend on the degree of oxidation of manganese.
Mn +2 O Mn +4 O2Mn2 +7 O 7
basic amphoteric acidic
Manganese(II) oxide
Manganese (II) oxide is obtained by reducing other manganese oxides with hydrogen or carbon (II) monoxide:
MnO 2 + H 2 MnO + H 2 O
MnO 2 + CO MnO + CO 2
The main properties of manganese (II) oxide are manifested in their interaction with acids and acid oxides:
MnO + 2HCl = MnCl 2 + H 2 O
MnO + SiO 2 = MnSiO 3
MnO + N 2 O 5 = Mn(NO 3) 2
MnO + H 2 = Mn + H 2 O
3MnO + 2Al = 2Mn + Al 2 O 3
2MnO + O 2 = 2MnO 2
3MnO + 2KClO 3 + 6KOH = 3K 2 MnO 4 + 2KCl + 3H 2 O
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Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.
All the necessary theory. Quick solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.
The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.
Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. A basis for solving complex problems of Part 2 of the Unified State Exam.
