Question

Проставьте степень окисления с обьяснениями только.1.NaCrO22.Cr2(SO4)33.HBrO4.HBrO35.BrF36.Na2MnO47.KMnO48.Na2SeO49.Na2SeO310.O211.OF212.CH413.H2O14.Mg3N215.(PO4)-316.(P2O7)-417.Al2Se3

Answer 1

1.Na(+)Cr(+3)O(-2)2
2.Cr(+3)2(S(+2)O(-2)4)3
3.H(+)Br(+)O(-2)
4.H(+)Br(+5)O(-2)3
5.Br(+3)F(-)3
6.Na(+)2Mn(+6)O(-2)4
7.K(+)Mn(+7)O(-2)4
8.Na(+)2Se(+6)O(-2)4
9.Na(+)2Se(+4)O(-2)3
10.O(0)2
11.O(-2)F(+)2
12.C(-4)H(+)4
13.H(+)2O(-2)
14.Mg(+2)3N(-3)2
15.(P(+5)O(-2)4)-3
16.(P(+5)2O(-2)7)-4
17.Al(+3)2Se(-2)3

Answer 2

1. NaCrO2

The compound NaCrO2 is sodium chromate (IV). The oxidation state of sodium (Na) is +1, and the oxidation state of oxygen (O) is -2. To determine the oxidation state of chromium (Cr), we can set up an equation:

(+1) + x + 2(-2) = 0

Simplifying the equation, we get:

+1 + x - 4 = 0

Solving for x, we find that the oxidation state of chromium in NaCrO2 is +3 [[1]].

2. Cr2(SO4)3

The compound Cr2(SO4)3 is chromium (III) sulfate. The oxidation state of sulfur (S) in sulfate is +6, and the overall charge of the sulfate ion is -2. To determine the oxidation state of chromium (Cr), we can set up an equation:

2x + 3(6) = 0

Simplifying the equation, we get:

2x + 18 = 0

Solving for x, we find that the oxidation state of chromium in Cr2(SO4)3 is +3 [[2]].

3. HBrO4

The compound HBrO4 is perbromic acid. The oxidation state of hydrogen (H) is +1, and the oxidation state of oxygen (O) is -2. To determine the oxidation state of bromine (Br), we can set up an equation:

(+1) + x + 4(-2) = 0

Simplifying the equation, we get:

+1 + x - 8 = 0

Solving for x, we find that the oxidation state of bromine in HBrO4 is +7 [[3]].

4. HBrO3

The compound HBrO3 is bromic acid. The oxidation state of hydrogen (H) is +1, and the oxidation state of oxygen (O) is -2. To determine the oxidation state of bromine (Br), we can set up an equation:

(+1) + x + 3(-2) = 0

Simplifying the equation, we get:

+1 + x - 6 = 0

Solving for x, we find that the oxidation state of bromine in HBrO3 is +5 [[4]].

5. BrF3

The compound BrF3 is bromine trifluoride. The oxidation state of fluorine (F) is -1. Since the overall charge of the compound is 0, the sum of the oxidation states must be 0. Let's assume the oxidation state of bromine (Br) is x. We can set up an equation:

x + 3(-1) = 0

Simplifying the equation, we get:

x - 3 = 0

Solving for x, we find that the oxidation state of bromine in BrF3 is +3 [[5]].

6. Na2MnO4

The compound Na2MnO4 is sodium permanganate. The oxidation state of sodium (Na) is +1, and the oxidation state of oxygen (O) is -2. To determine the oxidation state of manganese (Mn), we can set up an equation:

2(+1) + x + 4(-2) = 0

Simplifying the equation, we get:

2 + x - 8 = 0

Solving for x, we find that the oxidation state of manganese in Na2MnO4 is +6 [[6]].

7. KMnO4

The compound KMnO4 is potassium permanganate. The oxidation state of potassium (K) is +1, and the oxidation state of oxygen (O) is -2. To determine the oxidation state of manganese (Mn), we can set up an equation:

(+1) + x + 4(-2) = 0

Simplifying the equation, we get:

+1 + x - 8 = 0

Solving for x, we find that the oxidation state of manganese in KMnO4 is +7 [[7]].

8. Na2SeO4

The compound Na2SeO4 is sodium selenate. The oxidation state of sodium (Na) is +1, and the oxidation state of oxygen (O) is -2. To determine the oxidation state of selenium (Se), we can set up an equation:

2(+1) + x + 4(-2) = 0

Simplifying the equation, we get:

2 + x - 8 = 0

Solving for x, we find that the oxidation state of selenium in Na2SeO4 is +6 [[8]].

9. Na2SeO3

The compound Na2SeO3 is sodium selenite. The oxidation state of sodium (Na) is +1, and the oxidation state of oxygen (O) is -2. To determine the oxidation state of selenium (Se), we can set up an equation:

2(+1) + x + 3(-2) = 0

Simplifying the equation, we get:

2 + x - 6 = 0

Solving for x, we find that the oxidation state of selenium in Na2SeO3 is +4 [[9]].

10. O2

Oxygen gas (O2) is a diatomic molecule, and each oxygen atom has an oxidation state of 0. Since there is no change in oxidation state, the oxidation state of oxygen in O2 is 0.

11. OF2

The compound OF2 is oxygen difluoride. The oxidation state of fluorine (F) is -1. Since the overall charge of the compound is 0, the sum of the oxidation states must be 0. Let's assume the oxidation state of oxygen (O) is x. We can set up an equation:

x + 2(-1) = 0

Simplifying the equation, we get:

x - 2 = 0

Solving for x, we find that the oxidation state of oxygen in OF2 is +2 [[10]].

12. CH4

The compound CH4 is methane. The oxidation state of hydrogen (H) is +1. Since the overall charge of the compound is 0, the sum of the oxidation states must be 0. Let's assume the oxidation state of carbon (C) is x. We can set up an equation:

x + 4(+1) = 0

Simplifying the equation, we get:

x + 4 = 0

Solving for x, we find that the oxidation state of carbon in CH4 is -4 [[11]].

13. H2O

The compound H2O is water. The oxidation state of hydrogen (H) is +1, and the oxidation state of oxygen (O) is -2. Since the overall charge of the compound is 0, the sum of the oxidation states must be 0. Let's assume the oxidation state of oxygen is x. We can set up an equation:

2(+1) + x = 0

Simplifying the equation, we get:

2 + x = 0

Solving for x, we find that the oxidation state of oxygen in H2O is -2 [[12]].

14. Mg3N2

The compound Mg3N2 is magnesium nitride. The oxidation state of magnesium (Mg) is +2. Since the overall charge of the compound is 0, the sum of the oxidation states must be 0. Let's assume the oxidation state of nitrogen (N) is x. We can set up an equation:

3(+2) + 2x = 0

Simplifying the equation, we get:

6 + 2x = 0

Solving for x, we find that the oxidation state of nitrogen in Mg3N2 is -3 [[13]].

15. (PO4)-3

The polyatomic ion (PO4)-3 is phosphate. The overall charge of the ion is -3. To determine the oxidation state of phosphorus (P), we can set up an equation:

x + 4(-2) = -3

Simplifying the equation, we get:

x - 8 = -3

Solving for x, we find that the oxidation state of phosphorus in (PO4)-3 is +5 [[14]].

16. (P2O7)-4

The polyatomic ion (P2O7)-4 is diphosphate. The overall charge of the ion is -4. To determine the oxidation state of phosphorus (P), we can set up an equation:

2x + 7(-2) = -4

Simplifying the equation, we get:

2x - 14 = -4

Solving for x, we find that the oxidation state of phosphorus in (P2O7)-4 is +5 [[15]].

17. Al2Se3

The compound Al2Se3 is aluminum selenide. The oxidation state of aluminum (Al) is +3. Since the overall charge of the compound is 0, the sum of the oxidation states must be 0. Let's assume the oxidation state of selenium (Se) is x. We can set up an equation:

2(+3) + 3x = 0

Simplifying the equation, we get:

6 + 3x = 0

Solving for x, we find that the oxidation state of selenium in Al2Se3 is -2 [[