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2 years ago

Determine The Oxidation Number for

Chemistry

1 answer:

bixtya [17]2 years ago

7 0

Answer:

A is +4

B is +4

C is +6

D is +3

E is +4

Explanation:

In order to learn how to find the oxidation number of an atom in a given compound, it is important to learn what oxidation numbers are. The oxidation number of an atom is a number that represents the total number of electrons lost or gained by it.

Calculating Oxidation Numbers

An oxidation number can be assigned to a given element or compound by following the following rules.

Any free element has an oxidation number equal to zero.

For monoatomic ions, the oxidation number always has the same value as the net charge corresponding to the ion.

The hydrogen atom (H) exhibits an oxidation state of +1. However, when bonded with an element with less electronegativity than it, it exhibits an oxidation number of -1.

Oxygen has an oxidation of -2 in most of its compounds. However, in the case of peroxides, the oxidation number corresponding to oxygen is -1.

All alkali metals (group 1 elements) have an oxidation state of +1 in their compounds.

All alkaline earth metals (group 2 elements) exhibit an oxidation state of +2 in their compounds.

In the compounds made up of two elements, a halogen (group 17 elements) have an oxidation number of -1 assigned to them.

In the case of neutral compounds, the sum of all the oxidation numbers of the constituent atoms totals to zero.

When polyatomic ions are considered, the sum of all the oxidation numbers of the atoms that constitute them equals the net charge of the polyatomic ion.

Thus, the oxidation number of an atom in a given compound can be calculated with the steps mentioned above.

How to Calculate Oxidation Number

Hydrochloric Acid (HCl)

As per the rules discussed above, the oxidation state of a group 17 element (halogen) in a diatomic molecule is -1. It is also discussed that hydrogen always exhibits an oxidation number of +1 unless it is paired with a less electronegative element.

Since chlorine is more electronegative than hydrogen, an oxidation number of +1 can be assigned to the hydrogen atom in HCl.

Therefore, the oxidation number of hydrogen is +1 and the oxidation of chlorine is -1 in HCl. These values can be verified by adding these oxidation numbers. Since the total is zero, which is the value of the oxidation number corresponding to a neutral molecule, the values are verified.

Carbon Dioxide (CO2)

According to the rules to calculate oxidation number, which can be found in the previous subsection, the oxidation number of oxygen in its compounds (excluding peroxides) is -2.

Since there are two oxygen atoms in carbon dioxide, the total of the oxidation numbers corresponding to each oxygen is -4.

Since the CO2 molecule is neutral, the carbon atom must exhibit an oxidation state of +4 (the sum of all the oxidation numbers in a neutral molecule is zero).

Therefore, the oxidation state of oxygen was found to be -2 and the oxidation number of carbon is +4 in a carbon dioxide molecule.

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85 POINTS PLEASE HELP ASAP

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1. Answer is: there are 1.41·10²³ molecules of oxygen.
1) calculate amount of substance for oxygen gas:
m(O₂) = 7.5 g; mass of oxygen.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 7.5 g ÷ 32 g/mol.
n(O₂) = 0.234 mol.
2) calculate number of molecules:
N(O₂) = n(O₂) · Na.
N(O₂) = 0.234 mol · 6.022·10²³ 1/mol.
N(O₂) = 1.41·10²³.
Na - Avogadro constant.

2. Answer is: the percent yield for the reaction is 61.77%.
Balanced chemical reaction: 2HgO(s) → 2Hg(l) + O₂(g).
m(HgO) = 4.37 g.
n(HgO) = n(HgO) ÷ M(HgO).
n(HgO) = 4.37 g ÷ 216.6 g/mol.
n(HgO) = 0.02 mol.
From chemical reaction: n(HgO) : n(Hg) = 2 : 2 (1 :1).
n(Hg) = n(HgO) = 0.02 mol; amount of substance.
m(Hg) = n(Hg) ·M(Hg).
m(Hg) = 0.02 mol · 200.6 g/mol.
m(Hg) = 4.047 g.
yield = 2.5 g ÷ 4.047 g · 100%.
yield = 61.77%.

3. Answer is: 68.16 grams of the excess reactant (oxygen) remain.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 79.9 g.
n(O₂) = 79.9 g ÷ 32 g/mol.
n(O₂) = 2.497 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
0.489 mol : n(O₂) = 4 : 3.
n(O₂) = 3 · 0.489 mol ÷ 4.
n(O₂) = 0.367 mol.
Δn(O₂) = 2.497 mol - 0.367 mol.
Δn(O₂) = 2.13 mol.
m(O₂) = 2.13 mol · 32 g/mol.
m(O₂) = 68.16 g.

4. Answer is: there are 0.603 moles of ammonia.
m(NH₃) = 10.25 g; mass of ammonia.
M(NH₃) = Ar(N) + 3Ar(H) · g/mol.
M(NH₃) = 14 + 3·1 · g/mol.
M(NH₃) = 17 g/mol; molar mass of ammonia.
n(NH₃) = m(NH₃) ÷ M(NH₃).
n(NH₃) = 10.25 g ÷ 17 g/mol.
n(NH₃) = 0.603 mol; amount of substance (ammonia).

5. Answer is: the empirical formula mass of P₂O₅ is 141.89.
Empirical formula gives the proportions of the elements present in a compound.
Atomic mass of phosphorus is 30.97 g/mol.
Atomic mass of oxygen is 15.99 g/mol.
In phosphorus (V) oxide there are two atoms of phosphorus and five atoms of oxygen:
EFM(P₂O₅) = 2·30.97 g/mol + 5·15.99 g/mol = 141.89 g/mol.

6. Answer is: there are 1.108·10²⁴ molecules of water.
n(H₂O) = 1.84 mol; amount of substance (water).
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 1.84 mol · 6.022·10²³ 1/mol.
N(H₂O) = 11.08·10²³.
N(H₂O) = 1.108·10²⁴.
Na - Avogadro constant (number of particles (ions, atoms or molecules), that are contained in one mole of substance).

7. Answer is: iron (Fe) is the limiting reactant.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 45.8 g.
n(O₂) = 45.8 g ÷ 32 g/mol.
n(O₂) = 1.431 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
For 1.431moles of oxygen we need:
1.431 mol : n(Fe) = 3 : 4.
n(Fe) = 1.908 mol, there is no enough iron.

8. Answer is: there are 0.435 moles of C₆H₁₄.
N(C₆H₁₄) = 2.62·10²³; number of molecules.
n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.
n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.
n(C₆H₁₄) = 0.435 mol; amount of substance of C₆H₁₄.
Na - Avogadro constant or Avogadro number.

9. Answer is: 3.675 moles of carbon(II) oxide are required to completely react.
Balanced chemical reaction: Fe₂O₃(s) + 3CO(g) ⟶ 2Fe(s) + 3CO₂(g).
n(Fe₂O₃) = 1.225 mol; amount of substance.
From chemical reaction: n(Fe₂O₃) : n(CO) = 1 : 3.
1.225 mol : n(CO) = 1 : 3.
n(CO) = 3 · 1.225 mol.
n(CO) = 3.675 mol.

10. Answer is: there are 2.158 moles of barium atoms.
N(Ba) = 2.62·10²³; number of atoms of barium.
n(Ba) = N(Ba) ÷ Na.
n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.
n(Ba) = 2.158 mol; amount of substance of barium.
Na - Avogadro constant or Avogadro number.

11. Answer is: 6.26·10²³ carbon atoms are present.
n(C₂H₆O) = 0.52 mol; amount of substance.
N(C₂H₆O) = n(C₂H₆O) · Na.
N(C₂H₆O) = 0.52 mol · 6.022·10²³ 1/mol.
N(C₂H₆O) = 3.13·10²³.
In one molecule of C₂H₆O there are two atoms of carbon:
N(C) = N(C₂H₆O) · 2.
N(C) = 3.13·10²³ · 2.
N(C) = 6.26·10²³.

12. Answer is: the empirical formula is C₂H₄O.
If we use 100 grams of compound:
1) ω(C) = 51% ÷ 100% = 0.51.
m(C) = ω(C) · m(compound).
m(C) = 0.51 · 100 g.
m(C) = 51 g.
n(C) = m(C) ÷ M(C).
n(C) = 51 g ÷ 12 g/mol.
n(C) = 4.25 mol.
2) ω(H) = 9.3 % ÷ 100% = 0.093.
m(H) = 0.093 · 100 g.
m(H) = 9.3 g.
n(H) = 9.3 g ÷ 1 g/mol.
n(H) = 9.3 mol
3) ω(O) = 39.2 % ÷ 100%.
ω(O) = 0.392.
m(O) = 0.392 · 100 g.
m(O) = 39.2 g.
n(O) = 39.2 g ÷ 16 g/mol.
n(O) = 2.45 mol.
4) n(C) : n(H) : n(O) = 4.25 mol : 9.3 mol : 2.45 mol /2.45 mol.
n(C) : n(H) : n(O) = 1.73 : 3.795 : 1.

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