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AURORKA [14]
3 years ago
11

Identify the oxidation state of Ba 2 + . +2 Identify the oxidation state of S in SO 2 . +2 Identify the oxidation state of S in

SO 2 − 4 . +2 Identify the oxidation state of Zn in ZnSO4 .
Chemistry
1 answer:
cupoosta [38]3 years ago
3 0

Answer:

The oxidation state of Ba in cation Ba²⁺ is +2

The oxidation state of S in SO₂, is +4

The oxidation state of S in anion sulfate (SO₄⁻²) is +6

The oxidation state of Zn in the Zinc sulfate, is +2

Explanation:

We define oxidation state as the number which can be negative or positive that  

indicates the number of electrons that the atom has accepted or transferred.

All the elements in ground state has 0 as oxidation state.

This numbers are very important for redox reaction which are balanced by the ion electron method.

When the elements gain electrons, the element is being reduced so the oxidation state decreases.

When the elements release electrons, the element is oxidized so the oxidation state increases.

We have to think, that global charge of a compound is 0, for example in the ZnSO₄.

The sulfate anion has a global charge of -2 because it has released 2 protons, it came from the sulfuric acid (H₂SO₄). As the global charge is -2, oxygen acts with -2, and the anion has 4 atoms so the global charge of O is -8. Definetly S, has +6 as oxidation state.

In the SO₂, oxygen acts with -2 and there are 2 atoms in the compound, so the global charge is 0 and the global charge for O  is -4. Therefore S must act with +4.

Ba²⁺ is an element of group 2 and has a tendency to form a cation, so it can release electrons for that purpose.  At least, it can release 2 e⁻, that's why the oxidation state is +2. It can complete the octet rule and it will be isoelectronic with Xe.

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Explanation:

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           2NH_{4}NO_{3} \rightarrow 2N_{2} + O_{2} + 4H_{2}O

It is given that mass of ammonium nitrate is 86.0 kg.

As 1 kg = 1000 g. So, 86.0 kg = 86000 g.

Hence, moles of NH_{4}NO_{3} present will be as follows.

      Moles of NH_{4}NO_{3} = \frac{mass given}{\text{molar mass of NH_{4}NO_{3}}}

                                  = \frac{86000 g}{80.043 g/mol}

                                  = 1074.42 mol

Therefore, moles of N_{2}, O_{2} and H_{2}O produced by 1074.42 mole of NH_{4}NO_{3} will be as follows.

  Moles of O_{2} = \frac{1}{2} \times 1074.42 mol

                                = 537.21 mol

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                                = 1074.42 mol

Moles of H_{2}O = \frac{4}{2} \times 1074.42 mol

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Therefore, total number of moles will be as follows.

          537.21 mol + 1074.42 mol + 2148.84 mol

        = 3760.47 mol

According to ideal gas equation, PV = nRT. Hence, calculate the volume as follows.

                       PV = nRT

                     1 atm \times V = 3760.47 mol \times 0.0821 L atm/mol K \times 580 K[/tex]         (as 307^{o}C = 307 + 273 = 580 K)

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Answer:

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