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AlladinOne [14]
3 years ago
5

What is the oxidation state of each element in the species Mn(ClO4)3?

Chemistry
2 answers:
gayaneshka [121]3 years ago
7 0
The oxidation state of the compound Mn (ClO4)3 is to be determined in this problem. For oxygen, the charge is 2-, the total considering its number of atoms is -24. Mn has a charge of +3. TO compute for Mn, we must achieve zero charge overall hence 3+3x-24=0 where x is the Cl charge. Cl charge, x is +7. 
Ronch [10]3 years ago
4 0

Answer:

Mn: 3+

Cl: 7+

O: 2-

Explanation:

1) Compound given: Mn [ClO₄]₃

2) Initially you only know the oxidation state of O, since it is always 2-, except when it form peroxides, which is not the case.

3) So, you do not know the oxidation states neither of the Mn nor of the Cl, and you need some more information.

You might start from the ion [ClO₄] but you do not know its charge.

This ion comes from one of the oxoacids formed by Cl. Those are four different acids. These are them:

i) Oxidation state 1+: Cl₂O + H₂O → H₂Cl₂O₂ = HClO ⇒ ion ClO⁻

ii) Oxidation state 3+: Cl₂O₃ + H₂O → H₂Cl₂O₄ = HClO₂ ⇒ ion ClO₂⁻

iii) Oxidation state 5+: Cl₂O5 + H₂O → H₂Cl₂O₆ = HClO₃ ⇒ ion ClO₃⁻

iv) Oxidation state 7+: Cl₂O₇ + H₂O → H₂Cl₂O₈ = HClO₄ ⇒ ion ClO₄⁻

Finally, we have that our ion is ClO₄⁻ and the oxidation state of Cl is 7+.

4) Now you just have to find the oxidation state of Mn, for which you make a balance of charges:

Mn [ClO₄]₃

Since, the ion ClO₄⁻ has 1 negative charge, and there are 3 ions the total negative charge is 3-. Since the compound is neutral, you conclude that Mn has oxidation state 3+.

That according to this balance: 1(3+) + 3(1-) + 3 - 3 = 0.

5) Summarizing, the oxidation states are:

Mn: 3+

Cl: 7+

O: 2-



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What NaCl NaCl concentration results when 249 mL 249 mL of a 0.850 M 0.850 M NaCl NaCl solution is mixed with 667 mL 667 mL of a
tamaranim1 [39]

Answer:

The answer to your question is 0.54M

Explanation:

Data

Final concentration = ?

Concentration 1 = 0.850 M

Volume 1 = 249 ml = 0.249 l

Concentration 2 = 0.420 M

Volume 2 = 0.667 M

Process

1.- Calculate the number of moles in both solutions

Number of moles 1 = Molarity 1 x Volume 1

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Number of moles 2 = Molarity 2 x Volume 2

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                                = 0.280

Total number of moles =  0.212 + 0.280

                                      = 0.492

2.-Calculate the final volume

Final volume = Volume 1 + Volume 2

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