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konstantin123 [22]
2 years ago
12

In the following equation, ______ is being oxidized and ______ is being reduced.

Chemistry
1 answer:
Mashcka [7]2 years ago
6 0

oxidation \: number \: of \: oxygen =  \\ before \: rxn =  - 2 \\ after \: rxn =  - 2

oxidation \: number \: of \: hydrogen = \\ before \: rxn =  + 1 \\ after \: rxn =  \\ 2x - 2 = 0 \\ x =  + 1

oxidation \: number \: of \: carbon =  \\ before \: rxn =  \\ x  - 6 =  - 2 \\ x = 4 \\ after \: rxn =  \\ x - 4 = 0 \\ x = 4

<h2>Option A</h2>

oxidation \: numbers \: remain \: constant \\ so \: none \:a re \: undergoing \: oxidation \: \\ nor \: reduction \:

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Answer:

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8 0
2 years ago
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Explanation:

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Read 2 more answers
The force of attraction between a divalent cation and a divalent anion is 1.64 x 10-8 N. If the ionic radius of the cation is 0.
Elden [556K]

The radius of the anion is 7.413 nm

<h3>How to calculate the force of attraction between charges</h3>

The force of attraction (F) is given by the formula:

  • F = (1/4π∈r²)(Zc*e)(Za*e)

where:

∈ = permittivity of free space = 8.85*10⁻¹⁵ F/m

Zc = charge on the cation = +2

Zc = charge on the anion = -2

e = charge on an electron = 1.602 * 10⁻¹⁹ C

r = interionic distance

r = rc + ra

where rc and ra are the radius of the cation and anion respectively

F = 1.64 * 10⁻⁸ N

Therefore based on the equation of force of attraction:

1.64 *10⁻⁸ = [1/4π(8.85*10⁻¹⁵)r²](2 * 1.602*10⁻¹⁹)²

r² = 5.63 * 10⁻¹⁷

r = 7.50 nm

Since r = rc + ra

where rc = 0.087 nm

thus, ra = r - rc = 7.50 - 0.087

ra = 7.413 nm

Therefore, the radius of the anion is 7.413 nm

Learn more about ionic radius at: brainly.com/question/2279609

6 0
2 years ago
Which set of these comparisons is incorrect? (more stable means it has a more negative energy.)
Lady bird [3.3K]
I think your anwser ahould be b!
4 0
3 years ago
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