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
Answer:
Option-A [<span>Cl (g) + e</span>⁻ <span> → Cl</span>⁻<span> (g)] is the correct answer.
Explanation:
First electron affinity is the amount of heat evolved when an electron is added to a neutral atom.
Such problems often come in questions related to Born-Haber Cycle topic. Let suppose Mg and Cl</span>₂ is reacted to form<span> MgCl</span>₂ crystal.
Then the Cl₂ gas is first atomized as follow,
Cl₂ → 2 Cl
After atomization an electron is added to neutral chlorine atoms, as,
Cl + e⁻ → Cl⁻
So, option A is correct choice.
<u>Answer;</u>
Matter is converted to energy
<u>Explanation;</u>
- Nuclear fission is a type of nuclear reaction in which a nuclei of an atom splits or breaks down into two or more smaller nuclei.
- It involves subdivision of heavy nucleus of an atom such as Uranium into two or more fragments which has almost the same equal mass.
- Nuclear fission is accompanied by release of large amount of energy, thus we can say that matter is converted to energy.
Answer:
pH of the buffer is 10.10
Explanation:
trimethylamine is a weak base that, in presence with its conjugate base, trimethylammonium ion, produce a buffer.
To determine the pH of the buffer we use H-H equation for weak bases:
pOH = pKb + log [Conjugate acid] / [Weak base]
<em>pKb is -log Kb = 4.20</em>
<em />
pOH = 4.20 + log [N(CH₃)₃] / [NH(CH₃)₃]
Replacing the concentrations of the problem:
pOH = 4.20 + log [0.20M] / [0.40M]
pOH = 3.90
As pH = 14 -pOH
<h3>pH of the buffer is 10.10</h3>
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