Answer:
it has a huge electronegativity difference between it's atoms
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
You start by finding the mol of each
59.9g C x (mol C / 12.01 g C) = 4.98 mol C
8.06g H x (mol H / 1.00 g H) = 8.06 mol H
32.0 g O x (mol O / 16.0 g O) = 2 mol O
So when you set it up you have
C4.98 H8 O2
You divide each by the smallest mol. The smallest mol is 2
C2 H4 O2.5
However you can’t have half a mol in the empirical formula. If the value ends in 0.5, you multiply everything by 2
You’re left with
C2H8O5
The EMPIRICAL formula for lucite is C2H8O5
Note empirical is not the same as chemical formula.
Answer:
The correct answer is: d-orbital
Explanation:
The transition metals are the chemical elements that belong to the B-group of the periodic table, from group 1B to 8B. The B-group is located between the groups IIA and IIIA in the periodic table.
The general electronic configuration of the chemical elements belonging to this group is (n-1) d¹⁻¹⁰ ns⁰⁻², because of the <u>partially filled d-subshell</u> in the ground state or excited state. <u>Therefore, the transition metals are refereed as the </u><u>d-block elements</u><u>.</u>
