The atoms that would be expected to be diamagnetic in the ground state is magnesium
The magnetism of an atom refers to its electronic configuration. A diamagnetic atom is an atom whose electrons are all paired.
A paired electron is an electron that occurs in pairs in its orbital shell.
At their respective ground state, the electronic configuration of the given elements are as follows:
The electronic configuration of magnesium is 1s²2s²2p⁶3s². As such its a diamagnetic atom.
The electronic configuration of Potassium is 1s²2s²2p⁶3s²3p⁶4s¹. Hence, Potassium has one unpaired electron in its outermost shell.
The electronic configuration of Chlorine is 1s²2s²2p⁶3s²3p⁵. Hence, Chlorine has one unpaired electron in its outermost shell.
The electronic configuration of Cobalt is 1s²2s²2p⁶3s²3p⁶3d⁷4s². Hence, the unpaired electrons of Cobalt in its outermost shell are three.
Therefore, the atoms that are diamagnetic in the ground state is magnesium.
Learn more about diamagnetic atoms here:
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Answer:
Global warming stresses ecosystems through temperature rises, water shortages, increased fire threats, drought, weed and pest invasions, intense storm damage and salt invasion, just to name a few.
Explanation:
Global warming stresses ecosystems through temperature rises, water shortages, increased fire threats, drought, weed and pest invasions, intense storm damage and salt invasion, just to name a few.
Answer:
Calcium
Explanation:
The symbol for the element with a mass number of 27 is actually Al for Aluminum instead of Co for Cobalt. Mass number refers to atomic mass, not atomic number.
Aluminum has an atomic mass of 26.982 or 27.
The letters to unscramble are I, M, C, Al, U, and C.
The mystery element is Calcium.
Hope that helps.
So we have: 1 C2H5OH + _ O2 -> 2 CO2 + _ H2O Step 3: The only molecule on the right that is left is H2O. We start with 6 hydrogen atoms<span> on the left.</span>
Answer:
The mean free path = 2.16*10^-6 m
Explanation:
<u>Given:</u>
Pressure of gas P = 100 kPa
Temperature T = 300 K
collision cross section, σ = 2.0*10^-20 m2
Boltzmann constant, k = 1.38*10^-23 J/K
<u>To determine:</u>
The mean free path, λ
<u>Calculation:</u>
The mean free path is related to the collision cross section by the following equation:
where n = number density
Substituting for P, k and T in equation (2) gives:
Next, substituting for n and σ in equation (1) gives:
