The question is incorrect, the correct question is;
Which of the following ground-state electron configurations represents the atom that has the
lowest first-ionization energy?
a) 1s2
b) 1s22s2
c) 1s22s22p6
d) 1s22s22p63s23p1
e) 1s22s22p63s23p3
The correct ground state configuration that represents the atom that has the lowest first ionization energy is 1s² 2s² 2p⁶ 3s² 3p¹.
The first ionization energy is the energy required to remove an electron from the outermost shell of an atom.
Ionization energy decreases down the group as number of shells increases but increases across the period as nuclear charge increase.
As the number of shells increases, the degree of shielding or screening decreases it easier to remove the outermost electron.
The elements whose ground state electronic configurations were shown are;
Helium - 1s²
Beryllium - 1s² 2s²
Neon - 1s² 2s² sp⁶
Aluminum - 1s² 2s² 2p⁶ 3s² 3p¹
Phosphorus - 1s² 2s² 2p⁶ 3s² 3p³
Aluminium (1s² 2s² 2p⁶ 3s² 3p¹) is a metal so it has the lowest first ionization energy since metals are highly electropositive.
Learn more: brainly.com/question/17783060
Therefore, 1 mole<span> of gold weighs </span>196.9665<span> grams. So, in 2.8 grams of gold we would have:</span>
(2.8 gram)(1 mole/196.9665<span> gram) = 0.0142 mole.</span>
From Avogadro's number, we know that there are approximately 6.02 x 1023<span>atoms/mole.</span>
Answer:
20.6
Explanation:
I don't know how to show the work for it without pencil and paper but go from there, Good Luck!
The outcome of a experiment is the result
Waves depends on density and elasticity of the medium
