Answer: -
IE 1 for X = 801
Here X is told to be in the third period.
So n = 3 for X.
For 1st ionization energy the expression is
IE1 = 13.6 x Z ^2 / n^2
Where Z =atomic number.
Thus Z =( n^2 x IE 1 / 13.6)^(1/2)
Z = ( 3^2 x 801 / 13.6 )^ (1/2)
= 23
Number of electrons = Z = 23
Nearest noble gas = Argon
Argon atomic number = 18
Number of extra electrons = 23 – 18 = 5
a) Electronic Configuration= [Ar] 3d34s2
We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.
So more the first ionization energy, less is the atomic radius.
X has more IE1 than Y.
b) So the atomic radius of X is lesser than that of Y.
c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.
Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.
Answer:
The different types of energy include thermal energy, radiant energy, chemical energy, nuclear energy, electrical energy, motion energy, sound energy, elastic energy and gravitational energy.
Explanation:
Hope this helps :D
<span>To compute 4.659×104−2.14×104, the first step is the factorization. That is as follows:4.659×104−2.14×104= 10^4.(4.659−2.14), the next step is to compute 4.659−2.14=2.51, so 10^4.(4.659−2.14)=2.51x10^4=2.51x 10000 (because10^4=10000), the last calculus is 2.51x 10000=25100, the final answer is 25,000.Hope this helps. Let me know if you need additional help!</span>
Answer:
I am pretty sure the correct answer is a reproductive system
Answer:
ZnS
Explanation:
1. Number of Zn atoms
4 internal atoms = 4 Zn atoms
2. Number of S atoms
8 corners × ⅛ S atom/corner + 6 faces × ½ S atom/face = 1 S atom + 3 S atoms = 4 S atoms
3. Empirical formula
The atomic ratio is
4Zn:4S = 1Zn:1S
The empirical formula is ZnS.
