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.
The property of potential energy that distinguishes it from kinetic energy are Shape and position
The equilibrium constant expression for KSP of Sr3(PO4)2 is
KSP={(Sr^2+)^3 (PO4^3-)^2/ Sr3(PO4)2}
Explanation
write the ionic equation for Sr3(PO4)2
Sr3(PO4)2 → 3Sr^2+ + 2 PO4^3-
KSP is given by (concentration of the products raised to their coefficient /concentration of reactants raised to their coefficient)
Answer:
P=12.16 atm
Explanation:
Using the formula of ideal gas law:
PV = nRT
P= nRT/V
n= number of moles
R= Avogadro constant = 0.0821
T= Temperature in K => ºC + 273.15 K
P= (1.50 moles)(0.0821)( 296.15 K)/ 3.00L
P= 12.15
