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.
PH is a measure of hydrogen ion concentration, a measure of the acidity or alkalinity of a solution.
We are given the starting amount of Al metal to be used. This will be the starting point of the calculations. We do as follows:
74.00 g ( 1 mol / 26.98 g ) ( 1 mol Al2O3 / 2 mol Al ) ( 101.96 g / 1 mol ) = 139.83 g Al2O3 produced
Hope this answers the question. Have a nice day.
Answer:
residue is whatever remains after something else has been removed while filtrate is the liquid or solution that has passed through a filter, and which has been separated from the filtride.
filtration id done by placing a filter paper on the beaker or container then pour the filtride then let it settle and it will pass through and you will have the fitrate and residue
Explanation:
