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.
Carbon atomic number ⇒ 6
Carbon mass number ⇒ 12.
Carbon atomic number - Carbon mass number = number of neutrons.
12 - 6 = 6 neutrons.
Proton charge ⇒ +1
The total charge of the nucleus of a carbon atom ⇒⇒⇒ +6.
So the naswer is (3) +6
The moles of oxygen gas (O2) that is needed is 4 moles
Explanation
2H2 +O2 → 2H2O
The moles of O2 is determined using the mole ratio of H2:O2
that is from equation above H2:O2 is 2:1
If the moles of H2 is 8 moles therefore the moles of O2
= 8 moles x 1/2 = 4 moles
Scientists use scientific notation to simplify numbers, basically. When dealing with really big numbers or really small numbers, the usage of scientific notation prevents them from having to write a bunch of zeroes.
Hope that helped you!
