Answer:
Option c: Possible electron energy states are quantized within an atom.
Explanation:
The Bohr's Model of the hydrogen atom consisted of the movements of the electrons around the positively-charged nucleus in circular orbits that have a certain energy state. The energy of that orbit is given by:
<em>Where:</em>
E(n): is the energy of an electron in a particular orbit
R: is the Rydberg constant
h: is the Plank constant
c: is the speed of light
n: is a positive integer which corresponds to the number of the orbit
The ground state energy of a electron in the hydrogen atom is equal to -13,6 eV.
Bohr's Model aims to propose that the electron is restrictedly to occupy a certain region in the atom.
Therefore, the conclusion of Bohr after observing emission spectrum lines is that "possible electron energy states are quantized within an atom", so the correct option is c.
I hope it helps you!
Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06
When, it donates electons.
as for example take
NaCl ( sodium chloride)
it's an ionic compund,
that means it is formed by donating or gaining electrons
Na is writen first than, it must be electropositive i.e it has donated electons which made it positive and the clorine gains electron so it's electronegative.
Na is positive because
as we know it's atomic number is 11 that means it has 11 protons and 11 electrons
now, when it donate electon it has, greater number of protons whose change is +ve so the atom becomes overall positively charged ion or cation.
and something same happens in clorine and because it gains one electron and the number of electrons increase in it by 1 whise charge is -ve so, the atom becomes negatively charged ion or anion which has a -1 charge.
