The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as
Mg(IO3)2
MgI2O6
As we know that
Mg has +2
O has -2
So,
(+2) + 2I + 6 (-2)=0
2 + 2I - 12 =0
10+ 2I =0
10 = 2I
I =5
In order to see which species has the strongest dispersion forces, you need to calculate their molar mass, because the higher the molar mass, the stronger the dispersion forces.
Since E. C8H18 has the highest molar mass, its dispersion forces are also the strongest ones.
Answer:
The atomic mass of phosphorus is 29.864 amu.
Explanation:
Given data:
Atomic mass of phosphorus = ?
Percent abundance of P-29 = 35.5%
percent abundance of P-30 = 42.6%
Percent abundance of P-31 = 21.9%
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass / 100
Average atomic mass = (29×35.5)+(30×42.6) + (31×21.9) /100
Average atomic mass = 1029.5 + 1278 + 678.9/ 100
Average atomic mass = 2986.4 / 100
Average atomic mass = 29.864 amu.
The atomic mass of phosphorus is 29.864 amu.
The number of protons in an atom of krypton is 36.
<u>Explanation:</u>
The number written above the symbol of Kr in the periodic table represents the atomic number of the element . Then the number written in the below represents the mass number.
So the mass number of krypton is 84 and the atomic number of krypton is 36. As it is known that number of protons and electrons will be equal in a stable element. Thus, the krypton atom will be having 36 protons.
Answer:
The balanced reaction is:-

expression is:-
![K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20C_6H_5COOH%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BC_6H_5COO%5E-%5D%7D)
Explanation:
Benzoate ion is the conjugate base of the benzoic acid. It is a Bronsted-Lowry base and the dissociation of benzoate ion can be shown as:-

The expression for dissociation constant of benzoate ion is:
![K_{b}=\frac {\left [ C_6H_5COOH \right ]\left [ {OH}^- \right ]}{[C_6H_5COO^-]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20C_6H_5COOH%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BC_6H_5COO%5E-%5D%7D)