Answer:
Lithium looses one electron.
The charge -
- Cation
Nitrogen gains three electrons -
- Anion
Boron losses three electron.
The charge -
- Cation
Explanation:
Lithium looses one electron.
The charge -
- Cation
Nitrogen gains three electrons -
- Anion
Boron losses three electron.
The charge -
- Cation
The complete table is as follows.
The three sub particles of an atom are protons, electrons and neutrons.
<u>Explanation:</u>
Neutrons, electrons, and protons are the three sub atomic particles of an atom. Electrons are negatively charged, protons are the positively charged and neutrons are neutral. Electrons are having 9.1×
kg, protons and neutrons are having a mass of 1.67×
kg.
As protons possess positive charges, so the atomic number of protons will be 1 , similarly, the atomic number of electrons will be -1 and neutron is zero. So, electrons, protons and neutrons combine together to form the atom.
As the electrons present with negative charge and protons with positive charge, along with this the protons number are equal to electrons number, this confirm the neutral nature of atoms.
Answer:
At the nucleus point of a substance, the particles have enough kinetic energy to break free from their fixed positions.
Answer : nucleus
Answer:
The percent ionization is 0,16%
Explanation:
The percent ionization is defined as the number of ions that exist in a substance.
![PI=\frac{[A-]}{[HA]} x100](https://tex.z-dn.net/?f=PI%3D%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D%20x100)
First, we find the [A-] using the ka equation
HA ⇄ 
[H+] = [A-]
![Ka=\frac{[H+][A-]}{[HA]}\\ \\](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%2B%5D%5BA-%5D%7D%7B%5BHA%5D%7D%5C%5C%20%5C%5C)
since the ionization constant is very small we can assume that the final concentration of [HA] is the same
![Ka=\frac{[H+]^{2} }{[HA]} \\\\](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%2B%5D%5E%7B2%7D%20%7D%7B%5BHA%5D%7D%20%5C%5C%5C%5C)
![[H+]=\sqrt[2]{Ka.[HA]} \\\\](https://tex.z-dn.net/?f=%5BH%2B%5D%3D%5Csqrt%5B2%5D%7BKa.%5BHA%5D%7D%20%5C%5C%5C%5C)
![[H+] =\sqrt{(2,610^{-7} )(0,1)} = 1,61210^{-4}](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%5Csqrt%7B%282%2C610%5E%7B-7%7D%20%29%280%2C1%29%7D%20%20%3D%201%2C61210%5E%7B-4%7D)
Now we calculate the percent ionization using these values

PI=0,16%