Answer: The pH of a 4.4 M solution of boric acid is 4.3
Explanation:
at t=0 cM 0 0
at eqm 
So dissociation constant will be:
Give c= 4.4 M and 
= ?
Putting in the values we get:
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D4.4%5Ctimes%200.000011%3D4.8%5Ctimes%2010%5E%7B-5%7DM)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[4.8\times 10^{-5}]=4.3](https://tex.z-dn.net/?f=pH%3D-log%5B4.8%5Ctimes%2010%5E%7B-5%7D%5D%3D4.3)
Thus pH of a 4.4 M 
solution is 4.3
Each column is called a group<span>. The elements in each </span>group have<span> the same number of electrons in the outer orbital. Those outer electrons are also called valence electrons.</span>
Answer:
The attractive forces must be overcome are :
Explanation:
For the compound to dissolve the attractive forces existing between atoms of the compound must be reduced
<u>CsI is ionic compound </u><em>and its molecules are held together by ionic(electrostatic) force . These force must be weakened for its dissolution</em>
Forces in HF <em>:</em>
<em>1 .Hydrogen Bonding : In HF strong intermolecular Hydrogen Bonding exist between the electronegative F and Hydrogen</em>
2. Dipole - dipole : <em>HF is polar . So it is a permanent dipole and has dipole diople interaction</em>
Lewis Structure is drawn in following steps,
1) Calculate Number of Valence Electrons: # of Valence electrons in Mg = 2
# of Valence electrons in I = 7
# of Valence electrons in I = 7
---------
Total Valence electrons = 16
2) Draw Mg as a central atom surround it by two atoms of Iodine.3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.
Total Valence electrons 16
- Four electrons - 4
----------
12
4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.
The final lewis structure formed is as follow,
