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
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
It’s soft which makes It low energy
Roughness of the surface and weight of the object.
