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
It will be the Displacement method
the first is hydroxide. I believe d is the second one but I'm not 100% positive so you may want to get a second opinion
C(5) + O2(g)
Hope this helped
Answer:
I. Increasing pressure will allow more frequent successful collision between particles due to the particles being closer together.
II. Rate of reaction increases due to more products being made; as increased pressure favours the exothermic side of the equilibrium.
III. Increasing temperature provides particles lots of (Kinetic) energy, for more frequent successful collision due to the particles moving at a faster rate than before. However, favouring the endothermic side of the equilibrium due to lots of energy required to break and form new bonds.
IV. Rate of reaction increases due to increase temperature favouring both directions of the equilibrium - causing products to form faster.
Hope this helps!
