Answer:
The concentrations are :
![[HAsc^-]=0.000702 M](https://tex.z-dn.net/?f=%5BHAsc%5E-%5D%3D0.000702%20M)
![[Asc^{2-}]=5.92\times 10^{-8} M](https://tex.z-dn.net/?f=%5BAsc%5E%7B2-%7D%5D%3D5.92%5Ctimes%2010%5E%7B-8%7D%20M)
The pH of the solution is 3.15.
Explanation:

Initial
c 0 0
Equilibrium
c-x x x
![K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}](https://tex.z-dn.net/?f=K_%7Ba1%7D%3D%5Cfrac%7B%5BHAs%5E-%5D%5BH%5E%2B%5D%7D%7B%5BH_2Asc%5D%7D)


Solving for x:
x = 0.000702 M
![[HAsc^-]=0.000702 M](https://tex.z-dn.net/?f=%5BHAsc%5E-%5D%3D0.000702%20M)

Initially
x 0 0
At equilibrium ;
(x - y) y y
![K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}](https://tex.z-dn.net/?f=K_%7Ba2%7D%3D%5Cfrac%7B%5BAs%5E%7B2-%7D%5D%5BH%5E%2B%5D%7D%7B%5BHAsc%5E-%5D%7D)


Putting value of x = 0.000702 M


![[Asc^{2-}]=5.92\times 10^{-8} M](https://tex.z-dn.net/?f=%5BAsc%5E%7B2-%7D%5D%3D5.92%5Ctimes%2010%5E%7B-8%7D%20M)
Total concentration of ![[H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%2By%3D0.000702%20M%2B5.92%5Ctimes%2010%5E%7B-8%7D%20M%3D7.0206%5Ctimes%2010%5E%7B-4%7D%20M)
The pH of the solution :
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)

Answer:
B. Equal to 7.
Explanation:
Hydrobromic acid is a strong acid that decreases pH and ammonia is a strong base that increases pH.
As the initial pH of water is 7,0 the addition of 35.0mL of 0.400M HBr will produce a pH less than 7,0. But, the same effect of decreasing pH is reverted for the addition of 35.0mL of 0.400M HNO3.
That means the net effect of the two addition is to have a pH:
B. Equal to 7.
I hope it helps!
Answer:
heated crude oil enters a tall fractionating column , which is hot at the bottom and gets cooler towards the top. vapours from the oil rise through the column. vapours condense when they become cool enough. liquids are led out of the column at different heights.
Explanation: