Explanation:
The given reaction equation will be as follows.
![[FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}]](https://tex.z-dn.net/?f=%5BFeSCN%5E%7B2%2B%7D%5D%20%5Crightleftharpoons%20%5BFe%5E%7B3%2B%7D%5D%20%2B%20%5BSCN%5E%7B-%7D%5D)
Let is assume that at equilibrium the concentrations of given species are as follows.
![[Fe^{3+}] = 8.17 \times 10^{-3}](https://tex.z-dn.net/?f=%5BFe%5E%7B3%2B%7D%5D%20%3D%208.17%20%5Ctimes%2010%5E%7B-3%7D)
M
![[SCN^{-}] = 8.60 \times 10^{-3}](https://tex.z-dn.net/?f=%5BSCN%5E%7B-%7D%5D%20%3D%208.60%20%5Ctimes%2010%5E%7B-3%7D)
M
![[FeSCN^{2+}] = 6.25 \times 10^{-2}](https://tex.z-dn.net/?f=%5BFeSCN%5E%7B2%2B%7D%5D%20%3D%206.25%20%5Ctimes%2010%5E%7B-2%7D)
M
Now, first calculate the value of 
as follows.
![K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BFe%5E%7B3%2B%7D%5D%5BSCN%5E%7B-%7D%5D%7D%7B%5BFeSCN%5E%7B2%2B%7D%5D%7D)
= 
= 
Now, according to the concentration values at the re-established equilibrium the value for ![[FeSCN^{2+}]](https://tex.z-dn.net/?f=%5BFeSCN%5E%7B2%2B%7D%5D)
will be calculated as follows.
![11.24 \times 10^{-4} = \frac{8.12 \times 10^{-3} \times 7.84 \times 10^{-3}}{[FeSCN^{2+}]}](https://tex.z-dn.net/?f=11.24%20%5Ctimes%2010%5E%7B-4%7D%20%3D%20%5Cfrac%7B8.12%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%207.84%20%5Ctimes%2010%5E%7B-3%7D%7D%7B%5BFeSCN%5E%7B2%2B%7D%5D%7D)
![[FeSCN^{2+}] = 5.66 \times 10^{-2}](https://tex.z-dn.net/?f=%5BFeSCN%5E%7B2%2B%7D%5D%20%3D%205.66%20%5Ctimes%2010%5E%7B-2%7D)
M
Thus, we can conclude that the concentration of in the new equilibrium mixture is 
M.
Answer:
Final [B] = 1.665 M
Explanation:
3A + 4B → C + 2D
Average rection rate = 3[A]/Δt = 4[B]/Δt = [C]/Δt = 2[D]/Δt
0.05600 M/s = 4 [B]/ 2.50 s
[B] = 0.035 M (concentration of B consumed)
Final [B] = initial [B] - consumed [B]
Final [B] = 1.700 M - 0.035 M
Final [B] = 1.665 M
PH scale is used to determine how acidic, basic or neutral a solution is
pH can be calculated using the H₃O⁺
ph can be calculated as follows
pH = - log[ H₃O⁺]
[H₃O⁺] = 1 x 10⁻⁹
pH = - log [1 x 10⁻⁹]
pH = 9
pH of solution is 9
It’s extremely bad quality I really can’t read it
