<span>CH4 + 4 Cl2 → CCl4 + 4 HCl
(4.00 mol CH4) x (1/1) x (0.70) = 2.80 mol CCl4
(4.00 mol CH4) x (4/1) x (0.70) = 11.2 mol HCl
CCl4 + 2 HF → CCl2F2 + 2 HCl
(2.80 mol CCl4) x (2/1) x (0.70) = 3.92 mol HCl
11.2 mol + 3.92 mol = 15.1 mol HCl from both steps</span>
Answer:
![[H^+]=0.00332M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.00332M)
Explanation:
Hello,
In this case, considering the dissociation of valeric acid as:

Its corresponding law of mass action is:
![Ka=\frac{[H^+][C_5H_9O_2^-]}{[HC_5H_9O_2]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BC_5H_9O_2%5E-%5D%7D%7B%5BHC_5H_9O_2%5D%7D)
Now, by means of the change
due to dissociation, it becomes:

Solving for
we obtain:

Thus, since the concentration of hydronium equals
, the answer is:
![[H^+]=x=0.00332M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.00332M)
Best regards.
The last row going across
The question here is solved using basic chemistry. CaCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of CaCl2 that dissolves.
CaCl2(s) --> Ca+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.65 mol CaCl2/1L × 2 mol Cl⁻ / 1 mol CaCl2 = 1.3 M
The answer to this question is [Cl⁻] = 1.3 M