Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:
Let's say the initial concentration of the acid is c and the change in concentration x.
Then, equilibrium concentration of acid = (c-x)
and the equilibrium concentration for each of the product would be x
Equilibrium expression for the above equation would be:
![\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}](https://tex.z-dn.net/?f=%5CKa%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BHCOO%5E-%5D%7D%7B%5BHCOOH%5D%7D)
From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,
Let's solve this for x. Multiply both sides by 0.12
taking square root to both sides:
Now, we have got the concentration of ![[H^+] .](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20.)
![[H^+] = 0.00465 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%200.00465%20M)
We know that, ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
Answer:
1125mL
Explanation:
this can be done using general gas law
Answer:
(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
(b) ₁₀Ne: 1s² 2s² 2p⁶
---
(a) 3
(b) 6
(c) 7
Explanation:
We can state the ground-state electron configuration for each element following Aufbau's principle.
(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
(b) ₁₀Ne: 1s² 2s² 2p⁶
Second part
(a) Al belongs to Group 13 in the Periodic Table. It has 13-10=3 electrons in the valence shell.
(b) O belongs to Group 16 in the Periodic Table. It has 16-10=6 electrons in the valence shell.
(c) F belongs to Group 17 in the Periodic Table. It has 17-10=7 electrons in the valence shell.
<h2>0.2 atm </h2>
Explanation:
The new pressure can be found by using the formula for Boyle's law which is
Since we're finding the new pressure
We have
We have the final answer as
<h3>0.2 atm</h3>
Hope this helps you
