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:
Writing down background information can help you to remember what you learned and can be used for notes in the future when taking an assessment.
Explanation:
Can I get brainliest? It's for a challenge
Answer:
Food in the stomach that is high in protein will dilute absorption of alcohol in your stomach :) therefore always drink with food in your stomach, just drinking alchol without food will have serious consequences on helath.
Explanation:
The given reaction Mg + O₂ → MgO is a type of Synthesis reaction. Hence option (1) is correct.
<h3>What is Synthesis reaction ?</h3>
Synthesis reactions are reactions that occur when two different atoms or molecules interact to form one molecule or compound.
It is also known as combination reaction
In the given reaction Mg and O combines with each other to form one product i.e, MgO. Hence, The given reaction Mg + O₂ → MgO is a type of Synthesis reaction. Thus, option (1) is correct.
Learn more about chemical reaction here ;
brainly.com/question/27948961
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B or two relative <span>abundance of each isotope</span>