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.
HOPBr₂ = <span>dibromophosphinous acid
hope this helps!
</span>
Answer:
NAD (Nicotinamide adenine dinucleotide)
Explanation:
NAD+ (or NADH in its reduced form) is important in cellular respiration as it acts as an electron carrier, providing a shuttle system for the movement of electrons. This drives a proton pump and generates ATP via oxidative phosphorylation. Niacin is a form of vitamin B3 that we get from our diet and is a precursor for NAD.
Answer:
Carbon
Explanation:
Carbon is the most important element to life. Without this element, life as we know it would not exist. As you will see, carbon is the central element in compounds necessary for life.
<span>M(NO</span>₃<span>)</span>₂<span> fully separates into M</span>²⁺<span> and NO</span>₃<span> </span>²⁻<span> </span><span>and M(OH)</span>₂<span> partially separates
as <span>M</span></span>²⁺<span><span> and 2OH</span></span>⁻
<span>M(NO</span>₃<span>)</span>₂<span><span> </span>→
M</span>²⁺<span> + 2NO</span>₃²⁻
<span>0.202 M 0.202 M</span>
<span> M(OH)</span>₂<span>(s) ↔ <span>M</span></span>²⁺<span><span> (aq) + 2OH</span></span>⁻<span><span>(aq)</span></span>
<span>I - -</span>
<span>C -X +X +2X</span>
<span>E X 2X</span>
<span>Ksp = [M</span>²⁺<span> (aq)] [OH</span>⁻<span>(aq)]</span>²
4.45 * 10∧-12 = (0.202
+ X ) (2X)²
Since X is very small, (0.202 + X ) = 0.202
<span>4.45 * 10<span>-12 </span>= 0.202 *
4X</span>²
<span> X = 2.347 </span>× 10∧-6 M
Hence
the solubility of <span>M(OH)2
is 2.347 </span>× 10∧-6 M
