Strawberry, bread, bananas/ lemonade, milk, water/ nitrogen, oxygen, carbon dioxide
The Answer is D. a polymer is a substance that has a molecular structure consisting of a large number of similar units bonded together
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.
Ok first, we have to create a balanced equation for the dissolution of nitrous acid.
HNO2 <-> H(+) + NO2(-)
Next, create an ICE table
HNO2 <--> H+ NO2-
[]i 0.230M 0M 0M
Δ[] -x +x +x
[]f 0.230-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .230 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.230-x ≈ 0.230
4.5x10^-4 = x^2/0.230
Then, we solve for x by first multiplying both sides by 0.230 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.01M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.01M/0.230M = .0434 or
≈4.34% dissociation.
Not increases it DECREASES because the gas will have too much energy to dissolve in the solution
