Answer:
0.40 L
Explanation:
Calculation of the moles of
as:-
Mass = 51.24 g
Molar mass of
= 171.34 g/mol
The formula for the calculation of moles is shown below:
Thus,

Volume = 1.20 L
The expression for the molarity is:


Thus,
Considering
Given that:
So,
<u>The volume of 0.24925M stock solution added = 0.40 L
</u>
51.86 grams would be in the container.
One ounce is an equivalent of 28.34 grams, so times that by 12.2
.
12.2 * 28.34 = 345.75.
Put the percentage into decimal form, so 15% would now be 0.15.
0.15 * 345.75 = 51.86.
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,

pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
Answer:
The death is lower
Explanation:
im gonna get a warning watch
Answer: d. both (b) and (c)
Explanation:
Atoms are too small to be seen on microscopes, so we model them based on theories.