Sir William Alexander Clarke Bustamante.
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The mass and volume has direct proportional to each other.
<h3>What is the relationship between mass and volume?</h3>
The volume of the object is directly proportional to its mass which means that if the volume increases the mass of the object increases also increase and vice versa. Due to this direct relationship, the line of regression will go straight between X and Y axis.
So we can conclude that mass and volume has direct proportional to each other.
Learn more about volume here: brainly.com/question/1972490
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Answer:
The answer to your question is below
Explanation:
Data
Substance = NaCl
moles of NaCl = 49
volume = 150 ml
Process
Molarity is a unit of concentration that makes a relation of the moles of a substance and the volume.
Molarity = moles / volume (L)
1.- Convert 150 ml to L
1000 ml ------------------ 1 L
150 ml ----------------- x
x = (150 x 1) / 1000
x = 0.15 L
2.- Substitution
Molarity = 49 / 0.15
Molarity = 326. 7
I have a doubt if the number of moles is 49 moles or 49μmoles
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)
![\frac{[Deprotonated]}{[Protonated]}=0.01](https://tex.z-dn.net/?f=%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D%3D0.01)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
![\frac{[Protonated]}{[Deprotonated]}=100](https://tex.z-dn.net/?f=%5Cfrac%7B%5BProtonated%5D%7D%7B%5BDeprotonated%5D%7D%3D100)
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
