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
P * V = n R T
<span>1 * 2.50 = n * 0.082 * 273 = 2. 50 / ( 0.082 * 273 ) = 0.11 mol </span>
<span>mass = mole number * molecular mass </span>
<span>mass = 0.11 * ( 16 * 2 ) = 3.52 g </span>
Get on mathpapa is shows you the answer and how to explain it
Answer:
a. 473mL.
b. 79.38kg
c. 24.47lb of fat
d. 42.5g of N.
Explanation:
a. A qt is equal to 946mL. 0.500qt are:
0.500qt * (946mL / 1qt) = 473mL
b. 1lb is equal to 0.4536kg, 175lb are:
175lb *(0.4536kg / 1lb) = 79.38kg
c. The fat in kg of the athlete is:
74kg * 15% = 11.1kg of fat. In pounds:
11.1kg * (1lb / 0.4536kg) = 24.47lb of fat
d. The mass of nitrogen in the fertilizer is:
10.0oz * 15% = 1.5oz of N
1 oz is equal to 28.35g. 1.5oz are:
1.5oz * (28.35g / 1oz) = 42.5g of N
