Answer:
pH = 3.215
Explanation:
From the given information;
Using the equation for the dilution of a stock solution:
Since moles of C5H5N = moles of HNO3
Then:




The reaction between C5H5NH and H2O is as follows:
⇄ 




Now, the next step is to draw out the I.C.E table.
⇄ 
I 0.06333 0 0
C - x x x
E 0.06333 -x x x

![K_a = \dfrac{[C_5H_5N][H_3O^+] }{[C_5H_5N^+H] } \\ \\ 5.8824 \times 10^{-6} = \dfrac{x^2}{0.06333 - x}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cdfrac%7B%5BC_5H_5N%5D%5BH_3O%5E%2B%5D%20%7D%7B%5BC_5H_5N%5E%2BH%5D%20%7D%20%20%5C%5C%20%5C%5C%20%205.8824%20%5Ctimes%2010%5E%7B-6%7D%20%3D%20%5Cdfrac%7Bx%5E2%7D%7B0.06333%20-%20x%7D)
Assuming x < 0.06333

Then

![[H_3O^+] = x = 6.1035 \times 10^{-4} \ M \\ \\ pH = -log (6.1035 \times 10^{-4}) \\ \\ \mathbf{\\ pH = 3.215}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D%20%20x%20%3D%206.1035%20%5Ctimes%2010%5E%7B-4%7D%20%5C%20M%20%20%5C%5C%20%5C%5C%20%20pH%20%3D%20-log%20%286.1035%20%5Ctimes%2010%5E%7B-4%7D%29%20%5C%5C%20%5C%5C%20%5Cmathbf%7B%5C%5C%20pH%20%3D%203.215%7D)
Answer: D. Digestion increases as the volume of the enzyme increases.
Explanation:
Enzymes speed up reactions by reducing the activation time of a reaction. If there are more enzymes therefore, the reaction will move faster.
Pepsin is an enzyme that prefers acidic conditions so it can work with dilute HCL. It can also work in temperatures of 40 °C without getting denatured. As more Pepsin is added therefore, the reaction will move faster and digestion will increase.
Answer:
Smooth Muscle
Explanation:
In the digestive tract it's called the muscularis mucosa.
The answer is heavy/dense