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) 64
Explanation:
We have given number of chromosome pair that is haploid number = 6
We have to calculate the number of possible gamete in absence of recombination
The number of garnets is given by
where n is haploid number
In question we have given haploid number, that is n=6
So the number of gametes without recombination 
So option (d) is the correct answer
The answer is going to be somatic
Muscle tissues signaling muscle tissues to contract.
The correct answer is d. Please give me brainlest I hope this helps let me know if it’s correct or not thanks I appreciate it