Answer:
The pH of the final solution is 7.15
Explanation:
50 mL of 2.0 M of
and 25 mL of 2.0 M of
were mixed to make a solution
Final volume of the solution after dilution = 200 mL
Final concentration of ![K_2HPO_4, [K_2HPO_4] = \frac{50 mL\times 2 M}{200 mL} = 0.5 M](https://tex.z-dn.net/?f=K_2HPO_4%2C%20%5BK_2HPO_4%5D%20%3D%20%5Cfrac%7B50%20mL%5Ctimes%202%20M%7D%7B200%20mL%7D%20%3D%200.5%20M)
Final concentration of![KH_2PO_4, [KH_2PO_4] = \frac{25 mL\times 2 M}{200 mL} = 0.25 M](https://tex.z-dn.net/?f=KH_2PO_4%2C%20%5BKH_2PO_4%5D%20%3D%20%5Cfrac%7B25%20mL%5Ctimes%202%20M%7D%7B200%20mL%7D%20%3D%200.25%20M)
We use Hasselbach- Henderson equation:
![pH = pK_a+ log \frac{[salt]}{[acid]}pka of KH_2PO_4 = 6.85](https://tex.z-dn.net/?f=pH%20%3D%20pK_a%2B%20log%20%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7Dpka%20of%20KH_2PO_4%20%3D%206.85)
Substituting the values:

Therfore, the pH of the final solution is 7.15
First, assume the order of the given reaction is n, then the rate of reaction i.e. ![\frac{dx}{dt}=k\times[A]^{n}](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%3Dk%5Ctimes%5BA%5D%5E%7Bn%7D)
where, dx is change in concentration of A in small time interval dt and k is rate constant.
According to units of rate constant, the reaction is of second order.
(second order formula)
Put the values,

t= 587.9 s
Hence, time taken is 587.9 s
Answer:
It isnt changing
Explanation:
It stays basically the same becuase it is considered a constant
Answer:
Electron transport is the process by NADH + H+ and FADH2 are converted to NAD+ and FAD, donating electrons and hydrogen ions to oxygen
Explanation:
I think it's RbCl and CaO