I think the answer is A but I could be wrong
It is not a pure substance, because a solution are mixed chemicals in a way that the molecules are not bonded with one another. Thus, separating them from compounds and elements, which are pure substances.
Answer:
The correct answer is AMP+H2O→ Adenosine + pi
Explanation:
The above reaction is least energetic because there is no phosphoanhydride bond present with adenosine mono phosphate.Phospho anhydride bond is an energy rich bond.
As a result hydrolysis of AMP generates very little amount of energy in comparison to the hydrolysis of ATP and ADP.
Answer:
The value of dissociation constant of the monoprotic acid is
.
Explanation:
The pH of the solution = 2.46
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![2.46=-\log[H^+]](https://tex.z-dn.net/?f=2.46%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=0.003467 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.003467%20M)

Initially
0.0144 0 0
At equilibrium
(0.0144-x) x x
The expression if an dissociation constant is given by :
![K_a=\frac{[A^-][H^+]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D)

![x=[H^+]=0.003467 M](https://tex.z-dn.net/?f=x%3D%5BH%5E%2B%5D%3D0.003467%20M)


The value of dissociation constant of the monoprotic acid is
.
Answer : The concentration after 17.0 minutes will be, 
Explanation :
The expression for first order reaction is:
![[C_t]=[C_o]e^{-kt}](https://tex.z-dn.net/?f=%5BC_t%5D%3D%5BC_o%5De%5E%7B-kt%7D)
where,
= concentration at time 't' (final) = ?
= concentration at time '0' (initial) = 0.100 M
k = rate constant = 
t = time = 17.0 min = 1020 s (1 min = 60 s)
Now put all the given values in the above expression, we get:
![[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}](https://tex.z-dn.net/?f=%5BC_t%5D%3D%280.100%29%5Ctimes%20e%5E%7B-%285.40%5Ctimes%2010%5E%7B-3%7D%29%5Ctimes%20%281020%29%7D)
![[C_t]=4.05\times 10^{-4}M](https://tex.z-dn.net/?f=%5BC_t%5D%3D4.05%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration after 17.0 minutes will be, 