Answer:
See the answer below, please.
Explanation:
The equilibrium constant is defined as the relationship between products and reagents, each one elevated to their stoichiometric coefficients, in that of the given equation, the Kc is:
Kc= (NH4)^1/ (NH3)^1 x (HI)^1
NH4= products
NH3 and HI = reagents
d, one atom of oxygen and two atoms of hydrogen
Answer:
Both b and d can be correct
Explanation:
Generally, diffusion does not require energy (<em>making option a wrong</em>) because it is the movement of particles from a region of high concentration to a region of low concentration hence diffusion moves particles in the direction of a concentration gradient. An example of this is the passive transport (for instance, uptake of glucose by a liver cell).
However, in some cases, when diffusion is against the concentration gradient (i.e when particles move from a region of low concentration to a region of high concentration), diffusion will require energy in a case like this (<em>making option c wrong</em>). An example of this is active transport (transport of protein called sodium-potassium pump which involves pumping of potassium into the cell and sodium out of the cell).
The explanation above shows that diffusion can require energy to move particles (in or out) of the cell through the cell membrane.
Answer:
1.
was the
value calculated by the student.
2.
was the
of ethylamine value calculated by the student.
Explanation:
1.
The
value of Aspirin solution = 2.62
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=10^{-2.62}=0.00240 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2.62%7D%3D0.00240%20M)
Moles of s asprin = 
Volume of the solution = 0.600 L
The initial concentration of Aspirin = c = 

initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:



was the
value calculated by the student.
2.
The
value of ethylamine = 11.87


![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-2.13}=0.00741 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-2.13%7D%3D0.00741%20M)
The initial concentration of ethylamine = c = 0.100 M

initially
c 0 0
At equilibrium
(c-x) x x
The expression of dissociation constant :
:



was the
of ethylamine value calculated by the student.