Equation of decomposition of ammonia:
N2+3H2->2NH3
Euilibrium constant:
Kc=(NH3)^2/((N2)((H2)^3))
As concentration of N2=0.000105, H2=0.0000542
so equation will become:
3.7=(NH3)^2/(0.000105)*(0.0000542)^3
NH3=√(3.7*0.000105*(0.0000542)^3)
NH3=7.8×10⁻⁹
So concentration of ammonia will be 7.8×10⁻⁹.
Answer:
c
Explanation:
hydrogen Bond exist between ammoia, hydrogen fluoride and water so the answer is c
Answer:
we need the food chain to answer the question.
<u>Answer:</u> The cell potential of the cell is +0.118 V
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u> 
<u>Reduction half reaction (cathode):</u> 
In this case, the cathode and anode both are same. So,
will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}]_{diluted}}{[Cl^{-}]_{concentrated}}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%5D_%7Bdiluted%7D%7D%7B%5BCl%5E%7B-%7D%5D_%7Bconcentrated%7D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
= 0.0222 M
= 2.22 M
Putting values in above equation, we get:


Hence, the cell potential of the cell is +0.118 V
The answer is C, the volume decreases. that is wrong