Explanation:
It is given that r = 0.283 nm. As 1 nm =
.
Hence, 0.283 nm = 
- Formula for coulombic energy is as follows.

where, e =
C
= 

= 
- As 1 eV =

So, 1 J = 
Hence, U = 
= 8.9 eV
- Also, 1 J =

=
kJ/mol
Therefore, U =
kJ/mol
= 
Answer is: <span>
The reaction will not be spontaneous at any temperature.
</span>
<span>Gibbs free energy
(G) determines if reaction will proceed spontaneously.
ΔG = ΔH - T·ΔS.
ΔG - changes in Gibbs free energy.
ΔH - changes in enthalpy.
ΔS - changes in entropy.
T is temperature in Kelvins.
When ΔS < 0 (negative entropy change) and ΔH > 0
(endothermic reaction), the process is never spontaneous (ΔG> 0).</span>
Answer:
The
for the reaction
will be 4.69.
Explanation:
The given equation is A(B) = 2B(g)
to evaluate equilibrium constant for 
![K_c=[B]^2[A]](https://tex.z-dn.net/?f=K_c%3D%5BB%5D%5E2%5BA%5D)
= 0.045
The reverse will be 
Then, ![K_c = \frac{[A]}{[B]^2}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5BA%5D%7D%7B%5BB%5D%5E2%7D)
= 
= 
The equilibrium constant for
will be


= 4.69
Therefore,
for the reaction
will be 4.69.
Answer:
the net force us 0.............
Explanation:
please mark brain
Answer:
"A hydraulic turbine converts the energy of flowing water into mechanical energy. A hydroelectric generator converts this mechanical energy into electricity.
Explanation:
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