Explanation:
The given chemical reaction is:


The relation between Eo cell and Keq is shown below:

The value of Eo cell is:
Br- undergoes oxidation and I2 undergoes reduction.
Reduction takes place at cathode.
Oxidation takes place at anode.
Hence,

F=96485 C/mol
n=2 mol
R=8.314 J.K-1.mol-1
T=298K
Substitute all these values in the above formula:

Answer:
Keq=6.13x10^33
Hope this helps! 159.2086<span>Have a Good Day!</span>
[CO] = 1 mol / 2L = 0.5 M
[
According to the equation:
and by using the ICE table:
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
initial 0.5 0.5 0 0
change -X -X +X +X
Equ (0.5-X) (0.5-X) X X
when Kc = X^2 * (0.5-X)^2
by substitution:
1.845 = X^2 * (0.5-X)^2 by solving for X
∴X = 0.26
∴ [CO2] = X = 0.26
Answer: Hydrosphere, discontinuous layer of water at or near Earth's surface. It includes all liquid and frozen surface waters, groundwater held in soil and rock, and atmospheric water vapour. Earth's environmental spheresEarth's environment includes the atmosphere, the hydrosphere, the lithosphere, and the biosphere.
Explanation:
I think it is the gas and the heat in it