Primary Consumers
Herbivores
Trophic level II Members
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
Polarity is classified by the difference in two or more atoms that is the electronegativity value (EN)
En for Fluorine =4.0, Bromine=2.96 Chlorine =3.16 Hydrogen=2.1
for the molecules
HF=4.0-2.I=1.9
HCl=3.16-2.1=1.06
Hbr=2.96-2.1=0.86
Hi=2.66-2.1=0.56
H2=2.1-2.1=0
from the EN value above HF has the mast highest value of EN hence it is the most polar.
A. and D. would be the best pick for this sort of experiment, but maybe (unlikely) B. because you could see how they could react in certain situations, how they react to danger but I suggest A.
Hope this helps you ☁︎☀︎☁︎
