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
Answer:
2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3
Explanation:
First of all, let us balance the equation to give;
2Sb(OH)3 (s) + 3Na2S (aq) = Sb2S3 + 3NaOH
Now, we can observe the presence of positive Sodium ions (Na+) and negative hydroxyl ions (OH-) on both left and right sides of the equation.
Now, the two ions will cancel out. These ions are not really involved in the overall reaction and thus do not require being written in the overall equation. Hence, the overall net ionic reaction can now be written as:
2Sb^(+3) (aq) + 3S^(-2) (aq) = Sb_2•S_3
Malleable, shiny and good conductors
A B E
Answer:
-2
Explanation:
7 x 1 - 2 x 1 + 1 x 1 + 3C = 0 (no charge)
6 + 3C = 0
C = -2
