4HCl + O₂ → 2Cl₂ + 2H₂O
mole ratio of HCl : O₂ is 4 : 1
∴ if moles of HCl = 2.3 mol
then mol of O₂ = 2.3 mol ÷ 4
= 0.575 mol
mass of O₂ = moles of O₂ × molar mass of O₂
= 0.575 mol × (16 × 2) g/mol
= 18.4 g
I believe the answer is Exothermic
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:
3
Explanation:
left side has 2 N so right side must have a 2 which means 6 H on right side so to get 6 on left you have a coef. of 3 to make 6 H
A scientist would most likely need to update her model when it no longer supports the latest results.