Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
So,
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
Answer:
The answer is the third one which is 24 (glucose)
Answer: When the reaction reaches equilibrium, the cell potential will be 0.00 V
Explanation:
Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.
The equilibrium is dynamic in nature and the reactions are continuous in nature. Rate of forward reaction is equal to the rate of backward reaction.
The standard emf of a cell is related to Gibbs free energy by following relation:
The Gibbs free energy is related to equilibrium constant by following relation:
For equilibrium
Thus
Thus When the reaction reaches equilibrium, the cell potential will be 0.00 V
If the question includes x only than it’s value would be 1
Answer: 625 grams
Explanation:
14 goes into 56, 4 times.
10,000÷2=5,000 - first half life
5,000÷2=2,500 - second half life
2,500÷2=1,250 - 3rd half life
1,250÷2= 625 - 4th half life
Note: don't type "grams" after 625, just type "625".