Answer:
For this experiment we are going to take plate 1 as the control plate, so, in it there will be just E. coli in LB/agar; in plate 2, we are going to put E. coli in LB/agar and some ampicillin. Then, we have to wait for the E. coli colonies to form. After a while, the E. coli growth can be compared on both plates and determine if ampicillin affects or not the E. coli colonies.
Explanation:
If the ampicillin affects negatively E. coli colonies, we are going to observe that in plate 1 (control plate) there are E. coli colonies growing, but in plate 2, there is no E. coli colonies or, at least, there is a fewer number of colonies on it. If ampicillin doesn't affect E.coli, plate 1 (control) and plate 2 (ampicillin experiment) are going to be similar in number of colonies.
Answer:
The answer to your question is it is not at equilibrium, it will move to the products.
Explanation:
Data
Keq = 2400
Volume = 1 L
moles of NO = 0.024
moles of N₂ = 2
moles of O₂ = 2.6
Process
1.- Determine the concentration of reactants and products
[NO] = 0.024 / 1 = 0.024
[N₂] = 2/1 = 2
[O₂] = 2.6/ 1= 2.6
2.- Balanced chemical reaction
N₂ + O₂ ⇒ 2NO
3.- Write the equation for the equilibrium of this reaction
Keq = [NO]²/[N₂][O₂]
- Substitution
Keq = [0.024]² / [2][2.6]
-Simplification
Keq = 0.000576 / 5.2
-Result
Keq = 1.11 x 10⁻⁴
Conclusion
It is not at equilibrium, it will move to the products because the experimental Keq was lower than the Keq theoretical-
1.11 x 10⁻⁴ < 2400
There is an increase in the number of collisions between particles and the walls of the container<span>. b. There is an increase in the </span>temperature of the gas. If<span> the volume of a </span>container<span> of </span>gas<span> is reduced, what will </span>happen to the pressure inside<span> the </span>container
I would say they design and build things.
