Answer:
Kc =![\frac{[8.326x10-3]^{1} }{[1.113x10-2]^{1}[1.490x10-2]^{1} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B8.326x10-3%5D%5E%7B1%7D%20%7D%7B%5B1.113x10-2%5D%5E%7B1%7D%5B1.490x10-2%5D%5E%7B1%7D%20%20%7D)
Kc = 50.2059
Explanation:
1. Balance the equation
2. Use the Kc formula
Remember that pure substances, like H2 are not included on the Kc formula
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.
