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:
I don't know.
Explanation:
I actually don't know the answer so I wrote the answer is "I don't know" or simply "I dunno".
B. 3.0 mol·L⁻¹ NaCl
Explanation:
Freezing point is a colligative property: it depends only on the number of particles in solution.
The for freezing point depression ΔT_f is
ΔT_f = iK_fb
where
i = the number of moles of particles available from one mole of solute
K_f = the molal freezing point depression constant
b = the molal concentration of the solute
All your solutions are aqueous NaCl. They differ only in their concentrations.
Thus, the most concentrated solution will have the greatest freezing point depression and the lowest freezing point.
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Did you find the answer? Because I am on that question right now for my chem homework
