Answer:
The new concentration is 0.125 M.
Explanation:
Given data:
Initial volume V₁ = 125.0 mL
Initial molarity M₁ = 0.150 M
New volume V₂ = 25 mL +125 mL = 150 mL
New concentration M₂ = ?
Solution:
M₁V₁ = M₂V₂
0.150 M × 125 mL = M₂ × 150 mL
M₂ = 0.150 M × 125 mL / 150mL
M₂ = 18.75 M.mL/150 mL
M₂ = 0.125 M
The new concentration is 0.125 M.
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.
Potassium metal + Chlorine Gas -------->
Potassium Chloride<span>
The chemical equation using symbols and formula is
<span>K (s) + </span></span><span><span><span><span>Cl</span>2</span> </span><span>(g) ---------> 2KCl (s)</span></span>
