The correct response is A. Only the Fe is unbalanced.
Decrease because loss of electrons.
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:
pH = 9.48
Explanation:
We have first to realize that NH₃ is a weak base:
NH₃ + H₂O ⇔ NH₄⁺ + OH⁻ Kb = 1.8 x 10⁻⁵
and we are adding this weak base to a solution of NH₄NO₃ which being a salt dissociates 100 % in water.
Effectively what we have here is a buffer of a weak base and its conjugate acid. Therefore, we need the Henderson-Hasselbach formula for weak bases given by:
pOH = pKb + log ( [ conjugate acid ] / [ weak base ]
mol NH₃ = 0.139 L x 0.39 M = 0.054 mol
mol NH₄⁺ = 0.169 L x 0.19 M = 0.032 mol
Now we have all the information required to calculate the pOH ( Note that we dont have to calculate the concentrations since in the formula they are a ratio and the volume will cancel out)
pOH = -log(1.8 x 10⁻⁵) + log ( 0.032/0.054) = 4.52
pOH + pH = 14 ⇒ pH = 14 - 4.52 = 9.48
The solution is basic which agrees with NH₃ being a weak base.
