Answer: The correct answer is: [B]:
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" organic acid and amines " .
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<u>Note</u>: Choice B: "organic acid and amines" ;
is the only answer choice that contains "amines" (hint: <u> amin</u><u>o acid</u> / <u>amin</u><u>e)</u> ; which are "proteins" .
As such; Choice "B" is the <u><em>only</em></u> correct answer choice.
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Hope this helps!
Best wishes to you!
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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.
<span><span>Iron, Wrought 1482 - 1593- 2900</span><span>Iron, Gray Cast1127 - 1204 - 2200</span><span>Iron, Ductile1149</span></span>
Answer:
k = -0.0525 s⁻¹
Explanation:
The equaiton for a first order reaction is stated below:
ln[A]=−kt+ln[A]₀.
[A] = 5.50 x 10⁻³ M
[A]₀ = 7.60 x 10⁻² M
t = 85.0 - 35.0 = 50.0 s
The rate constant is represented by k and can be calculated substituting the values given above:
k = (ln[A]₀ - ln[A])/t
k = (ln5.50 x 10⁻³ M - ln7.60 x 10⁻² M)/50.0s
k = -0.0525 s⁻¹
Answer:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Explanation:
