The answer is 25g. Let me know if you need an explanation
If you notice in the graph for antibiotic A, the number of bacteria actually INCREASES as time increases after the antibiotic was given. In the second graph, the amount of bacteria increases just a little bit (likely as the antibiotic sets in) and then decreases until no bacteria is left at all. This means that the antibiotic was the most successful because not only did the amount of bacteria decrease over time, but also all of the bacteria were eventually killed.
The last graph is shown as no antibiotic given. This is a graph showing the control group. There is always a control group in an experiment where nothing is done to the group. This is used to compare the results in the end of the experiment.
An example would be that they can reproduce just like anything else that is living
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
