Answer:
A.The concentration of water is greater outside the cell than inside the cell.
Explanation:
The contractile vacuole of certain organisms functions to regulate water flow in and out of the cell. It does this by storing excess water that comes into the cell. In the case of this organism with a filled up contractile vacuole, it means water is flowing into the cell.
Naturally, water will flow into a living cell when an osmotic gradient i.e. difference in concemtration, has been created between intracellular and extracellular solutions. Osmosis involves movement of substances from a region of high water concentration to a region of low water concentration. This means that if water is flowing into the cell, which is stored by the contractile vacuole, the concentration of water must be greater outside the cell than inside.
Answer: D
Explanation:
A reducing agent is a species that reduces other compounds, and is thereby oxidized. The whole compound becomes the reducing agent. In other words, of a compound is oxidized, then they are the reducing agent. On the other hand, if the compound is reduced, it is an ozidizing agent.
Since we have established that a reducing agent is the compound being oxidized, we know that A is not our answer. An oxidized compound is losing electrons. Choice A states exactly this.
For B, this is true as we have established this already.
C is also correct. Since a reducing agent loses electrons, it becomes more positive. This makes the oxidation number increase.
D would be our correct answer. It is actually a good oxidizing agent is a metal in a high oxidation state, such as Mn⁷⁺.
Answer:
The correct answer is the first option. No, the structures above cannot undergo a Fischer esterification reaction to form an ester.
Explanation:
The reaction that will take place can be found in the attached file. The reaction does not require any catalyst and it cannot undergo a Fischer esterification reaction to form an ester.
Answer:

Explanation:
Hello,
In this case, we work on a dilution process in which we can state that the moles remain the same after the dilution process. In such a way, we can write:

That in terms of molarities and volumes is:

Whereas
is the initial molarity (3.00 M) of the stock solution,
the molarity of the diluted solution (0.289 M),
the aliquot of the stock (concentrated) solution and
the volume of the diluted solution (100 mL), thus, we compute
as required:

Best regards.