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3 years ago

You note that cells placed in a solution crenate. What could you definitively conclude about the solution?

Chemistry

2 answers:

lys-0071 [83]3 years ago

6 0

Answer:it means the cell is an hypertonic or high concentration solution

Explanation: osmosis is defined as movement of water from a region of high concentration to a region of low concentration of water.

Now being in a hypertonic solution means the sorrounding solution in a high concentration of the solute which means water is in low in concentration , so osmotic pressure move water from the cell (which has a relatively higher water concentration )to it's sorrounding which has a lower concentration of water, resulting in the shrinking or crenation of the cells.

Likurg_2 [28]3 years ago

4 0

Answer:Conclusion: The solution is hypertonic in nature.

Explanation:

when a cell is exposed to a hypertonic solution,cell will crenate,that is contraction of cell occurs.This is because water moves from the cell to the surrounding via osmosis as result cell shrinks and an abnormal notchings occurs in its membrane.

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Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

$Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V$

$Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive): $Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Notice that the overall cell potential upon summing is:

$E_{cell}=1.66 V + 1.50 V=3.16 V$

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

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3 years ago