# Achef chops vegetables into a bowl of water. would you expect the vegetable slices to gain or lose water?

Release water and plasmolysis
Explanation: The potential of water decreases if the concentration of a solution increases. Now, if we put a plant cell in a solution with a lower water potential, it means that the solution contains more solute than the plant cell. In this case, the cell will lose water and shrink. This state is called the plasmolyzed state. The more the water potential becomes negative, the more solute is added to it.

The correct answer is that water will enter the cell because the solution is hypotonic to the cell. Explanation: The solute potential of the cell is -1.2 bar, while the solute potential of the beaker containing the cell is -3.8 bar. thus, it can be stated that the osmotic pressure of the cell is greater than that of the solution present in the beaker, hence the solution in the beaker is hypotonic with respect to the solute potential of the cell. As a result, water will move into the cell.

The water potentials (Ψ) of the cell and its environment are the same. Explanation: When a cell is kept in hypotonic environments such as distilled water, osmotic movement of water to the cell occurs. The ingress of water causes the cell to swell and turgor. Water potential is mainly determined by solute and pressure potentials. Here, the solute potential of the cell and distilled water were different, leading to differences in their respective water potential values ​​which, in turn, served as a driver for endosmosis. When the cell is completely turgid, the solute concentration of the cell and the surrounding distilled water becomes equal. Under these conditions, the water potential of the cell and distilled water are the same.

The water potential of pure water in an open container is zero because there is no solute and the pressure in the container is zero.

The water potential difference ∆P is 7 bar. ∆P = 7 bar Explanation step by step: Given; The water potential of a plant cell was calculated at -4 bar P1 = -4 bar The water potential of another cell was found to be -11 bar. P2 = -11 bar The water potential difference ∆P is; ∆P = P1 – P2 By substituting the given values; ∆P = -4 – (-11) bars ∆P = -4 +11 bars ∆P = 7 bars

Water will leave the cell because the solution in the beaker has a lower water potential than the cell is the response. Option C is therefore correct. Explanation: The way water moves from a point outside the cell to another point inside the cell is called water potential. It depends on two factors, namely the solute potential and the pressure potential. The solute potential has an inverse relationship with the water potential. If the solute potential is higher than the water potential, it will be lower and vice versa. Therefore, in this experiment mentioned above, the solute potential of the cell is lower than that of the solution, so the water comes out of the beaker according to the rule.

The correct answer is A. Water potential is a measure of the water desirability of a biological or chemical compartment. Indeed, any substance in aqueous solution (ion or molecule) exerts an attractive force on water molecules. The more concentrated the solution, the stronger the force of attraction and the less water molecules have the power to leave it. The observed water potential Ψ therefore represents the water outflow potential of a given compartment. The higher it is (less negative) in a compartment, the more water tends to come out of it. In contrast, water tends to enter compartments with low water potential (very negative).
Thus, in a plant cell, the total water potential is zero due to a positive force exerted by the cell wall so that pure water does not enter inside the cell.

A chef cuts vegetables in a bowl of water. I would expect the chopped vegetables to gain water because the solute concentration in the vegetable is low causing more water to enter the pores of the vegetable. This process is known as osmosis.

The water potentials (Ψ) of the cell and its environment are the same. Explanation: When a cell is kept in hypotonic environments such as distilled water, osmotic movement of water to the cell occurs. The ingress of water causes the cell to swell and turgor. Water potential is mainly determined by solute and pressure potentials. Here, the solute potential of the cell and distilled water were different, leading to differences in their respective water potential values ​​which, in turn, served as a driver for endosmosis. When the cell is completely turgid, the solute concentration of the cell and the surrounding distilled water becomes equal. Under these conditions, the water potential of the cell and distilled water are the same.

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