An animal cell with a solute potential of -1.2 bars is placed into an open beaker containing a solution with a solute potential of -3.8 bars. what is the resulting movement of water in this example? a) water will move into the cell because the solution is hypotonic to the cell. b) water will not move into nor out of the cell because the solution and the cell are isotonic. c) water will move out of the cell because the beaker solution has a lower water potential than the cell. d) water will move into the cell because the cell has a higher water potential than the solution in the beaker.

The correct answer is A water will move into the cell because the solution i hypotonic to the cell.

Explanation:

The solute potential of the cell is -1.2 bars whereas the solute potential solution of the beaker containing the cell is -3.8 bars.

      so it can be stated that the osmotic pressure of cell is greater than that of the solution present in the beaker as a result the solution of the beaker is hypotonic to the respect of the solute potential of the cell.

       As a result water will move into the cell.

yes you are doing it right great gob

explanation:

when a reversible reaction is occurring in both directions at the same rate; the products and reactants do not change in concentration. a chemical reaction where the compound is broken into its constituent parts.