Question

data table: a control (con+kcl) b experimental 0% c experimental 5% d experimental 10% e experimental 15% beaker solution % sugar 10% 10% 10% 10% 10% dialysis tube solution % sugar 10% 10% 10% 10% 10% initial mass 17.59 8.75 11.24 10.71 18.05 final mass 17.60 10.40 12.10 10.57 15.66 amt. of mass gained (+) or lost (-) 0.07 1.65 0.86 0.14 2.45 on your beaker drawing, now draw the tubes in the beakers and show the behaviors of the water in the different concentrations of sugar using arrows to show which direction the water is moving. now turn off the lab and go plug your computer back into the computer cart. analysis: in the lab, what are 4 variables that remained constant throughout the lab (what stayed the same) in the lab, which variable is the dependent variable (what you measured in the experiment) in the lab, which variable is independent (the one thing changed by the scientist) if the percent concentration is greater in the fluid outside a cell than inside a cell, how will the water flow? this means the solution would be which tube(s) did this happen to? if the percent concentration is equal both inside and outside a cell, how will the water flow? this means the solution would be which tube(s) did this happen to? if the percent concentration is greater in the fluid inside a cell than outside a cell, how will the water flow? this means the solution would be which tube(s) did this happen to? which dialysis tube(s) had little or no change in mass after the 24 - hour period? explain why this happened.

Explanation:

1. Constant variables: These are factors that are kept the same throughout an experiment to ensure that only the independent variable affects the dependent variable. Possible constant variables here could be the type of beaker, the type of dialysis tube, the temperature of the environment, and the time - duration of the experiment (24 - hour period mentioned).
2. Dependent variable: This is the variable that is measured as a result of the changes made to the independent variable. Here, it is the mass of the dialysis tube (change in mass is recorded).
3. Independent variable: The variable that the scientist manipulates. In this case, it is the percent concentration of sugar in the beaker solution.
4. Water flow when outside concentration is greater: Water will flow out of the cell (dialysis tube) due to osmosis. This is a hypertonic solution. To determine which tubes this happened to, we would need to compare the inside and outside sugar concentrations for each tube.
5. Water flow when concentrations are equal: There will be no net movement of water. This is an isotonic solution.
6. Water flow when inside concentration is greater: Water will flow into the cell (dialysis tube). This is a hypotonic solution.
7. Tubes with little or no change: Tubes where the inside and outside sugar concentrations are similar will have little or no change in mass as there is no significant osmotic gradient for water movement.

Answer:

1. Possible constant variables: Type of beaker, type of dialysis tube, temperature, time - duration of experiment.
2. Dependent variable: Mass of the dialysis tube.
3. Independent variable: Percent concentration of sugar in beaker solution.
4. Water will flow out of the cell (dialysis tube). Hypertonic solution. (Determine tubes by comparing concentrations).
5. No net movement of water. Isotonic solution. (Determine tubes by comparing concentrations).
6. Water will flow into the cell (dialysis tube). Hypotonic solution. (Determine tubes by comparing concentrations).
7. Tubes with similar inside and outside sugar concentrations. Little or no osmotic gradient for water movement.