movement will flow into the bag. Also since the dialysis bag’s weight heavily depends on the amount of water‚ the mass will go up when water is gained. The class average for amount of mass gained was 22.55%. A bag that lost mass was the bag with 0% concentration‚ which means it has 100% water. The outside of the bag has 10.57% sucrose and 89.43% water. This makes the outside of the bag hypertonic to the inside‚ resulting in a net flow of water the outside in order to reach equilibrium. By losing water
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or solvent‚ concentrations with salt‚ or solute‚ concentrations in tissues and cells (Campbell). Organisms must perform this process of regulation because of the “Fick’s Law of Diffusion”‚ also called osmosis‚ which states that‚ “The flow of solvent from a solution of lower solute concentration to one of higher solute concentration” will occur. This means that in an organism water will cross a semi-permeable layer‚ tissues in cell walls in this case‚ from an area of lesser concentration of salts to
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however‚ the 5% salt concentration did not result in the least weight gain (g). Table 1 clearly highlights that the 3% salt concentration resulted in the least weight gain (g). The distilled water resulted in the most weight gain (g) because the solution is hypotonic – a hypotonic solution results in the water diffusing into the cell because the molecules of the substance are moving from an area of higher concentration (outside of the cell) to an area of lower concentration (inside the cell) (Hammel
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1. The hypothesis if water is added to the cup then the mass and volume will decrease because water will go into the tubing to decrease solute concentration was supported by the data found in Table 1 and 2 because the cup holding the dialysis tubing with the protein solution’s mass decreased by 7.61 grams and the volume decreased by 17.3 milliliters while the mass of the cup with the tubing holding glucose solution also decreased by 1.19 grams and the volume decreased by 2.2 milliliters (Emma Ricks
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Exercise 1 – Potato Change in Mass versus Sucrose Concentrations Introduction: This exercise involves estimating the osmotic concentration of potato tuber cells by using a change in mass method. The null hypothesis states that there will be no change of mass of the potato disks after they have been incubated in any sucrose solution. This means that the concentration of sucrose that the potatoes are in will no effect the movement of water in or out of the potato cells. However‚ the alternative hypothesis
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therefore water will diffuse into the cell by a process of osmosis. The opposite however will occur when the water potential outside the cell is lower than that inside the cell (i.e. with hypotonic solutions‚ such as a solution with a sucrose concentration of 1m and 0.75m). This is due to the water potential outside the cell being less than that inside the cell and therefore water moves out of the cell by osmosis. With a larger water potential gradient‚ more particles will move in order to balance
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Introduction: This lab helps further our understanding of essential principles behind diffusion and osmosis. Permeability‚ concentration gradients‚ plasmolysis‚ water potential‚ and equilibrium were also concepts that were delved into in this lab. Understanding how diffusion and osmosis works is essential to understanding biology. Each time a cell has something move into or out of it‚ some sort of principle studied in this lab is occurring. Diffusion‚ osmosis‚ and passive and active transport are
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Diffusion Lab Report Emily Cocq Research Question: Will there be a difference in the experimented potato cores after being subjected to high concentrations of sugar? Introduction: Water can move through protein channels in cell membrane/cell walls. The water will move due to the water potential of the cells. If there is a greater concentration of solutes (chemicals) inside the cell than outside the cell and the chemicals can not move‚ then water will respond by moving into the cell. This
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Data collection Quantitative Data Raw Data Table 1: Table showing the mass of the amount of unknown acid X measured in grams (±0.001g) Table 2: Table of reading of the burette initially filled with 25mL of 0.201moldm-3 sodium hydroxide (NaOH) to titrate 25mL (±0.03mL) of unknown acid X in mL (±0.05mL) after each titre. Reading on the burette initially filled with 25mL of 0.201moldm-3 NaOH (±0.05mL) First titre 21.3 Second titre 18.2 Third titre 15.2 Fourth titre 12.0 Qualitative
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In order to fully understand the concept of this experiment‚ one must first be introduced to diffusion. Diffusion is a spontaneous process by which molecule particles move from one area that is highly concentrated to another area in which its concentration is lower. Cells contain fluids and are surrounded by fluids; in order for a cell to function it is required to be in a balanced state. The progress in which a cell is in its balanced state is called equilibrium. Diffusion is a functioning way for
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