Solutions

Discussions Bag A stimulates cell whose contents are hypotonic to its environment. The situation is because the concentration of the dissolved substances in the cell is less than the surrounding media. In this case, the cell contents contain 1% sucrose solution that is less concentrated than the 50% surrounding media. Therefore, a large concentration gradient develops between the two media. Consequently, because of the high concentration gradient, the bag loses water faster and this leads to its…

soaked through the bag of the syrup. This created a solution of water and syrup 2. The plant takes up water by osmosis and starts to swell. The cell wall prevents it from bursting. Animal cells don’t have a cell wall so the cells will burst in a hypotonic solutions. 3. A wilted plant would have vacuoles are unreplenished. The lack of water causes the cell walls to lose their structure and the plant as a whole wilts. 4. In hypertonic solution, the cell will shrink and…

are three different possibilities that can happen. The first one is that a dilute solution. That happens when the solution has less solute than the maximum amount that it is able to dissolve. The second one is saturated. That happens when the amount of solute is equal to the amount as its solubility. The third one happens “if there is more solute than is able to be dissolved, the excess solute separates from the solution.” Said by Antoinette Mursa of UCD and Prof. Kenneth W. Busch.…

The more precisely the cubes can be cut, the better Spill 96 cm3 of water in the beaker Add 4 grams of the NaOH solution in the beaker that contains the water Put the four agar cubes into the solution Turn on the timer as soon as the cubes are in the beaker Do not take out the cubes for the following 10 minutes. In the meanwhile, observe the temperature Once the 10 minute has been reached, take out the…

located in the mitochondrial membrane and aids in cell respirations and enzyme generation. In order to conduct this experiment, students cut a liver and through a process created six fractions. Each of the six fractions was moved to a cuvette with solutions containing different volumes and recoding the absorbance every five seconds for four minutes. Once the absorbance was obtained for each, we determined that cuvette E (0.0012) had the least activity per milliliter while cuvette B had the most…

Schiff’s reagent to a clean and dry test tube. Also, 4 drops of the unknown were added to the test tube and mixed with a pipette. The Schiff’s test was determined to be negative. Then, the Bisulfite test was conducted by adding 1 mL of Bisulfite solution to a clean, dry test tube. 8 drops of the unknown sample were added to the test tube and it was capped. After 10 minutes, the Bisulfite test was determined to be negative. After, idoform test for water-insoluble unknowns was performed. 2 mL of 1…

innate protein particles or aggregates.However, there is a mechanistic difference between the two aspects. In the case of a pure solubility issue, i.e. a protein becomes insoluble in a certain solution (no changes in protein structure and integrity) and a solid phase is formed, a strong dilution of the system may lead to the dissolution of the protein. In such a circumstance, the term “protein associates” may more appropriately describe the formed protein solid .In this instance, protein…

better blank to use in this experiment. The better blank solution would have been the one with active enzyme because it would cause the spectrophotometer to measure the absorbance of every component in the blank except the substrate, creating a 100% transmittance. This allow the spectrophotometer to read everything that was not in the blank when the actual sample is inserted in because it would “blank” out all the components in the “blank” solution and only read the “uncommon” component. With…

1) Create 6 different 10cm3 concentrated solutions, using Table A 2)…

Cecilia Wong Nature of the task and the research question: Two centimetre piece of potato is bathes into different salt solutions (0%, 5%, 10%, 15%, 25%, 35%) over a period of time, to determine the effect of different salt solutions on osmosis. To investigate the effect of increasing the concentration of sodium chloride (0%, 5%, 10%, 15%, 25%, 35%) on the process of osmosis, by calculating the mass change of the potato core (g, ±0.01g) at room temperate of lab (25.0˚C) Table 1: Raw data Table…