Lab 2 Essay Example | Topics and Well Written Essays - 1250 words

Lab 2 - Essay Example Bases cause the litmus to turn blue. Strength of a base is dependent on concentration of OH- ions produced. On the pH scale, basicity of a substance increases from 7 to 14: 7 being neutral and 14 being strongest base (one that generates maximum OH- ions when dissolved). Name two acid and bases that you often use. Acids: Milk, Orange Juice Bases: Soap, Tooth paste Cube Dimensions Surface Area (cm2) Volume (cm3) Surface Area : Volume Time Required for Complete Color Change Distance of Diffusion 1 cm X 1 cm X 1 cm 6 1 6:1 3 min 0.50 cm 1 cm X 2 cm X 2 cm 16 4 4:1 2 min 50 sec 0.30 cm 1 cm X 1 cm X 6 cm 36 6 6:1 2 min 42 sec 0.48 cm How did the surface area effect the diffusion of the cube? What about the volume? What about the surface area to volume ratio? Which of these had the greatest effect on the diffusion of the cube? Time required for complete color change is found to be approximately same for all three cubes (average time: 2 minutes 50 seconds) which concludes that rate of diffu sion of vinegar into the agar is same irrespective of shape and size. Further, volume had no impact on the time required for color change. No regular trend has been observed for distance of diffusion of vinegar inside the agar with respect to surface area or volume, when each is considered alone. Ideally, greater distance of diffusion should have been found in Cube B ( 1 cm X 2 cm X 2 cm) as compared to Cube A ( 1 cm X 1 cm X 1 cm) as it has the greater surface area (larger surface area enhances amount of diffusion); however, the irregular pattern reveals that there is a third factor that has a direct bearing on the distance of diffusion. Studying the results of the experiment, it becomes evident that surface area to volume ratio is the single parameter that majorly affects the amount of diffusion of vinegar into the agar. Cube A ( 1 cm X 1 cm X 1 cm) and Cube C (1 cm X 1 cm X 6 cm) having same surface area to volume ratio yield the same distance of diffusion. On the other hand, cub e B ( 1 cm x 2 cm x 2cm) tails at 0.3 cm. Even though greater amount of vinegar diffuses in to cube B than in cube A due to greater surface area, a larger time period is required for it to reach the center. In conclusion, the experiment demonstrates that depth of diffusion is lesser for materials having smaller surface areas to volume ratio than that with higher surface area to volume ratio. How does this experiment demonstrate the need for larger cells to divide? A cell meets its requirements for growth, repair and removal of waste materials through its cell membrane, the amount of transfer directly relating to its surface area. More so, the extent of movement of substances into a cell-towards its center- depends upon its size and volume. An increase in cell size is accompanied with an increase in requirements of the cell. More nutrients need to be diffused in and a greater amount of waste materials require to be diffused out. Consequently, a mechanism needs to be established that would allow a rapid transfer of substances through the cell membrane. As the cell is grows uncontrollably without division, the surface area to volume ratio decreases. This limits the distance to which substances diffuse into the cell, posing a restriction to the functionality of various components of the cell-especially those located towards the center. To counter this, it