effect of surface area to volume ratio on the rate of osmosis of Solanum tuberosum L._ BACKGROUND A cell needs to perform diffusion in order to survive. Substances‚ including water‚ ions‚ and molecules that are required for cellular activities‚ can enter and leave cells by a passive process such as diffusion. Diffusion is random movement of molecules in a net direction from a region of higher concentration to a region of lower concentration order to reach equilibrium. Diffusion does not require
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Table of results Surface area to volume ratio Time taken for HCL acid to diffuse (seconds) 14 38 9 90 7.3 139 6.5 178 6 185 3. Do your results support your hypothesis‚ use date from your table and graph to support your answer‚ you should also identify any anomalous results. My hypothesis was ‘as the surface-area-to-volume ratio decreases the diffusion rate increases’ my results support my hypothesis‚ because in my table of results when the surface area to volume ratio is at 14 the diffusion
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Mathematics Volume of Solids Formulae for Volume of Solids Cube | Cuboid | Triangular Prism | Cylinder | Cone | Pyramid | Sphere | AnyPrism | s3 | lwh | ½bhl | Πr2h | 1/3πr2h | 1/3Ah | 4/3πr3 | Ah | A = area of the base of the figure s = length of a side of the figure l = length of the figure w = width of the figure h = height of the figure π = 22/7 or 3.14 1. Compute the volume of a cube with side 7cm. Volume of cube: s3 s = 7cm s3 = (7cm x 7cm x 7cm) = 343cm3 2. Compute
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eukaryotic cells can either exist as a single celled organism or be found in multicellular organisms. The unicellular and multicellular organisms are linked to cell size and surface area to volume ratio. The experiment for cell size and diffusion was set to see how and how much water can go to the cells. This movement of water is called Osmosis. Osmosis is the movement of water molecules from an area of low concentration (lots of water) to an area of high concentration (little water) through a semi
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AREA (i) The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonal of the rhombus is 22 cm‚ find the length of the other diagonal. (ii) The floor of a rectangular hall has a perimeter 250m. If the cost of paining the four walls at the rate of Rs 10 per m2 is Rs 1500. Find the height of the hall. (iii) A room is half as long again as it is broad. The cost of carpeting the room at Rs
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rate of reaction and the surface area. The larger the surface are‚ the faster the rate of reaction seems to be. However‚ although the grain with the largest surface area (the smallest grain) reached the highest point within the shortest amount of time‚ its end result was still lower then the medium sized grain. The explanation for this result is relatively easy. The rate of a chemical reaction can be increased by increasing the size of the surface area to volume ratio of the solid reactant. Cutting
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An investigation of Factors that Affect Diffusion Biology The aim of this investigation is to prove the effect of increasing size on the efficiency of diffusion. Diffusion is the process that cells use to obtain oxygen‚ water and food. Also‚ how they lose waste substances‚ for example‚ urea and carbon dioxide. Basically‚ Diffusion is when particles move from an area of high concentration to an area of low concentration. The surface area to volume ratio of the cell is an important factor in
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Investigating Ratios of Areas and Volumes In this portfolio‚ I will be investigating the ratios of the areas and volumes formed from a curve in the form y = xn between two arbitrary parameters x = a and x = b‚ such that a < b. This will be done by using integration to find the area under the curve or volume of revolution about an axis. The two areas that will be compared will be labeled ‘A’ and ‘B’ (see figure A). In order to prove or disprove my conjectures‚ several different values for n will
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factors that can affect the reaction rate are surface area‚ temperature‚ and concentration and pressure. Surface area is the exposed‚ outer layer of a solid. Increased surface area increases reaction rate because more particles on the surface of a solid come in contact with the particles of another substance. For example‚ if you place a donut stick into water‚ the donut will react slowly with the acid. This is because the acid only comes in contact with the particles on the surface of the donut. But
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Surface area Surface area is the measure of how much exposed area a solid object has‚ expressed in square units. Mathematical description of the surface area is considerably more involved than the definition of arc length of a curve. For polyhedra (objects with flat polygonal faces) the surface area is the sum of the areas of its faces. Smooth surfaces‚ such as a sphere‚ are assigned surface area using their representation as parametric surfaces. This definition of the surface area is based on methods
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