Diffusion and Osmosis NGSSS: SC.912.L.14.2 Relate structure to function for the components of plant and animal cells. Explain the role of cell membranes as a highly selective barrier (passive and active transport). SC.912.L.14.3 Compare and contrast the general structures of plant and animal cells. Compare and contrast the general structures of prokaryotic and eukaryotic cells. AA Background: (Source: www.explorelearning.com) Diffusion is the process in which there is a net movement of
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1 Diffusion is the movement of molecules (or ions) from a high concentration to low concentration. Diffusion is form of passive transport as energy does not need to be generated. Diffusion is complete when the concentration of molecules is equal on either side of the membrane. Diffusion rate can be influenced by many factors such as: Concentration gradient across the the membrane. Permeability of the membrane to the diffusing substance. Temperature. Surface area of the membrane. Question 2 2.1
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Diffusion is a natural act that takes place in the human body in order to utilize and absorb important concentrations into the body. In order to understand how diffusion works‚ it is important to understand how temperature plays a role. In connection with diffusion‚ a certain temperature must be obtained to begin the process. Introduction Diffusion is very important in the body for the movement of substances. An example would be the movement of oxygen from the air into the blood and carbon dioxide
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Introduction: Diffusion and osmosis are passive processes of transport. Passive transport involves no disbursement of energy by the cell. Diffusion movement is from high concentration to low concentration‚ which the driving force for this type of movement is kinetic energy particles themselves. Which crystal (Methylene blue‚ solid or Potassium Permanganate KMnO4-purple) will move further than the other due to the driving force (kinetic energy)? My prediction is that Potassium Permanganate KMnO4-
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Biology Lab #1: Chemical Diffusion Purpose To discover how the size of a cell affects the diffusion of chemicals throughout the cell. Hypothesis I predict that as the cell size increases the diffusion depth and the diffusion rate will decrease. Equipment • Eye protection • 250 mL beaker • Timing device • Scoopula • Ruler • Scalpel • Sodium hydroxide solution • 3 different sized cubes of phenolphthalein agar • Paper towels Purpose
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1.0 Introduction 1.1 Aim and Assumptions: The aim of this experiment is to observe the rate of diffusion of diethyl ether vapour into stagnant air‚ and then determine the diffusivity. An error analysis of this calculation also had to be carried out. The following assumptions made were: * Diethyl Ether is an ideal gas * Partial pressure at the top of the tube was equal to zero * Mass transfer resistance between the liquid surface at the bottom of the tube is insufficient‚ compared with the
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Activity One: 1. The two major variables that affect the rate of diffusion: a. The composition of the lipid bilayer (eg. more cholesterol‚ less permeability to polar substances) b. The structure of the molecule undergoing diffusion (eg. steric conformation‚ size‚ polarity‚ amount and strength of hydrogen bonding) 2. Urea was not able to diffuse through the 20 MWCO because the pores of the membrane were too small for the urea to pass through. The molecular weight of urea is 60.06 g/mol‚ over
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AP Biology Osmosis and Diffusion Lab I. Introduction: Diffusion is vital to many life functions of a cell‚ it allow the transportation of vitally important nutrients and compounds without the expenditure of excess metabolic energy. To explain diffusion‚ it is as if a bottle of perfume is opened at one end of the room‚ then in a short amount of time a person at the other end of the room can detect the scent of the perfume; this is the process of diffusion. Diffusion is a movement from a higher
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Membrane Functions: Aim: To demonstrate the process of osmosis using a chicken’s egg Hypothesis: I predict that the egg’s mass after 10 min. in the 5% salt solution would increase due to and balancing out the solution. In the 10% salt solution‚ the egg’s mass would decrease since there is a high amount of salt. In the dilute water solution‚ the egg’s mass will increase‚ as the membrane would take in that water. Materials: 1- One fresh egg 2- One plastic spoon 3- One plastic fork
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Purpose: The purpose of this lab is to test the effects of osmosis on eggs in hypertonic solutions and hypotonic solutions. Hypothesis: If the corn syrup is a hypertonic solution‚ then the egg’s size will shrink. If the water is a hypotonic solution‚ then the egg will swell. Materials: Styrofoam cups (2 per group) Styrofoam cups to weigh down egg (2) Balance Paper towels Vinegar (enough to cover egg) Distilled water Corn syrup Fresh eggs (2 per group) Procedure: Day
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