percent. For the 0.2‚ 0.4‚ 0.6 0.8‚ and 1.0 Moles of sucrose solutions‚ the bag gained mass. The water had to diffuse across the membrane into the bag to space out the sucrose molecules because the beaker contained no sucrose. The sucrose was unable to diffuse across the dialysis tubing because the molecules were too large for the pores (as referenced in the previous paragraph). This part experiences osmosis because only the water‚ not the sucrose‚ diffused across the membrane to close the gap between
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3. Describe the differences between animal and plant cells. Although plant and animal cells both have mitochondria‚ cytoplasm‚ and ribosomes they differ in many ways. Animal cells do not have a cell wall and are round with irregular shapes. Centrioles are also present in all animal cells‚ as well as most eukaryotic cells. Plant cells only have centrioles if they are in the lower plant species. Both types of cells have vacuoles‚ but their functions are quite different. The purpose of vacuoles in
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cylinder to measure 1 mL of the 1.0 M sucrose solution and use the “DW” pipet to add 4 mL of distilled water to test tube “b”‚ which is now a .2 M solution. Then use the graduated cylinder to measure 2 mL of 1.0 M sucrose solution and add 3 mL of distilled water to test tube “c”‚ which this creates a .4 M solution. Measure 3 mL of 1.0 M sucrose and 2 mL of distilled water to test tube “d”‚ which then creates the .6 M solution. Measure 4 mL of the 1.0 M sucrose solution and 1 mL of distilled water
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of time it would take to get to the result of fermentation with different sugars. This fermentation rate was calculated with water displacement using pipettes to discover the span of time before the release of air bubbles known as Carbon Dioxide. Sucrose had the highest fermentation rate in comparison to all of the other sugars that were tested. The results confirm the capability of cells when it comes to cellular respiration despite the lack of oxygen. Introduction Adenosine triphosphate‚ known
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Phenol | C6H5OH | 43.0 | 7.40 | 181.0 | 3.56 | Use the previous formula and the constant from Table 1 to calculate the temperature at which a solution of 50 grams of sucrose (C12H22O11) in 400 grams of water will freeze. The molecular weight of sucrose is 12(12.01) + 22(1.01) + 11(16.00) = 342.34 g/mole so‚ the number of moles of sucrose is and the concentration of the solution in moles per kilogram of water is By taking the freezing point constant for water as 1.86 from Table and then substituting
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hypogaea L.‚ as Influenced by Plant Growth Regulators‚ Sucrose and pH1 Q. L. Feng‚ H . E. Pattee*‚ and H. T. Stalkef ABSTRACT Research on in vitro embryo culture in Arachis has the objective ofrescuinginterspecific hybridembryos which abort before they reach maturity. This study explored effects ofthe three exogenous plant growth regulators 1naphthaleneacetic acid (NAA)‚ gibberellic acid (GA‚)‚ and 6-benzylaminopurine (6-BAP); sucrose; and medium pH on in vitro fruit and embryo development
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Northern Kentucky University BIO 150L: Introduction to Biology I Instructor: Dr. Bethany Bowling Osmosis and Diffusion Report Estimating the Osmolarity of Plant Cells--Potato YAO ZHANG 03/26/2012 Introduction: It is undeniable that all cells have the kinetic energy. It will led the cells move randomly around to others. For this molecular movement‚ there are two results that might happen. Diffusion is one of them. Diffusion is the movement of molecules that between the
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ENZYME LAB REPORT 1. In your own words‚ define objective qualitative and objective quantitative data. DO NOT USE EXAMPLES AS YOUR SOLE DEFINITION. (3) There are two different types of information that can be obtained from research. The types of information that can be obtained are quantitative and qualitative data. Research results are considered qualitative when the results can be answered with a simple statement of yes or no. Qualitative data does not attempt to give a numerical value; instead
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use any of the other factors that we discussed with the others was because‚ me and my Partner disagreed on using any of the other factors‚ and that’s why I have chosen to stay with this factor. |Water Concentration in ml |Concentration of Sucrose in ml|BEFORE |AFTER | |50 |5ml |0.90 |0.80 | |50 |5ml
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concentration of molecules from one side of the membrane to another. Cell membranes will allow small molecules like oxygen‚ water‚ carbon dioxide‚ ammonia‚ glucose‚ amino acids‚ etc.‚ to pass through. Cell membranes will not allow larger molecules like sucrose‚ starch‚ protein‚ etc.‚ to pass through. Problem Statement(s): What is the movement of material
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