Extraction is a technique used for separating a compound from a mixture. An example is separating a water-insoluble organic compound from an aqueous mixture by extracting it into a water-insoluble organic solvent. These extractions are often part of the workup procedure for isolating and purifying the product of an organic reaction. Because trace amounts of water are often present at the end of an extraction process, a drying reagent is needed to ensure a dry product. The process of liquid-liquid
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APPROVALBOOK.COM Page 1 New Heritage Doll Company Capital Budgeting Solution NEW HERITAGE DOLL COMPANY CAPITAL BUDGETING SOLUTION New Heritage Doll Company Capital Budgeting Solution a great book which gives a great insight into the workings of a new heritage doll company capital budgeting solution. Clear descriptions of various systems within the new heritage doll company capital budgeting solution. Written from an american point of view but this doesn't really detract from a great book
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issues of the case and discuss your thoughts and opinions and more importantly why you think them!!!! If you want a really basic format that works, here is one: 1. Background & key facts 2. Main Problem Statement 3. Possible Solutions and Options 4. Your Recommended Solution & Analysis (WHY) Most cases will have questions that help you to tackle the case. I suggest that you maybe even read those questions before you read the case. This should help you in getting started. I do not want you to
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which the solution can be made. As the temperature increases, the particles of the solid move faster, which increases the chances that they will interact with more of the solvent particles. This results in increasing the rate at which a solution occurs. Temperature can also increase the amount of solute that can be dissolved in a solvent. Generally speaking, as the temperature is increased, more solute particles will be dissolved. For instance, when you add table sugar to water, a solution is quite
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2.04 AP Practice: Part 2 10/15 total points earned. Question 1 (Worth 5 points) The boxes described below have the same volume, but different dimensions and surface area. Use this data as the basis for a discussion of beneficial cell shapes. Essay Submission When the volume is too large relative to the surface area of the cell, diffusion cannot occur at sufficiently high rates to ensure this. Therefore cells size is wanted to be relatively small. Box A's surface area to volume ratio is
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The purpose of this lab was to get used to using lab machinery such as the spectrometer. Spectroscopy is the interaction of matter and electromagnetic radiation. Most of our lab was based on the idea that matter would absorb and then reflect the colors we wanted to see based on the dyes we used. We also tested how strong the dyes were by testing them on filter paper and seeing how far they would travel up the paper. During dry lab we put all the classes’ data together to get accurate absorbance averages
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Define the following terms and use them in your answers below. - Solubility: The greatest quantity of a solute that can be dissolved into a solvent - Soluble: When a substance can be dissolved into a solvent. Ex: sugar can be easily dissolved into coffee or tea. - Insoluble: When a substance cannot be dissolved into a solvent. Ex: oil cannot be dissolved into water. - Hydrophilic: Describes a substance that can easily be mixed or dissolved into water. Ex: salt - Hydrophobic: Describes
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background knowledge, I know that there are stomach acids in our stomach that help dissolve food, which exemplifies how this acid requires high solubility in order to whole and dissolve all the food we eat. Changing the potential hydration of an aqueous solution, “you can change the charge state of the solute” [4]. By changing the charge state of a solute, it is possible to convert molecules to having no net electrical charge and having decreased electrical charge “the solute has minimal solubility” [4]
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the solutions and collecting data, the solution most similar to the test solution in concentration is Solution A. According to my data, Solution B, C, and D all lost mass (grams) while Solution A’s mass increased. Informed by this data, the mass change should be positive in order to match the test solution. Due to the exposure of the hypotonic solution in Solution A, the potato increased from 2.71 ± 0.01 grams to 3.01 ± 0.01 grams. Through osmosis, the water molecules moves from the solution into
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The purpose of this experiment is to find the unknown concentration of two different proteins; the gamma globulin, and the bovine serum albumin. The measurement is done in the spectrophotometer by using the method of the Bradford Essay with the Coomassie Brilliant Blue G-250 dye, as a reagent. To identify the unknowns, first a colorimetric assay of both proteins with known concentrations is executed, and the data in plotted in two standard curves respectively. Finally, the unknown absorbance measurements
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