election. The voters are too evenly divided on the issue‚ and he would lose a substantial number of votes no matter what stand he took. Yet with pressure increasing from both sides‚ he has to do something. After much distress he thinks is an ideal solution to his dilemma. He has appointed a committee to study the problem and make some recommendations. To make sure that the committee’s work will not be completed before the election comes up it was important to pick the right people. Specifically‚ Sam
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September 17th at the beginning of class. Show your work. 1. Match the differential equation in (a)-(c) to a family of solutions in (d)-(f). The point of this exercise is not to solve the differential equations in a) - c). (a) y = y 2 (b) y = 1 + y 2 (c) yy = 3x (d) y = tan(x + C) (e) 3x2 − y 2 = C (f) y = −1/(x + C) 2. Find the value of k so that y = e3t + ke2t is a solution of y − 2y − 3y = 3e2t . 3. Solve the following differential equations and IVP’s. You may solve these equations implicitly
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consumption of CO2. All the oxygen within the leaf disks will be removed before being placed into either of the solutions. III. Question & Hypothesis 1. What solution will allow the leaf disks to undergo Photosynthesis first? 2. If the cup with bicarbonate solution allows the leaf disk to produce oxygen by the process of photosynthesis then the leaf disk in the bicarbonate solution will be the first to float within the 15 minutes allowed. IV. Methods‚ Constants‚ Controls‚ & Trials
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potato disks after they have been incubated in any sucrose solution. This means that the concentration of sucrose that the potatoes are in will no effect the movement of water in or out of the potato cells. However‚ the alternative hypothesis states that the mass of the potato disks will increase after they have been incubated in a hypertonic solution. The mass of the potato disks will decrease after they have been incubated in a hypertonic solution. After the results have been gathered‚ appropriate estimations
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region of high concentration (of water molecules) to a region of lower concentration (of water molecules). We define 3 types of solutions based upon osmotic activity. In an isotonic solution‚ the solute concentration is the same on both sides of the membrane so there is no net movement of water. ( See Figure 1) In a hypotonic solution‚ solute concentration is higher inside the cell than outside. So water moves from a high concentration outside the cell to a lower concentration
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Andrew Dickson Background When a plant cell is bathed in a solution of the same concentration (isotonic) as its intracellular environment‚ its mass and volume remain the same. This is because water enters and leaves the cells at the same rate. There is no net loss or gain of water by osmosis. Samples of cells can be placed in a range of solutions of different concentration. The cells will gain water by osmosis when placed in solutions which are more dilute (hypotonic) than the intracellular environment
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Experiment 1: Mixtures Aim: To develop an understanding of different types of mixtures including solutions and examine the different solubilities of some solutes and two solvents: water and ethanol Equipment Bunsen burner Tripod Gauze mat Beakers (2 x 100mL) Boiling chips Distilled water Sodium chloride Sodium chloride (solid) Glass rod Measuring cylinder (25mL) Ethanol Watch glass Thermometer 100% Orange Juice 100% Apple Juice 100% Prune Juice Solder Fruit Scone Charcoal Copper(II) Chloride Magnifying
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(p-tert-butylphenol)‚ and a neutral compound (acetanilide). Extraction is the process of selectively dissolving one or more of the compounds of a mixture into an appropriate solvent‚ the solution that contains these dissolved compounds is called an extract (Manion‚ 2004). Impurities that are present in the solution can be removed by extracting them from the original solvent into another solvent. This is done by mixing two immiscible (insoluble to one another) solvents (Manion‚ 2004). By mixing
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consists of two steps: (1) thoroughly mixing the solution with saturated aqueous sodium chloride (saturated salt solution) and discarding the aqueous layer (this is a preliminary drying step that removes most of the water) and (2)‚ adding a solid inorganic drying agent (this strongly binds to the remaining traces of water dissolved in the organic solvent)‚ allowing it to remain in contact with the solution for several minutes‚ then separating the solution from the solid drying agent. The solvent can
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current through a salt solution and measuring the resultant amperage‚ a pattern in the recordings will help find a soil sample’s salinity level. Variables Controlled Amount of solution (100 mL)‚ supplied voltage (4V)‚ amount of soil (50 gm) & water (150 mL) in soil sample’s solution. Independent Amps recorded by ammeter. Uncontrollable Impurities dissolved into water. Materials Power pack; copper electrodes; ammeter; 8 different concentrations of salt solution (NaCl) 0.1‚ 0.5‚ 1‚
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