osmosis and diffusion is that they both equalize the concentration of the two solutions into a membrane. Osmosis and diffusion both work together to move water molecules from an area of high concentration to an area of low concentration. The
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compound. Introduction The solubility product constant‚ Ksp‚ is a particular type of equilibrium constant. The equilibrium is formed when an ionic solid dissolves in water to form a saturated solution. The equilibrium exists between the aqueous ions and the undissolved solid. A saturated solution contains the maximum concentration of ions of the substance that can dissolve at the solution’s temperature. The equilibrium equation showing the ionic solid lead chloride dissolving in water is:
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the properties unique to solutions depend only on the number of dissolved particles and not their identity. Such properties are called colligative properties. The colligative property that will be examined in this experiment will be the freezing point depression as an example of a colligative property. Every liquid has a freezing point: the temperature at which a liquid undergoes a phase change from liquid to solid. When solutes are added to a liquid‚ forming a solution‚ the solute molecules disrupt
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being the highest point with a change of mass of 1.75% finishing with the 10% solution on a decline of -12.28%. From observing the scatter plot the data points were not far from the line of best fit‚ indicating the results were hardly affected by random errors. Resulting in fairly precise and reliable results. The hypothesis stated ‘As the concentration of sodium chloride increases the mass of the potato in the solution will decrease.’ This was supported as with the increase of sodium chloride concentration
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Write down the ionic chemical equation when the following pairs of solutions are mixed together. (Take note of their solubilities of the products formed) Question 1. riz Remember that all ionic compounds are soluble to an extent; those classi ed
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kg 0.29367OsmkgH2O=0.226gM*9.876*10-3kg solvent 0.29367OsmkgH2O=22.88 gM*kgH2O M=22.880.29367 M=77.91gmol Based on the molar mass of calculated‚ the identity of the unknown substance D is Glycine. 2) Finding the concentration of a known solution Table 1.2: Table used for the preparation of the calibration curve used to find the concentration of alcohol in a wine. 0.2mL of each sample were taken and placed in the Osometer. Ethanol sample (g EtOH/100mL) | Osmolality (mOsm/kg H2O) | 5
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tablespoons * Green - 8 tablespoons 3. Label each cup with the amount of sugar added. 4. Stir the water until the sugar is dissolved. It is crutial that all of the sugar is dissolved in each cup. You may need to supersaturate the sugar water solution to get all of the sugar to dissolve. Place the cup in the microwave for 20-30 seconds to warm the water and dissolve more sugar. Continue stirring until all of the sugar is gone. 5. Start with the cup with the most sugar. Using a pipette‚ dropper
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NaCl‚ the ions become surrounded by the solvent‚ say H2O(water) molecules. The ions are said to be ’solvated’ as they become surrounded by the solvent‚ similar to a hoard of bees surrounding a nest. The solvation process helps stabilize the ions in solution and prevents cations and anions from recombining. Furthermore‚ because the ions and their shells of surrounding water molecules are free to move about‚ the
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of succinate broken down by the mitochondrial solution. We detected the amount of DPIP in the solution with a spectrophotometer and measuring the absorbance of light at the 600nm range. DPIP is a useful chemical to use in this experiment because it goes from a blue color when oxidized to a colorless liquid (Ogura‚ 281)‚ this is due to the hydrogen ions and electrons released during the transitional step between succinate and fumarate. The three solutions used contained the same amount of mitochondrial
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variable will be the sucrose concentration of water. This is controlled by using a specific amount of sugar that we pour into the distilled water. We will do the activity in the solutions with concentration of 0‚ 0.25‚ 0.50%. Dependent Variable The dependent variable will be the size of potato slices that goes into the solutions. This depends on the various environmental factors during making a slice of potato such as the thickness of core or the peel of potato. Thus it could differ during the experiment
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