and the final change in temperature for urea was 2.09 °C. To cause these results‚ ammonium chloride‚ which is a salt of a weak base‚ more than likely dissociated into ammonium ions and chlorine ions‚ resulting in twice as many more particles in solution than urea‚ which probably doesn’t dissociate much or at all‚ in the water/ice
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them. When they are heated this water evaporates‚ the crystals become dehydrated and turn white. Equation for this Reaction. CuSo4 5H2O ---> CuSo4 + 5H2O Solute- Solid which is dissolving Solvent- liquid in which the solute dissolves Solution- the solute and solvent mixed together Sodium hydroxide- Soluble Copper Oxide- insoluble Calcium hydroxide- Insoluble Silver iodide- insoluble Aluminium nitrate- soluble Most ionic substances will dissolve in water but covalent substances
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as medicine where those substances might be used as medications Crystallization is based on the principles of solubility: compounds (solutes) tend to be more soluble in hot liquids (solvents) than they are in cold liquids. If a saturated hot solution is allowed to cool‚ the solute is no longer soluble in the solvent and forms crystals of pure compound. Impurities are excluded from the growing crystals and the pure solid crystals can be separated from the dissolved impurities by filtration. Chemical
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dissolve in 100ml at 40 C because the temperature would have to be at least 69 C for it to dissolve completely. 8. In order to make a saturated solution at 55 C you must have 70 g of KNO3 9. You would need about 14g at the min. to grow a crystal. For best results the temp. Should be heated up to 20 C to get that KNO3 to crystallize. Certain solutions grow different things due to the amount of chemicals and heat is applied. 10. Conclusion: In conclusion we found that the more KNO3 you add the
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ZOOL 1 Lecture 3 1 2 ¡ ¡ ¡ ¡ Chemical substances that cannot be broken down to simpler forms by ordinary chemical reactions Atom Atomic number Atomic mass Compound Molecule Macromolecule 3 ¡ ¡ ¡ ELEMENT Oxygen Carbon Hydrogen Nitrogen Calcium Phosphorus Potassium Sulfur
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1molecule = 2.8 x 1016 molecules 2.1 x 10-15 cm² 3. Moles of stearic acid in monolayer Average drops of stearic acid to make up the monolayer = 6 Average drops of stearic solution in 1 mL = 96 ml of stearic acid in monolayer = 6 / 96 = .0625 ml
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(Legend) Figure 1. The concentration-response curve to ACh in the absence of a competitive antagonist. Each dose of ACh was added to an organ bath containing a section of Guinea Pig ileum and Tyrode’s solution at physiological temperature. The response to ACh was measured by the isotonic contractions of the ileum until the maximum response was reached for the dose‚ using an isotonic force transducer with metal counter-weight of approximately 0.5g. A wash
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the 0.400M solution‚ the potato decreased in mass‚ this was due to the reason that the solution was hypertonic. There was a higher concentration of solute and lower concentration of water in the solution than there was in the potato cells. This led the water to travel down its concentration gradient‚ which was from a higher concentration (in the cell) to a lower concentration (in the solution)‚ which meant the potato cells lost water‚ therefore the potato lost mass. The 0.300M solution was also a
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Using Sucrose Solutions to Determine Osmolarity of Potato Tubers Based on Weight By Thomas Pelikan Biology 200A Section 004 Kimberly Schmidt October 2‚ 2012 Abstract: In this experiment we were trying to determine the osmolarity of potato tubers by weighing them before and after incubating them in solutions of sucrose with varying molarities. To find the osmolarity we took a potato and used a cork borer to obtain seven samples of potato tubers. We then prepared seven beakers with concentrations
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to measure the concentration of a specific chemical species in a solution. This was conducted in 2 stages. The first stage involved making a solution of potassium iodide. The potassium iodide was then diluted to several different concentrations that were ran in potentiometer to find the mV of each solution which would then be used to find the PH of the solution. After analyzing a sample of each concentration an unknown solution was compared. Along with the the unknown tap water and deionized water
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