to determine the equilibrium constant for the reaction Fe3+(aq) + HSCN(aq) –>FeSCN2+(aq) + H+(aq). The equilibrium constant expression Kc for Reaction is kc=FeSCN2+[H+]Fe3++[HSCN] Procedure *Preparation of the Beer’s law plot Prepare five solutions of FeSCN2+(aq) of known concentrations between 1x10-5M and 1x10-4M by diluting various volumes of 4.62x10-4 HSCN. Calculate the Final concentration FeSCN2+ for Beer’s law Using the C1V1=C2V2 Initial Concentration (M) | Initial Volume(ml) | Final
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conduct my experiment‚ I measured 10g of ascorbic acid powder and mix it within 1 litre of water. I believed that by having the solution prepared before putting in the d-block elements would save time overall. Unfortunately‚ the data was all over the place‚ this is because the later I conduct the experiment‚ there is a higher chance of oxidation occur in the ascorbic acid solution as it have a direct contact with oxygen in the atmosphere. In order to keep constancy‚ I had to reduce the amount of ascorbic
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buffer was made. We began this experiment by measuring seven constant amounts of 1ml of 0.1% catechol using a 1 mL pipet into each seven cuvettes. The catechol is our substrate solution. Next‚ different amounts of phosphate buffer ph.6 of 0.5ml‚ 1.0ml‚ 1.5ml‚ 2.0 ml 2.5ml‚ 3.0ml and 4.0ml were added to each of the solution of the 0.1% catechol. The purpose of adding phosphate buffer pH6 is to maintain the particular ph6 for the polyphenol-oxidase (enzyme)‚
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CBI Holding Company‚ Inc. Case Solution I. Summary A CBI Holding Company was a New-York based parent company for several wholly-owned subsidiaries. These marketed an extensive line of pharmaceutical products that were purchased from drug manufactures‚ warehoused in storage facilities and then resold to retail pharmacies‚ hospitals long-term care facilities and related entities. CBI’s chairman and president Robert Castello was the seat of the CBI‘s troubles and principal implementer of
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are avoiding put-downs‚ connecting solutions‚ listening for data‚ and summarizing ideas. Being able to overcome these challenges will bring about successful group discussions. Challenge 1: Avoiding put-downs are extremely important. Put-downs occur when members of the group ridicule the ideas that are presented by other members. When this happens‚ it can slow down the productivity of the group‚ and can stop other members for voicing their own ideas or solutions. One way to solve this problem is
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cups with two hundred fifty milliliters of water. Then‚ mixing teaspoons of salt in water one at a time‚ until you’re able to submerge a raw egg in the solution and have it float up. My hypothesis is‚ salt does infact change water’s density‚ and if the egg floats it’s proof of density change in the water‚ because it must be less dense than the solution in order to float. The procedure is as follows‚ I filled two cups with two hundred fifty milliliters of water and mixed four teaspoons of salt in one
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lose weight as the potatoes would have lower water potential when placed in a solution ‚ and in this can be taken from the theory of osmosis which states that “the passive diffusion of water molecules across a selectively permeable membrane down a concentration gradient” (Taylor and Bruenn‚ 2009)‚ moving water from a high to a low concentration‚ so in this case water would move from the potato tuber to the sucrose solution reducing the weight of potatoes. Pospíšilová (1970) also states that the water
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concentration gradient the diffusion that did occur happened a lot faster and diffused more efficiently compared to no iodine solution and just water. This is due to the molecular collisions speeding up the experiment in the time period given because of the Soule molecules found within the given volume. This experiment can also refer to hypo‚ hyper and isotonic due to the two solutions being separated by the dialysis tubing (which acts as a semipermeable membrane) and diffusing from
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using Osmosis. I predict that the potato in the higher concentration of salt solution will lose an increasingly high amount of weight. I think this because the higher the concentration in salt‚ the more water particles will move through the potato cell’s semi-permeable membranes to try and balance the salt concentration. For example‚ if you placed a larger piece of potato into a beaker of an almost 100% salt solution; the potato would completely shrivel up because it would lose all of the water
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solute particles is greater than the other solution. Core B was isotonic meaning that it had an equal balance throughout. When two environments are isotonic it means that the total molar concentration of dissolved solutes is the same in both of them (equal). When cells are in an isotonic solution‚ movement of water out of the cell is exactly balanced by movement of water into the cell. Core C was the most flexible which made it be hypotonic. A hypotonic solution expresses that the total concentration
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