The ionic strength of a solution containing 0.3 M sucrose and 0.6 M sodium Phenobarbital (1:1) electrolyte, and 0.1 M sodium acetate (1:1) is: a. 0.6 b. 0.7 c. 0.2 d. 0.3 e. 0.5
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- To prepare a stock solution of EDTA, 73 grams of EDTA were mixed to 500 mL of distilled water. What is the: A. Molar concentration of EDTA stock solution is Blank 1M Using the prepared EDTA stock solution above, 25 mL of 200 mM working solution has to be made. Calculate for the following: B. Volume of EDTA stock solution to be diluted is Blank 2mL C. Volume of distilled water is Blank 3mLAn equimolar solution of ammonium sulfate, sodium chloride, and magnesium nitrate has an ionic strength of 0.15M. What is molar concentration of the solution wiht respect to each solute? a. 0.010 b. 0.015 c. 0.020 d. 0.025A sodium thiosulfate solution can be standardized by using it to titrate the iodine liberated by the action of excess KI on a known weight of primary standard K2Cr2O7: Cr2O72- + 6I- + 14H+ → 2 Cr3+ + 3I2 + 7H2O I2 + 2S2O32- → 2I- + S4O62- Calculate the molar concentration of the sodium thiosulfate solution if 31.47 mL of this solution were required to titrate a sample prepared using 0.2177 g pure K2Cr2O7.
- a) The value for Ψ in root tissue was found to be -3.3 bars. If you take the root tissue and place it in a 0.1 M solution of sucrose at 20°C in an open beaker, what is the Ψ of the solution, and in which direction would the net flow of water be? b) NaCl dissociates into 2 particles in water: Na+ and Cl-. If the solution in question 4 contained 0.1M NaCl instead of 0.1M sucrose, what is the Ψ of the solution, and in which direction would the net flow of water be? c) A plant cell with a Ψs of -7.5 bars keeps a constant volume when immersed in an open-beaker solution that has a Ψs of -4 bars. What is the cell’s ΨP?A stock salt solution is prepared by dissolving 5.69 g of table salt in water and diluting the mixture to 150 mL with water. A 15 mL of the stock salt solution is then taken and diluted to 150 mL with water. The concentration of second solution is ___ the strength of the stock salt solution.Calculate the ionic strength of a solution made by mixing 100.0 mL of 0.0200 M strontium nitrate solution, 100.0 mL of 0.0300 M potassium nitrate, and 0.625 grams of sodium thiosulfate (if you don’t know what the thiosulfate ion is, look it up). Assume complete dissociation and complete solubility of all species. Think carefully about the concentrations. Drawing a picture will help!
- Why is it less desirable to wash AgCl precipitate with aqueous NaNO3 than with HNO3 solution? (Select all that apply.) A) HNO3 evaporates during drying. B)HNO3 is nonvolatile and will lead to a high mass for the precipitate. C )NNO3 is and will lead to a high mass for the precipitate. D)NaNO3 evaporates during drying.Thallium(I) iodate (TlIO3) is only slightly soluble in water. Its Ksp at 25°C is 3.07 × 10-6 . Estimate the solubility of thallium(I) iodate in water in units of grams per 100.0 mL of water.Calculate the Ksp for hydroxide if the solubility of Mn(OH)2 in pure water is 7.18 × 10-1 g/L. Group of answer choices 2.10 × 10-6 5.50 × 10-11 8.07 × 10-3 7.18 × 10-1 5.25 × 10-7
- A 1L buffer solution needs to be prepared. The buffer will be composed of the following, 200mM tris, 400mM glycine, and 0.5% (w/v) glycerol. The stock reagents for tris (124.14g/mol) and glycine (75.07g/mol) are stored as separate lyophilized powders. Glycerol is stored as a liquid at 20% (w/v). Approximately how many mL of glycerol should be measured for this buffer?Shan wanted to extract 3.12g of a new compound from 150mL of water using chloroform. She determined that the solubility of the compound is 0.019g/L and 0.276g/L in water and in chloroform, respectively. Determine: Kd value of the compound between chloroform and water. Amount of the compound that would be extracted if Shan performed a one-step extraction using 75.0mL of chloroform Amount of the compound that would be extracted after the first step if Shan performed a two-step extraction with 37.5mL of chloroform each Amount of the compound that would be extracted after the second step Percent efficiency of the two-step extraction compared to the one-step extractionSeveral amount of unknown salt was added to a solvent. Initial amount of the solvent is 30 ml. When the mixture weighed 0.2202 lb, undissolved salts were observed. Upon separation of the salts, it was found to be around 5 g. What is the solubility of the unknown salt? Before separating the undissolved solute from the mixture, was the solution supersaturated and why? (Density of solvent = 1.02 g/ml)