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- A student performs a crystallization on an impure sample of biphenyl. The sample weighs 0.5 g and contains about 5% impurity. Based on his knowledge of solubility, the student decides to use benzene as the solvent. After crystallization, the crystals are dried and the nal weight is found to be 0.02 g. Assume that all steps in the crystallization are performed correctly, there are no spills, and the student lost very little solid on any glassware or in any of the transfers. Why is the recovery so low?You need to purify 0.8 grams of an impure sample of trans-cinnamic acid. After the purification is completed you isolate 0.57 grams of trans-cinnamic acid and record a melting point range of 129-132 °C. Cinnamic Acid has a solubility in ethanol of 13.9 g/100 ml at 10 °C and a solubility in water of 0.055 g/100 ml at 25 °C. (a) Explain why the recrystallization of trans-cinnamic acid must use a mixed solvent system. Please explain why the sample couldn't be successful recrystallized in ethanol alone. Calculations should be used to support your answer. (b) Calculate the % recovery and the % error for the melting point.How much drying agent, MgSO4, would be needed for drying of 20mL dichloromethane solution that contains 0.01% water, knowing that each mole of MgSO, can absorb 7 moles of H;0?
- When a Vitamin C (ascorbic acid; MM = 176.12 g mol-1) tablet is crushed, dissolved and titrated with 0.0340 M KIO3(aq) to a purple/blue endpoint (given by a starch indicator), the volume of KIO3 used is 29.80 mL. If 60 mg of ascorbic acid is the recommended dietary allowance (i.e., 100% of the RDA), then what is the % RDA for the Vitamin C in the tablet? KIO3(aq) + 5 KI + 6 H+ → 3 I2(aq) + 3 H2O I2 (aq) + ascorbic acid → 2 I- + dehydroascorbic acid19. A liquefied mixture of n-butane, n-pentane, and n-hexane has the following composition: n-C,H10 50%, n-C,H12 30%, and n-C,H14 20%. For this mixture, calculate: a) The weight fraction of each component. b) The mole fraction of each component. c) The mole percent of each component. d) The average molecular weight of the mixture.The mixture contains benzoic acid, 2-napthol, and naphthalene. Dissolve the mixture in the polar solvent ether. When you react the mixture with 10% NaHCO3(sodium bicarbonatea weak base in aqueous solution) reacts with benzoic acid and forms sodium benzoate an ionic compound (similar to NaCl) that dissolves in aqueous solution (heavier than ether in density) and goes to the bottom of the centrifuge tube or to the bottom of the separatory funnel forming two layers. Then remove the bottom aqueous layer and collect in a beaker labeled as bicarbonate extract. Acidify the bicarbonate extract in the beaker to give benzoic acid. The top ethereal layer contains in the centrifuge tube or separatory funnel 2-napthol and naphthalene. To this top layer add 10% NaOH (sodium hydroxideis a strong base in aqueous solution reacts with only 2-napthol, a weak acid forming 2-napthoxide an ionic compound soluble in aqueous solution (similar to NaCl) that goes to the bottom of the centrifuge tube or to…
- for the following, should i use fractional or simple distillation?separation of ethylene glycol (Tb = 195 oC) from acetone (Tb = 56 oC)5 g of cobalt carbonate and 13 mL of concentrated hydrochloric acid were used. In another glass, prepare a solution with 8.4 mL of anhydrous ethylenediamine and 2.8 mL of concentrated hydrochloric acid. They are mixed and added and 6.5 mL of peroxide were added to it.30% hydrogen, thenconcentrated hydrochloric acid is added, followed by60 mL of ethanol. 5.1234 g were obtained orange solid [Co (en) 3] Cl3 What is the percentage yield of the reaction. The reactions are shown in the images.How many grams of Ba(103)2 (487 g/mol) can be dissolved in 400 mL of water at 25°C. (Ksp of Ba(1O3)2 = 1.57 x 10-9) Ba(103)2 Ba2+ + 2103 O 0.298 g 0.122 g 0.221 g O 0.926 g O 0.143 g
- Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample. The stated should be 104.5, how??Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample.2. How many mL of a 0.250 M KCI solution must be diluted to 1.000 L so that the diluted solution (density = 1.00 g/mL ) is 400 ppm K* by weight ? (MW KCI = 74.55; AW K = 39.10)