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- 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??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 acidOral 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.
- Volume of acetic anhydride: 2.98 mL Mass of salicylic acid: 1.00 g Mass of aspirin recovered: 0.67 g Show that mmol of aspirin recovered divided by mmol of limiting reagent x 100% gives the same % yield (please explain) Thank you.The separation and purification processes given below and the method used against them are given. Which or which of these pairings are correct? Brewing the tea - Extraction Purification of water from impurities-Chromatography Separating olive pomace from oil while olive oil is being produced - Decantation Obtaining oil fractions - Ordinary distillation Obtaining essential oils and using them in perfume making - Water-steam distillation A. I, IV, V B. I, II, III, IV, V C. I, II, V D. I, II, III, V E. I, II, III, IVA 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?
- Q3/ A) 0.63 g of a sample containing Na CO,, NaHCO; and inert impurities is titrated with 0.2 M HCI, requiring 17.2 ml. to reach the phenolphthalein end point and a total of 43.5 mL to reach the modified methyl orange end point, How many grams Na CO, and NaHCO, are in the mixture?19. 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.Dissolve 0.25g NaClI, 0.20g NaHCO3. 0.35g KCI, and 1.5g of glucose in 100 ml distilled water using a 250 mL capacity beaker. Use the formula to create a 60 mL ORS. Based on the given procedure, calculate the total amount of glucose in the prepared solution. (MW: Na: 23, K: 39, CI: 35, H: 1, C: 12, 0: 16) O 83.33 mmol/L 66.91 mmol/L 90.40 mmol/L 71.11 mmol/L O 23.81 mmol/L O 47.30 mmol/L
- The boiling points for the compounds are 118 °C and 35 °C respectively. The solubility for both compounds is the same (8g/100g water). Explain this observation for (i) boiling point disparity; (ii) solubility similarity H-bonds form in diethyl ether; n-butyl alcohol forms H-bonds in water H-bonds form in n-butyl alcohol; diethyl ether forms H-bonds in water H-bonds in n-butyl alcohol; Both compounds form H-bonds in water O Both compounds form H-bonds; Both compounds form H-bonds in waterHexanoic acid was added to an immiscible biphasic solvent system, water and CCl4 at 20.0OC and the equilibrium concentrations of hexanoic acid were determined to be 3.66 g/L in H2O and 67.0 g/L in CCl4. Caluclate the distrubution coeffiecent (D2) of hexanoic acid in water with respect to CCl4.Calculate the amount of potassium chloride needed to prepare a 60 mL ORS based on the given procedure: Dissolve 0.25g NaCi, 0.20g NaHCO3, 0.35g KCI, and 1.5g of glucose in 100 ml distilled water using a 250 mL capacity beaker. (MW: Na: 23, K: 39, CI: 35, H: 1, C: 12, O: 16) 0.9 g O 0.12 g 0.15 g O 0.21 g