The distribution coefficient of an organic compound Y between hexane and water is 4.73. If 192.7 mg/L of Y is initially present in 80.0 mL water, what will be the extracted fraction of Y after four 10.00-mL portions hexane 0.175 0.825 none of the choices 0.844 0.156
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- 3. Which of the following statement/s is/are correct? 1. 6.00 N H2CO3 is equivalent to 3.00M H2CO3. II. 5.29 % (w/v) H2SO4 is equivalent to 0.539 N H2SO4 II. 1.15 N Mg(OH)2 is equivalent to 16.3 g Mg(OH)2 in 500 mL solution. (MH2S04 = 98.086 g/mol; MH2C03 = 62.026 g/mol; MMg(OH)2 = 58.326 g/mol) %3DQ3/ 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?if all the N in 10.0 mmols urea, CO(NH2)2, is converted to NH4HSO4, and if, with excess NaOH, the NH3 is evolved and caught in 50.0 mL of HCl (1.00 mL = 0.03000 g CaCO3), what volume of NaOH (1.00 mL = 0.3465 g H2C2O4. H2O) would be required for complete titration? using Kjeldahl's Method
- If all the N in 10.0 mmols urea, CO(NH2)2, is converted to NH4HSO4, and if, with excess NaOH, the NH3 is evolved and caught in 50.0 mL of HCl (1.00 mL = 0.03000 g CaCO3), what volume of NaOH (1.00 mL = 0.3465 g H2C2O4. H2O) would be required for complete titration?(72 l l ..:01:EP is متعد د الخيارات What weight is needed to prepare solution of 0.15 N from Na2CO3 in 300 ml of distilled water. Na = 23, C = 12, 0 = 16 %3D 2.385g 3.285g 5.82ml 2.385ml6.3017g sample was weighted Analytically and made up volumetrically to 500mL. A 25mL alliqout of this solution required a total valume of 41.32mL of 0.0987MHCl to reach bromocresol green endpoint Another 25ml alligout of this salution was treated with 25mL of 0.1014M NaOH react with excess of BaCl₂ solution und left over NaOH required 21.83mL of 0.0987MHCl to reach the phenolpthelin end point. Determine the percentage of Na₂CO3 and. NaHCO3
- If all the N in 10.0 mmols urea, CO(NH2)2, is converted to NH4HSO4, and if with excess NaOH the NH3 is evolved and caught in 50.0 mL of HCl (1.00 mL ≈ 0.03000 g CaCO3), what volume of NaOH (1.00 mL ≈ 0.3465 g H2C2O4.2H2O) would be required for complete titration?A 0.3005g sample containing Na2CO3 and NaOH requires 11.5 mL of 0.1055 M HCl to reach the phenolphthalein end point. Methyl orange is then added and 7.9 mL more of 0.1055 M HCl is required to reach the end point. Calculate the % composition of Na2CO3 and NaOH in the sample. a. 5.15% NaOH and 29.55% Na2CO3 b. 5.05% NaOH and 29.55% Na2CO3 c. 5.15% NaOH and 29.40% Na2CO3 d. 5.05% NaOH and 29.40% Na2CO3How many milliliters of a thimerosal 25% stock solution must be used to prepare 100 ml of thimerosal topical solution of 1:1000 ratio strength? A diphenhydramine elixir contains 12.5 mg drug in one teaspoon. The volume of oral vehicle that is to be added to 100 ml of this elixir to reduce its strength by one half its original strength is: When preparing an ounce of 1% hydrocortisone cream the technician made an error and added 200 mg of hydrocortisone in the preparation. Calculate the additional amount of hydrocortisone powder that should be added to the 1 ounce cream to make the strength 1%. Onabotulinumtoxin A (BOTOX) is available in a 100 units vial for reconstitution. 2.5 ml of sterile preservative free normal saline was added to this vial and reconstituted. How many units of botox is present in 0.1 ml of the dilution? (ignore any volume changes during reconstitution)
- A mixture contains Na2CO3, NaOH and inert matter. A sample weighing 1.500 g requires 28.85 ml of 0.500N HCI to reach a phenolphthalein endpoint, and an additional 23.85 to reach a methyl orange endpoint.What are the percentages of Na2C03 and NaOH?Apple seeds contain amygdalin, a substance that releases cyanide into theblood stream. if 3.5 kg of apple seeds contain 750 milligram of amygdalin, theconcentration of cyanide in ppm will be.. * Non of these O 214.28 0.0046 O 4.66 O 0.214When 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 acid