An oil sample had its iodine number determined using the pyridinium tribromide method. The sample and reagents used are as follows: Oil sample: 0.20 mL, density: 0.958 g/mL Pyridinium tribromide: 0.486 g Na2S203: 0.1 M, endpoint 42.17 mL %3D What is the approximate Bromine Number?
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- A wittig reaction experiment: Week 1 -》 preparation of phosphonium salt Materials : triphenylphosphine (5.3g),methyl Bromoacetate 3.36g (2.1ml) , ethanol 30ml . The yield obtained was 8.93g . Calculate the percentage yield Week 2 -》 Formation of the tlide and wittig reaction Materials :Phosphonium salt (5g), napthyl-2- carboxwaldehyde( 2.65g), 20 ml of water and 5 nl of Nahco3. Yield obtained was 0.51g .Calculate the percentage yield Week 3: Solvent free wittig reaction Materials: Benzyltriphenylphosphonium chloride (0.5g ), 4 - Bromoabenzaldehyde (0.24g), Potassium phosphate (tribasi c) 0.275g .The yield obtained was 1.21g. Calculate the percentage yieldSome people have an allergic reaction to the food preservative sulfite (SO32−). Sulfite in wine was measured by the following procedure: To 50.0 mL of wine were added 5.00 mL of solution containing (0.8043 g KIO3 + 5 g KI)/100 mL. Acidification with 1.0 mL of 6.0 M H2SO4 quantitatively converted IO3− into I3−. The I3− reacted with SO32− to generate SO42−, leaving excess I3− in solution. The excess I3− required 12.86 mL of 0.04818 M Na2S2O3 to reach a starch end point. lO_3^- + 8l^- + 6H^+ rightarrow 3l_3^- + 3H_2O l_3^- + SO_3^2 - + H_2O rightarrow 3l^- + SO_4^2- + 2H^+ l_3^- + 2S_2O_3^2- rightarrow 3l^- + S_4O_6^2- Find the concentration of sulfite in the wine. Express your answer in mol/L and in mg SO32− per liter._____mol/L HINTS: Don't use scientific notation, do put zero before decimal point, carefully evaluate significant figures.What is Solubility of phenazopyridine in water?acidic media? basic media ?and what is most common solvent?
- Five white, 500-mg uncoated ascorbic acid (AA) tablets with an average weight of 0.6152-g were pulverized in a mortar. A sample of the powdered ascorbic acid weighing 0.4700-g was placed in an iodine flask and was dissolved in 50-mL H2SO4 then 5-g of KBr was added to the resulting solution. The solution was titrated with 47.81-mL of 0.09640 N STD. KBrO3 to reach a faint yellow endpoint then 3-g KI and 5-mL Starch TS. The blue color solution is then titrated with 2.73-mL of 0.09123 N STD. Na2S2O3 to reach the disappearance of the blue iodostarch complex. MW: KBrO3 = 167.0 ; KIO3 = 214.0 ; Na2S2O3 = 158.11 ; C6H8O6 = 176.12 Compute the milligrams of pure AA per tablet from the assay. 293.3 mg 502.5 mg None of the choices 383.9 mg(9.29 × 105/8)-20.81 = with correct sig figsPlease show all the calculations. I have these information available: 32.00 mL of 1 butanol and 20.00 mL glacial acetic acid were in the 100 mL distilling flask. Initial Aliquot:1 mL and add around 20 mL of water.Volume of NaoH required for titration: 67.3 mL After Aliquot:1 mL and add around 20 mL of water.Volume of NaOH required for titration: 23.4 mL Concentration of NaOH: 0.0972 M
- Calculate the final molarity of H2O2H2O2 if 1.3 mL1.3 mL of the 3.0% w/w H2O23.0% w/w H2O2 solution is diluted with 3.7 mL3.7 mL water, then added to 5.0 mL5.0 mL of a starch-iodide solution.A cyanide solution with a volume of 12.02 mL12.02 mL was treated with 20.00 mL20.00 mL of Ni2+Ni2+ solution (containing excess Ni2+Ni2+) to convert the cyanide into tetracyanonickelate(II): 4CN−+Ni2+⟶Ni(CN)2−44CN−+Ni2+⟶Ni(CN)42− The excess Ni2+Ni2+ was then titrated with 11.49 mL11.49 mL of 0.01257 M0.01257 M ethylenediaminetetraacetic acid (EDTA): Ni2++EDTA4−⟶Ni(EDTA)2−Ni2++EDTA4−⟶Ni(EDTA)2− Ni(CN)2−4Ni(CN)42− does not react with EDTA. If 39.02 mL39.02 mL of EDTA were required to react with 30.34 mL30.34 mL of the original Ni2+Ni2+ solution, calculate the molarity of CN−CN− in the 12.02 mL12.02 mL cyanide sample. [CN−]= MThe amount of ascorbicacid ,in aorange juice was determined by oxidzing the ascorbic acid to dehydroascorbic acid , with a known excess I3 , and back titrating the excess I3 with na2s3o4. A 5 ml sample of filtered orange juice was treated with 50ml of excess 0.01023mol/l. After the oxidation was complete, 13,72 ml of 0,07203 mol/l na2s204 was needed to reach the starch indicator end point. Report the concentration acid in miligrams per 100ml
- Procaine hydrochloride (MW = 273; L iso value = 3.37). Calculate its NaCl Equivalent 0.24 0.19 0.18 0.21Following the monograph procedure, a 724-mg of aspirin (MW+180 g/mol) dissolved in 18.5 ml of cold neutralized alcohol. This solution was then initially titrated with 0.101 N sodium hydroxide solution, then later neutralized with 0.104 sulfuric acid. 1. What is the milliequivalent weight consumed by the acidic titrant? a. 1.5392 g-meq b. 2.0907 g-meq c. 1.4948 g-meq d. 2.1528 g-meq 2. What is the milliequivalent weight consumed by the basic titrant? a. 5.8656 g-meq b. 1.5392 g-meq c. 5.6964 g-meq d. 1.4948 g-meq 3. What is the difference of milliequivalent weight consumed in the reaction? a. -4.1572 g-meq b. 4.3708 g-meq c. 0.5515 g-meq d. 4.1572 g-meqA 0.9134-g of KIO3 is dissolved in 500-mL distilled water. A 50.0-mL aliquot portion was run down into an Erlenmeyer flask and 2-g of KI and 2-mL of 6M HCl were added to it. Directly after, the solution was titrated with a 24.47-mL Na2S2O3 solution to a faint yellow solution. Starch TS was added to the resulting solution and the titration with the Na2S2O3 solution continued wherein an additional 1.33-mL was required to bring the blue color of the solution to disappear completely. MW: KIO3 = 214.0 g/mol ; Na2S2O3 = 158.1 g/mol Compute the M of the Na2S2O3 solution. 0.1047 M None of the choices 0.1107 M 0.09926 M
