An oil sample had its iodine number determined using the pyridinium tribromide method. The sample and reagents used are as follows: Oil sample: 0.19 mL, density: 0.954 g/mL Pyridinium tribromide: 0.472 g Na2S203: 0.1 M, endpoint 27.68 mL What is the estimated lodine Number?
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- You were assigned to assay a product sample of milk of magnesia. A 0.600-g sample was reacted with 25.00 mL 0.10590 N H2SO4 . The excess unreacted acid in the solution required 13.00 mL of 0.09500 N NaOH when titrated to reach the methyl red end point. Determine the dosage strength of the product in terms of % Mg(OH)2 content. Type your answer in 2 decimal places, numbers only.You're at the same pharmaceutical company, and they also make magnesium pills. You weigh out 5 tablets (3.2834 grams). And crush them with a mortar and pestle. You take a subsample of 384 mg and dissolve it in acid, making it to a final volume of 250mL. • You then take 50 mL and dilute it to 250mL volumetrically. • You further dilute the sample by taking 15 ml and diluting it to 250mL. • You run this on a varian AA with your standards and get the following response: Response (ABS) 0.546 1.134 2.135 1.634 ● ● Solution 1.0 ppm Mg 2.0 ppm Mg 4.0 ppm Mg Sample Calculate the amount of Mg in one pill. Line of best fit y = 0.5255x+0.0455R 400m N CO₂tBu OH CO₂Me Ph LDA THF, -78 °C silica gel 6 days room temperature R MeO₂C NCO₂Bu CHO CO,Me I + H₂O
- 1.What is the molarity of a solution made from placing 0.2200 g tartrazine powder in a 1L volumetric flaskand diluting it to the mark? Hint: the solution’s total volume is 1.000 L. 2. What is the concentration of a solution that takes 10.00 mL of a stock solution (4.12 × 10-4 M) and diluteit to 100.00 mL.aching and lear x m/takeAssignment/takeCovalentActivity.do?locator=assignment-take f6 6 Your one-time code-escobedoa X G how do you cover a ml to liter - X + Use the References to access important values if needed for this question. In our bodies, sugar is broken down with oxygen to produce water and carbon dioxide. According to the following reaction, how many grams of oxygen gas are necessary to form 0.705 moles carbon dioxide? C6H12O6 (s) + 602 (g) → 6CO₂(g) + 6H₂O(2) Submit Answer & g oxygen gas 7 Q. Search LD COGNOF fg Retry Entire Group * त 4+ 8 fg hp KAA 2 more group attempts remaining 9 f10 ►I O 104 f11 12 ✈ + ins prt sc ← delete backspace Next > home O num lock 海口 6:54 PM 5/18/2023 endIn the preparation of a standard stock solution for protein analysis, 50 mg of bovine serum albumin (BSA) was dissolved in deionized distilled water to a final volume of 10 mL. What is the weight of BSA in 0.60 mL stock solution?
- this type of analysis refers to the determination of the identity of the analyte in the sampleA student weighs out exactly 1.562 grams of p-aminophenol and exactly 40 mL of water. After dissolving p-aminophenol in HCl, 10 mL of acetic anhydride is added. The student follows the procedure and acetaminophen is prepared. What is the maximum amount of acetaminophen she could obtain from the reaction if the percentage yield is 76%?7. Sodium nitrite with antipyrine solution and hydrochloric acid forms? A) Emerald green coloration B) Blue staining C) Brown sediment. D) White sediment E) Blue-violet coloration
- A solution of potassium permanganate of unknown concentration was analyzed against a standard of various concentrations. Refer to the following data set to answer the succeeding questions (MM KMNO4 = 158.034 g/mol). Table 1. Data for Standard Calibration Curve KMN04 Standard Absorbance at 525 nm (grams/Liter) 0.008 0.10437 0.158 0.20197 0.316 0.45288 0.474 0.68232 0.632 0.89361 Table 2. Data for KMNO4 Solution of Unknown Conc. Sample of Unknown Conc. Absorbance at 525 nm Undiluted 0.95423 KMNO4 Diluted KMNO4 0.46514 (50% dilution) What is the concentration of the undiluted KMNO4 solution? 6.872 g/L 1.6872 g/L 0.4872 g/L 0.6872 g/LThe piperazine content of an impure commercial material can be determined by precipitating and weighing the diacetate: :NH HN: + 2CH CO,H → HẠN Piperazine FM 86.136 → HẠN NH,(CH,CO,), Acetic acid FM 60.052 Piperazine diacetate FM 206.240 In one experiment. 0.3126 g of sample was dissolved in 25 mL of acetone, and 1 mL of acetic acid was added. After 5 min, the precipitate was filtered, washed with acetone, dried at 1108°C, and found to weight 0.7121 g. Find the wt% of piperazine in the sample.A solution is made by the following steps. 1. 90.0 mg ± 0.2 mg of dye with molar mass 685.79 g/mol ± 0.02 g/mol is dissolved in a 100.00 mL ± 0.05 mL volumetric flask with water to make solution #1. 2. 25.00 mL of solution 1 is added 100.00 mL ± 0.05 mL volumetric flask with water to make solution #2. 3. 25.00 mL of solution 2 is added 100.00 mL ± 0.05 mL volumetric flask with water to make solution #3. A.) If you have available a 5.00 mL ± 0.01 mL, 10.00 mL ± 0.05 mL, and 25.00 mL± 0.15 mL. Describe how to use the pipettes and the directions above to make the most precise solution #3. Explain why this is the most precise method. B.) Determine the concentration ± absolute error of solution #3 in molarity. Explain the calculations you needed to do. C.) Using Beer's law (10% error), solution #3 is found to be 2.0 ✕ 10-5 M. Was solution #3 likely made within the expected error. Explain.