Calculate the mass in grams of Al (FM = 26.98) present in an unknown sample if 35.79 mL of 0.8765 M EDTA was required to reach the Calmagite endpoint. Group of answer choices 0.6607 g 0.8464 g 1.163 g 1.102 g
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- 1. Add 3 trailing zeros if your answer is a whole number 2. Encode the unit, attach it immediately to the preceding number, no spaces in between3. For decimal answers, round off to 4 decimal places How much volume (mL) is needed to standardize 0.05M of EDTA using 180mg of magnesium ribbon? MW: 24g/mol* How many grams of EDTA is required to prepare 250mL of a 0.05M solution? MW: 292g/mol* What is the Molarity of the EDTA solution if 35mL of the titrant was added to 0.2g of calcium carbonate? Round off to four decimal places*Five white, 500-mg uncoated ascorbic acid (AA) tablets with an average weight of 0.6100-g were pulverized in a mortar. A sample of the powdered ascorbic acid weighing 0.4610-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 46.73-mL of 0.0152 M 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 2.78-mL of 0.1047 M 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. None of the choices 349.7 mg 264.3 mg 462.7 mgA sample of anhydrous Na2CO3 (FM = 105.989) is suspected to be contaminated with either NaHCO3 (FM = 84.007) or NaOH (FM = 39.997) To verify the suspicion, a 0.7483 g sample was dissolved to prepare a 50.00 mL solution, and a 10.00 mL aliquot was taken to prepare a 100 mL solution, where 25.00 mL was analyzed using single flask method. If the sample requires 27.50 mL of standard 0.0125 N HCl to reach the phenolphthalein endpoint and another 28.40 mL to reach the bromcresol green endpoint, calculate the percentage composition (in %w/w) of all the basic components in the sample.
- A sample of anhydrous Na2CO3 (FM = 105.989) is suspected to be contaminated with eitherNaHCO3 (FM = 84.007) or NaOH (FM = 39.997) To verify the suspicion, a 0.7483 g sample was dissolved to prepare a 50.00 mL solution, and a 10.00 mL aliquot was taken to prepare a 100 mL solution, where 25.00 mL was analyzed using single flask method. If the sample requires 27.50 mL of standard 0.0125 N HCl to reach the phenolphthalein endpoint and another 28.40 mL to reach the bromcresol green endpoint, calculate the percentagecomposition (in %w/w) of all the basic components in the sample.A sample of anhydrous NaHCO3 (FM = 84.007) is suspected to be contaminated with either NaOH (FM = 39.997) or Na2CO3 (FM = 105.989). To verify the suspicion, a 0.7483 g sample was dissolved to prepare a 50.00 mL solution, and a 10.00 mL aliquot was taken to prepare a 100 mL solution, where 25.00 mL was analyzed using double flask method. If the sample requires 0.45 mL of standard 0.0125 N HCl to reach the phenolphthalein endpoint and 35.95 mL to reach the bromcresol green endpoint, calculate the percentage composition (in %w/w) of all the basic components in the sample.A sample of anhydrous NaHCO3 (FM = 84.007) is suspected to be contaminated with eitherNaOH (FM = 39.997) or Na2CO3 (FM = 105.989). To verify the suspicion, a 0.7483 g sample was dissolved to prepare a 50.00 mL solution, and a 10.00 mL aliquot was taken to prepare a 100 mL solution, where 25.00 mL was analyzed using double flask method. If the sample requires 0.45 mL of standard 0.0125 N HCl to reach the phenolphthalein endpoint and 35.95 mL to reach the bromcresol green endpoint, calculate the percentage composition (in %w/w) of all the basic components in the sample.
- A sample of an iron ore was prepared for Fe3+ analysis as following: 3.4g of the sample was added anddissolved in acid environment then diluted to 250 mL using volumetric flask. After that, 10 mL of the resultingsolution was transferred by pipet to a 50-mL volumetric flask and continue to be diluted. The scientists foundout that this solution gives the concentration of Fe3+ as 2.3 mg/L. Find the weight percentage of Fe3+ in theoriginal sample.A 5.00 mL tap water sample was measured out with a volumetric pipette, and added to a 25 mL Erlenmeyer flask. It was then titrated with a 0.0100 M Na2EDTA.2H2O solution, and found to take 0.635 mL of the EDTA solution to reach the blue endpoint. How many moles of the 0.0100 M Na2EDTA.2H2O were added in the titration reaction? Show calculation. How many moles of calcium ions were reacted? Show calculation.A 5.00 mL tap water sample was measured out with a volumetric pipette, and added to a 25 mL Erlenmeyer flask. It was then titrated with a 0.0100 M Na2EDTA.2H2O solution, and found to take 0.635 mL of the EDTA solution to reach the blue endpoint.
- 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 mg1. How much volume (mL) is needed to standardize 0.05M of EDTA using 180mg of magnesium ribbon? MW: 24g/mol2. How many grams of EDTA is required to prepare 250mL of a 0.05M solution? MW: 292g/mol3. What is the Molarity of the EDTA solution if 35mL of the titrant was added to 0.2g of calcium carbonate? Round off to four decimal placesplease help with 1 and 2, all information given Calculate the number of moles of Zn2+ in the 20 mL aliquot (show workings) 1.050/65.38= 0.016mol (0.016x20)/250= 0.00128mol Calculate the number of moles of EDTA in the average titre (show workings) Mol EDTA= 0.00122 Volume of EDTA consumed= 0.02106L 0.00122mol/0.02106L=0.0579mol 1. Calculate the concentration of the EDTA solution (show workings) 2. Calculate the concentration of the DILUTED EDTA solution (show workings) Sample No. 1 2 3 4 Final Burette reading (mL) 20.25 42.75 22.90 44.50 Initial Burette reading (mL) 0.05 20.25 1.50 22.90 Final-Initial Titre (mL) 20.20 22.50 21.40 21.60