Calculate the expected absorbance of a 20mg/L. solution of potassium permanganate at 525m. Note the Mt of potassium permanganate is 158 g/mol and assume the molar absorptivity at 525mm is 2100 M~1 cm*1. Round answer to 2
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- g 2.png Average absorbance, .129 +0.128 +0.127)/3 = 0.128 Chrome Concentration of Females of Fe volume Q Molarity of Fe(NO₂)) = 0.2M Volume of Fe(NO₂)); = 5.00mL Moles of Fe³+= Molarity x Molarity x volume(L) Moles of Fe³* = 0.2M x 0.0054 = 10³mole 10-mole 500 x 10-3=0.002M [Fe³+] = [SCN¯] = X = ✪ [Fe³] = 0.002mol/L nknown concentrations to determine [FeSCN]at equilibrium in solution 8: Y = 4290X - 0.068 72% (0.127 + 0.068) 4290 0.127=4290X - 0.068 25ml Fe³+ x 0.002 mol/L-¹ 100mL 100% = 4.5 x 10-5mol/L 20ml SCN X 0.002 mol/L-¹ 100mL = 5.00 x 10^4 mol/L^-1 = 4.00 x 10 mol/L-1 W Flask 8 Flask 9 Flask 10 Flask 11 3.png 4.png Part 3: Q8 Flask 8 Flask 9 Flask 10 Flask 11 Part 3: 09 Flask 8 Flask 9 Flask 10 Flask 11 [Fe³+ (mol/L) 5.00 x 10-4 5.00 x 10-4 5.00 x 10-4 5.00 x 10-4 Average absorbance 0.127 0.166 0.222 69% 0.308 100% [FESCN¹ (mol/L) 4.5 x 10- 5.4 x 10 6.76 x 10-5 8.76 x 10- [SCN (mol/L) 4.00 x 10-4 4 x 10-4 7 x 10-4 1x 10-³ 3 7.png Place the results from calculations 9 and 10 into…45.0 mL of an unknown FeCl3 solution is diluted to a total volume of 230.0 mL. The diluted FeCl3 solution measured an absorbance of 0.223. Calculate the molarity of the undiluted unknown FeCl3 solution if the molar extinction coefficient is 230.0 M-1⋅cm-1 and a path length of 1.00 cm. Group of answer choices 0.000190 M 0.00496 M 0.00593 M 10.0 M 0.00399 MThe nitrite in a series of standard solutions (mg/L, n = 5) are converted to azo dye and the slope of the calibration curve is 2.0 ppm1. A 10.00-mL mineral water sample is treated in the same way as standards and diluted to a final volume of 100.00-mL. It gives an absorbance of 0.80. The absorbance of blank solution under the same conditions is 0.10. Calculate ppm (mg/L) of NO2 (46 g/mol) and the molarity of NANO2 (69 g/mol) in the original sample
- M 23 (H) Inorganic Analytical Chemistry pard My courses CHEM 23 (H) FS-AY:20-21 Week 4-5: Quantitative Analysis by Gravimetry and Titration Quiz 4 Qu 10 The 500.0-mg sample of impure Na2 CO3 (FM: 105.96) required 22.00 mL of the HCI standard solution (the obtained concentration of HCI from the previous question, problem 5: Titrimetry short answer type) for complete neutralization. ut of 1. Give the balanced chemical equation between the sample, Na2CO3 and the standard, HCI. 2. What is the mass (g) of Na, CO3 in the sample? 3. Calculate the % (w/w) purity of a 500.0-mg sample of impure Na2 CO3 Note: Type your solution here or upload its pic in Jpeg or pdf format. Another option is to send it to me via email (subject: Fam name-Quiz 4) or messenger, privately. Do not forget to box the final answer and write your name. В I 1.I did an experiment: Molar concentration of Fe (NO3)3=.002 Molar concentration of NaSCN= .002 On my 1st trial: Volume of Fe(NO3)3 = 5 mL Moles of Fe3+, initial (mol) = 1 x 10-5 Volume of NaSCN (mL)= 1 mL Moles of SCN-,initial (mol)= 2 x 10-6 Absorbance was .08 Now trying to find Calculation of Kc I already found [FeNCS2+] equilibrium, from calibration curve (mol/L)= 2.25 x 10-5 Now I need to know: 1. moles FeNCS2+ at equilibrium (10mL) (mol) 2. moles Fe3+, reacted (mol) 3. moles Fe3+, equilibrium (mol) 4. [Fe3+] equilibrium, unreacted, 10 mL (mol/L) 5. moles SCN-, reacted (mol) 6. Moles SCN-, equilibrium (mol) 7. [SCN-] equilibrium (unreacted) 10 mL (mol/L) Feel free to just give me the formula to find these. Or to charge me 2 or 3 questions if I asked too much. Thank you!!!!!!4
- An unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO, free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 311 mL of the standard acid. What is/are the component/s of the sample? O Cannot be determined O NAHCO3 only O NAOH and Na,CO3 O NazCO3 only O NAHCO, and Na2CO3An unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO, free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 31.1 mL of the standard acid. What is/are the component/s of the sample? O NazCO3 only NaHCO3 only O NAHCO3 and Na>CO3 NaOH and NazCO3 O Cannot be determined45.0 mL of an unknown FeCl3 solution is diluted to a total volume of 230.0 mL. The diluted FeCl3 solution measured an absorbance of 0.223. Calculate the molarity of the undiluted unknown FeCl3 solution if the molar extinction coefficient is 230.0 M-1⋅cm-1 and a path length of 1.00 cm.
- What is the molar absorptivity and concentration of caffeic acid? Molar mass= 180g/mol the cuvette= 1cm absorbance= 1.98 Started by adding 0.0019g of caffeic acid in 25 mL methanol. This solution was diluted with methanol in 1:100mL, then again in 1:10 dilution.A 0.5131-g sample that contains KBr (MM: 119.0023) is dissolved in 50 mL of distilled water. Titrating with 0.04614 M AgNO3 requires 25.13 mL to reach the Mohr end point. A blank titration requires 0.65 mL to reach the same end point. Report the %w/w KBr in the sample.Calculate the total hardness of a sample water in ppm CaCO3, if 50 ml of the sample isdiluted to 250 ml and 50 ml of the diluted sample requires 30.00 ml of EDTA (0 1550 gCaCO3 = 25.00 mL EDTA) for titration 1Add file