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Can you help me find the mixtures and the average on the second page. The first page has the experimental data on it
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- The acid-base indicator HIn undergoes the following reaction in dilute aqueous solution: HIncolor1H++Incolor2 The following absorbance data were obtained for a 5.00 I0-4 M solution of HIn in 0.1 M NaOH and 0.1 M HC1. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells. 0.1 M NaOH A485 = 0.075 A625 = 0.904 0.1 M HC1 A485 = 0.487 A625 = 0.181 In the NaOH solution, essentially all of the indicator is present as In-; in the acidic solution, it is essentially all in the form of HIn. (a) Calculate molar absorptivities for In- and HIn at 485 and 625 nm. (b) Calculate the acid dissociation constant for the indicator ¡fa pH 5.00 buffer containing a small amount of the indicator exhibits an absorbance of 0.567 at 485 nm and 0.395 at 625 nm (1.00-cm cells). (c) What is the pH of a solution containing a small amount of the indicator that exhibits an absorbance of0.492 at 485 nm and 0.245 at 635 nm (1.00-cm cells)? (d) A 25.00-mL aliquot of a solution of purified weak organic acid HX required exactly 24.20 mL of a standard solution of a strong base to reach a phenolphthalein end point. When exactly 12.10 mL of the base was added to a second 25.00-mL aliquot of the acid, which contained a small amount of the Indicator under consideration, the absorbance was found to be 0.333 at 485 nm and 0.655 at 625 nm (1.00-cmcells). Calculate the pH of the solution and Ka for the weak acid. (e) What would be the absorbance of a solution at 485 and 625 nm (1.50-cm cells) that was 2.00 10-4 M in the indicator and was buffered to a pH of 6.000?The chromium in an aqueous sample was determined by pipetting 10.0 ml. of the unknown into each of five 50.0-mL volumetric flasks. Various volumes of a standard containing 12.2 ppm Cr were added to the flasks, following which the solutions were diluted to volume. Unknown,mLStandard, mLAbsorbancc 10.00.00.201 10.010.0 0.292 10.020.0 0.378 10.030.0 0.467 10.040.0 0.554 (a) Plot the data using a spreadsheet. (b) Determine an equation for the relationship between absorbance and volume of standard. (c) Calculate the statistics for the least-squares relationship in (b). (d) I)ctcrmine the conccnt ration oÍCr in ppm in the sample. (e) Find the standard deviation of the result in (d).An unknown sample of Yellow 6 solution has the absorbance equal to 0.96. The slope of the Beer's law calibration plot constructed using a 3.8E-5M stock solution of Yellow 6 is 0.0117 (%) 1. What is the molar concentration of the unknown sample (in mole L-¹)? Please give your answer to 3 significant figures (1 additional SF is used to minimize rounding errors). M Hint
- 4. The total absorbance of a solution is the sum of the absorbances of all different materials present. AT = Aa + Ap + Ac .. etc. At 427 nm, the molar absorbtivity of Yellow #5 is 2.73×10°M-1-cm1. At 427 nm the molar absorptivity of Red 40 is 7.49x103 M-1.cm-1. Calculate the total absorbance of a solution in which the concentration of Yellow #5 = 2.00x10-5 M and the concentration of Red 40 = 4.80x10-5 M, when measured at 427 nm in a 1.00 cm cell.The absorbance values at 250nm of 5 standard solutions, and sample solution of a drug are given below: Conc. (ug/ml) A 250 nm10 0.16820. 0.32930 0.50840. 0.66050 0.846Sample. 0.661Calculate the concentration of the sample6. Blue Blue dye stock solution 0.293 M Absorbance at 630 nm 0.00265 Calibration curve y = 0.0833x A solution is prepared by diluting 2.79 mL of the blue dye stock solution to 25.00 mL. The measured absorbance for the prepared solution is listed in the data table. (a) What is the theoretical molar concentration? [Blue]theoretical x 10 |M (b) What is the experimental molar concentration? [Blue]experimental x 10 M (c) What is the percent error? Percent error (blue) = %
- 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…A UVIVIS molecular absorption spectrometric determination of analyte X in aqueous solution gave the following calibration data (absorbances corrected for the blank). Estimate the concentration of Xin an aqueous sample that yielded an instrument response of 0.238 (A 1.00- mL cuvette was used for all absorbance measurement s) OA. 7.7 ppm OB. 8.0 ppm, 148 O c. 23.8 ppm O D.0.0308 ppb OE. 0.8 ppm Concentration (ppm X) 1.0 2.0 5.0 7.0 8.5 10.0 Absorbance 0.031 0.068 0.150 0.221 0.262 0.308 3:49 PM3. The lead in a swab sample, lead standards, together with a blank were made up in 5.00 mL volumetric flasks using 0.2 % HNO3. 20 μL aliquots of these solutions were injected into the spectrophotometer and the absorbance measured at 217 nm. The following results were obtained. lead / ppb Absorbance 0 0.0591 10.00 0.0858 50.00 0.1926 100.0 0.3260 150.0 0.4594 200.0 0.5929 Swab sample 0.3700 (a) Determine the amount of lead in the swab sample in μg.
- At 21 °C, a solution of 12 mL each of 0.00200 M Fe3+ and 0.001000 M SCN- was mixed and brought to 25.0 mL with 0.1 M HNO3. The absorbance at 450 nm was measured to be 0.350 (A450 = 0.350). This solution was then heated to 45 °C and the absorbance at 450 nm was measured to be 0.142 (A450= 0.142). Is the reaction: Fe3+(aq) + SCN -(aq) → FeSCN2+(aq) exothermic or endothermic? Explain.A 2.50 mL aliquot of a solution that contains 3.85 ppm Fe(III) is treated with an appropriate excess of KSCN and diluted to 100.0 mL. What is the absorbance of the resulting solution at 580 nm in a 0.50 cm cell? (Molar absorptivity of Fe (SCN)2+is 7.0 x 105L mol-1cm-1).Preparing a standard curve. 1. determine the absorbance max of [FeSCN]2+ it is 448.1 2. determine the concentration of FeSCN2+ in the stock solution. M1V1=M2V2 stock solution: ~0.200 M iron (iii) nitrate in 1M nitric acid intial volume: 0.3mL final volume: 10.3mL volume: 10mL ~0.002M potassium thiocyanate initial volume: 0.3mL final volume: 10.3mL volume: 10mL not sure if this is needed stock solutions: ~0.200M iron (iii) nitrate in 1M nitric acid: 0.207M ~0.002M iron (iii) nitrate in 1M nitric acid: 0.00209M ~0.002M potassium thiocyanate: 0.00193M