extinction coefficient = ε, path length = b, concentration C A sample containing an unknown concentration of FeCl3 is measured to have an absorbance of 0.43 at 334 nm in a 1.0 cm cuvet. What is the molar concentration of the solution
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extinction coefficient = ε, path length = b, concentration C
A sample containing an unknown concentration of FeCl3 is measured to have an absorbance of 0.43 at 334 nm in a 1.0 cm cuvet. What is the molar concentration of the solution
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- .Q1. To determine the concentrations (mol/L) of Co(NO3)2 (A) and Cr(NO3)3(B) in an unknown sample, the following representative absorbance data wereobtained.A (mol/L) B (mol/L) 510nm 575nm5×10−1 0 0.714 0.0970 6×10−2 0.298 0.757Unknown Unknown 0.671 0.330Measurements were made in 1.0 cm glass cells.i. Calculate the four molar absorptivities: ∈A(510), ∈A(575), ∈B(510) and ∈B(575).ii. Calculate the molarities of the two salts A and B in the unknown.The 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 ppm. 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 sampleA group of students made 5 standard solutions and measured their corresponding absorbance values at a wavelength of 525.0 nm to generate the graph below. Absorbance of Standard Solutions to Find Concentration of Unknown X Solution 0.50 y = 75.9x + 0.0045 R? = 0.9946 0.40 %3D 0.30 0.20 0.10 0.00 0.001 0.002 0.003 0.004 0.005 0.006 Concentration of X (mole/L) The colorless reactants V and W form unknown X from the reaction: 2V + W 5 X A solution is prepared by mixing together the following: Volume 0.0150 M V (mL) Volume 0.0150 M W (mL) 6.00 4.00 This solution reaches equilibrium, and then is placed in a 1.00 cm wide cuvet and inserted into the spectrometer, producing an absorbance reading of 0.275 at a wavelength of 525.0 nm. Calculate Keg for the above reaction, including units. Absorbance
- The molar absorptivity constant of a particular chemical is 2.31 M/cm. Determine the concentration of a solution made from this chemical that has an absorbance of 0.521 with a cell path length of 1.04 cm.An ethanol solution of 3.5 mg/100 ml of compound Y (150.0 g/mol) in a 1.00 cm quartz cell has an absorbance (A) of 0.972 at λmax=235 nm. What is its molar extinction coefficient? Report your answer to the correct number of significant figures. Do not include units in your answer.Caffeine, C8H10O2N4 H2O (MW = 212.2 g/mol) has been shown to have an average absorbance of 0.644 for a concentration of 1.783 mg per 100 mL at 272 nm. A sample of 3.658 g of a soluble coffee product was mixed with water to a volume of 500 mL and a 25 mL aliquot was transferred to a flask containing 25 mL of 0.1 M H2SO4. This was subjected to the prescribed clarification treatment and made up to 500 mL. A portion of this treated solution showed an absorbance of 0.666 at 272 nm. Assumbe b = 1.0 cm. Calculate the % caffeine (w/w) in the sample.
- In an analytical laboratory a chemical compound namely Paracetamol (151.163 g/mol) is determined in a sample. A sample weighing 0.0295 g was dissolved in a solvent and diluted the solution to 1 L. The solution has λ max at 243 nm (ε=2.6×104 cm-1 mol-1 L). The solution exhibits an absorbance of 0.638 in a 2 cm cell. Calculate the percentage paracetamol in the sample.6. 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) = %The standard curve was made by spectrophotographic analysis of equilibrated iron(III) thiocyanate solutions of known concentration. You are asked to analyze a Fe(SCN)2+ solution with an unknown concentration and an absorbance value of 0.392. The slope-intercept form of the equation of the line is y = 4538.1x +0.0077. The unknown was analyzed on the same instrument as the standard curve solutions at the same temperature. What is the Fe³+ concentration of the unknown solution? [Fe³+] = mol/L Absorbance Iron(III) thiocynate standard curve 1.0 V 0.5 0.0001 Fe³+ concentration (M) 0 0.0002
- A colored ion solution has a concentration of 0.200 M with a measured absorbance A = 0.880. Another ion solution made of the same chemicals has an absorbance A = 0.172. What is the concentration of this unknown sample solution?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.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…