2- In a soil with Da = 1.30 g/cm3 at a depth of 15 cm, an absorbance value of 0.80 was obtained using 15 mL of filtration. Calculate phosphorus in units of kg/ha.
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- (d) 3-13. Distis me Jum 4 A method of analysis yields weights for gold that are low by 0.4 mg. Calculate the caused by this uncertainty if the weight of gold in the sample is (a) 900 mg. (c) 150 mg. 1-7. The method described in Problem 3-6 is to be used for the analysis of ores that assay about 1.2% gold. What minimum sample weight should be taken if the relative error resulting from a 0.4-mg loss is not to exceed -0.2%? percent relative error (a) (b) (b) 600 mg. (d) 30 mg. 3-14. Wha stan stam 3-15. Сог mea (b) -0.5%? (a) -0.8%? (c) (d) 1.2%? 3.5 3-8. The color change of a chemical indicator requires an overtitration of 0.04 mL. Calculate the percent relative error if the total volume of titrant is (a) 50.00 mL. (c) 25.0 mL. 3-9. A loss of 0.4 mg of Zn occurs in the course of an analysis to determine that element. Calculate the per- cent relative error due to this loss if the weight of Zn in the sample is (a) 40 mg. (c) 400 mg. 3.1 3.1 (b) 10.0 mL. 3.3 2.5 (d) 40.0 mL. Fo (a (E (b) 175…4- A sample containing H2C204 had a purity equal to 90.50% (w/w). An unknown mass of this sample was dissolved in water and transferred to a 50.00 mL flask. An aliquot of 25.00 mL was transferred to an erlenmeyer flask and 50.00 mL of a 0.2000 mol L^-1 NaOH solution were added. %3 Excess NaOH was titrated with 0.09000 mol L^-1 HCI solution using 2.000 mL. Calculate the mass (g) of the sample used. Data: H = 1.008 C= 12.01 O = 16.00an absorbance of a solution with a pathlength of 1.00 cm is 0.544 and concentration is 1.40x10^ ^ -3 M , what is its molar absorptivity?
- The max of aspirin (molar mass 180g/mol) in neutral solution is 225 nm and & is 650 mol-¹Lcm ¹. An absorbance of 1.80 was measured for a solution of aspirin, with a 1 cm path length. Calculate the concentration of the aspirin solution.2-28. Describe the preparation of (a) 500 mL of 0.0650 M AGNO3 from the solid reagent. (b) 400 mL of a solution that is 0.0825 M in Kt. starting with solid K,Fe(CN)6 . (c) 2.00 L of 0.120 M HCIO, from the commercial reagent [60% HCIO4 (w/w), sp gr 1.60].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 sample
- A 10.0-mL aliquot of a standard manganese solution containing 62.5 mg/L of Mn is oxidized to MnO4- and diluted to 50.0 mL in a volumetric flask. The absorbance is measured in a 1.00-cm cell at 525 nm is 0.343. Next, a 850-mg steel sample containing manganese in dissolved in acid and diluted to 250 mL. A 50.0-mL aliquot is treated with KIO4 to oxidize Mn to MnO4- and diluted to 100.0 mL. If the absorbance at 525 nm is 0.468 in a 1.00-cm cell, calculate the weight percent of Mn in the steel.A solution of Tryptophan has an absorbance at 280 nm of 0.54 in a 0.5 cm length cuvette. Given the absorbance coefficient of trp is 6.4 x 103 Lmol-1cm-1. What is the concentration of solution? The options are as follows A) 0.17 x10-3 mol/L B)2.56x10-3mol/L C)0.17x10+3mol/L D)2.56x10+3mol/LA solution of Tryptophan has an absorbance at 280 nm of 0.54 in a 0.5 cm length cuvette. Given the absorbance coefficient of trp is 6.4 x 103 Lmol-1cm-1. What is the concentration of solution? The answers are as follows A)0.17x10-3mol/L B)2.57x10-3mol/L C)0.17x10+3mol/L D)2.57x10+3mol/L
- Calaculate the initial concentration of [Fe3+] and [SCN-] of a solution with 5 mL of 0.002 M Fe(NO3)3, 2 mL of 0.002 M KSCN, 3 mL of water, and an absorbance of 0.221.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 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 sampleThe 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 sample