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Chapter 18: Q10 P (page 456)

18-10. Color and absorption spectra. Color Plate 17 shows coalored solutions and their spectra. From Table 18-1, predict the color of each solution from the wavelength of maximum absorption. Do observed colors agree with predicted colors?

Short Answer

Expert verified

In Compound B and F the observed colors does not agree with the predicted colors. In Compound A,C,D and E the observed colors agree with the predicted colors.

Step by step solution

01

Definition

Spectrophotometry is a standard and inexpensive technique to measure light absorption or the amount of chemicals in a solution. It uses a light beam which passes through the sample, and each compound in the solution absorbs or transmits light over a certain wavelength.

02

Absorbance and the observed color in all compounds

The compound A has maximum absorbance at 760 nm. The predicted color from table 18-1 is green, and the observed color is green.

The compound B has maximum absorbance at 700 nm. The predicted color from table 18-1 is green, and the observed color is blue- green.

The compound C has maximum absorbance at 600 nm. The predicted color from table 18-1 is blue, and the observed color is blue.

The compound D has maximum absorbance at 530 nm. The predicted color from table 18-1 is violet, and the observed color is violet.

The compound E has maximum absorbance at 500 nm. The predicted color from table 18-1 is red or purple red, and the observed color is red.

The compound F has maximum absorbance at 410 nm. The predicted color from table 18-1 is green-yellow, and the observed color is yellow.

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Most popular questions from this chapter

(a) A 3.96×10-4Msolution of compound A exhibited an absorbance of 0.624 at 238nmin a 1.000 cmcuvet; a blank solution containing only solvent had an absorbance of 0.029at the same wavelength. Find the molar absorptivity of compound A.

(b) The absorbance of an unknown solution of compound A in the same solvent and cuvet was 0.375 at 238 nm. Find the concentration of A in the unknown.

(c) A concentrated solution of compound A in the same solvent was diluted from an initial volume ofto a final volume of 25.00mLand then had an absorbance of 0.733. What is the concentration of A in the concentrated solution?

Standard addition. Selenium from 0.108 g of Brazil nuts was converted into the fluorescent product in Reaction 18-15, and extracted into 10.0 mL of cyclohexane. Then 2.00 mL of the cyclohexane solution were placed in a cuvet for fluorescence measurement. Standard additions of fluorescent product containingSe/mL are given in the table. Construct a standard addition graph to find the concentration of Se in the 2.00-mL unknown solution. Find the wt%of Se in the nuts and its uncertainty and 95% confidence interval

You have been sent to India to investigate the occurrence ofgoiter disease attributed to iodine deficiency. As part of your investigation, you must make field measurements of traces of iodide I2in groundwater. The procedure is to I2 into an intensely colored complex with the dye brilliant green inthe organic solvent toluene.

(a)A 3.15×10-6Msolution of the colored complex exhibited an absorbance of 0.267 at 635nmin a 1.00 - cm cuvet. A blank solution made from distilled water in place of groundwater had an absorbance of 0.019 . Find the molar absorptivity of the coloredcomplex.

(b) The absorbance of an unknown solution prepared from groundwater was0.175. Find the concentration of the unknown.

Consider a molecule that can fluoresce from the S1 state and phosphoresce from the T1 state. Which is emitted at longer wavelength, fluorescence or phosphorescence? Make a sketch showing absorption, fluorescence, and phosphorescence on a single spectrum.

Gold nanoparticles (Figure 17-31) can be titrated with the oxidizing agent TCNQ in the presence of excess ofBr2-to oxidize Au(0)"toAuBr2-in deaerated toluene. Gold atoms in the interior of the particle are Au(0) . Gold atoms bound to C12H25S- (dodecanethiol) ligands on the surface of the particle are Au(I) and are not titrated.

The table gives the absorbance at $856 0.700 mL of 1.00×10-4MTCNQ+0.05M(C8H17)4N+Br-in toluene is titrated with gold nanoparticles (1.43 g/Lin toluene) from a microsyringe with a Teflon-coated needle. Absorbance in the table has already been corrected for dilution.

(a) Make a graph of absorbance versus volume of titrant and estimate the equivalence point. Calculate the Au(0) in 1.00 g of nanoparticles.

(b) From other analyses of similarly prepared nanoparticles, it is estimated that 25 % of the mass of the particle is dodecanethiol ligand. Calculate mmol of localid="1667559229564" C12H25Sin 1.00 g of nanoparticles.

(c) The Au(I) content of 1.00 g of nanoparticles should be 1.00 mass of Au(0) - mass of C12H25S. Calculate the micromoles of Au(I) in 1.00 g of nanoparticles and the mole ratio Au(I) :. In principle, this ratio should be 1: 1. The difference is most likely because C12H25Swas not measured for this specific nanoparticle preparation.
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