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stepan [7]
3 years ago
11

he reported molar absorptivity of Red Dye #3 is 1.217 L mol-1 cm-1. If a solution of Red Dye #3 displays an absorbance of 0.699

in a cuvette that is 1.526 cm in length, what is the concentration of the dye in solution? Report your response to three digits after the decimal. _____ M
Chemistry
1 answer:
Svetach [21]3 years ago
3 0

Answer:

Concentration of the solution is 0.376M

Explanation:

Lambert-Beer's law describes the absorbance of a solution that is directely proportional to concentration of the solution, molar absorptivity and path length. The law is:

A = ε×C×l

<em>Where A is absorbance (0.699 for Red Dye #3), ε is molar absorptivity (1.217L mol⁻¹cm⁻¹), C is concentration of solution and l is path length (1.526cm)</em>

Replacing:

0.699 = 1.217L mol⁻¹cm⁻¹×C×1.526cm

C = 0.3764 mol/L

<em>Concentration of the solution is 0.376M</em>

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Una muestra de 258.4 g de etanol (C2H5OH) se quemó en una bomba calorimétrica
Ilia_Sergeevich [38]

Answer:

Explanation:

Unclear question.

I infer you want a clear rendering, which reads;

A 258.4 g sample of ethanol (C2H5OH) was burned in a calorimetric pump using a Dewar glass. As a consequence, the water temperature rose to 4.20 ° C.

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Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
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