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Korvikt [17]
3 years ago
6

The best fit curve line to an absorbance vs. concentration plot for standard solutions of a dye has a slope of 0.1815 mL/mg and

intercept of 0.0477. A sample solution prepared of the same dye has an absorbance of 0.464 at the wavelength used to establish the Beer's law plot. What is the concentration of dye in the prepared solution?Select one:
Chemistry
1 answer:
omeli [17]3 years ago
6 0

Answer:

The concentration of dye is 2.29 mg/mL.

Explanation:

According to Beer's law the absorbance (A) of a solution is proportional to its concentration (c). The best fit curve line to an A vs. c plot has a slope of 0.1815 mL/mg and intercept of 0.0477. The resulting equation is:

A =  0.0477 + 0.1815 mL/mg . c

If a solution has an absorbance of 0.464, the concentration of dye will be:

0.464 =  0.0477 + 0.1815 mL/mg . c

c = 2.29 mg/mL

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134 grams of nitric acid is added to 512 grams of water. Calculate the molality of nitric acid.
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Answer:   4.15234 m

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Nitric Acid:  HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol

H = 1.008 g/mol

N = 14.007 g/mol

O3 = 3*15.999

134 g HNO₃ * \frac{mol}{63.012 g} = 2.126 mol

m = \frac{2.126  mol}{0.512  kg} = 4.15234 m

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Which of the following is the term for the heaviness on something's surface that exerts a force over an area? a. force b. pressu
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For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)
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Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = -12666.6 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

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