Countercurrent Flow

Chemistry

1 answer:

Nikitich [7]1 year ago

7 0

Answer: All options (I, II, III, IV, V)

Explanation: The answer is all options because they never reach equilibrium or intersect at a region as the both decrease at the same rate.

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Identify the precipitate(s) formed when solutions of na2so4(aq), ba(no3)2(aq), and nh4clo4(aq) are mixed.

Lostsunrise [7]

Answer : BaS $O_{4}$ will be the precipitate which will be formed.

Explanation : When all the three solutions namely; $NaSO_{4} + Ba(NO_{3})_{2} + NH_{4} ClO_{4}$ are mixed together a white precipitate of BaS $O_{4}$ is formed as a product in the solution along with the soluble by product of Ammonium nitrate which is $NH_{4} NO_{3}$

7 0

2 years ago

Crystal violet standard solutions were prepared and their absorbance at a certain wavelength was measured using a spectrophotome

ehidna [41]

Answer:

(a) 128459 L/mol·cm

(b) 5.11 x 10⁻⁶ M

Explanation:

This question was incomplete, it is missing the Excel plot and the line fit, please see attached image of the question.

Crystal violet standard solutions were prepared and their absorbance at a certain wavelength was measured using a spectrophotometer. The path length was 1 cm. The data was plotted using Excel and a line was fit to the points (see below). A sample of unknown concentration was found to have an absorbance of 0.652 at the same wavelength. (a) What is the molar absorptivity of this compound at the certain wavelength? (b) What is the concentration of the unknown sample solution?

(a) This problem we calls for the use of Beer´s law:

A = εcl

where A is the absorbance, ε is the molar absorptivity, c the concentration, and l the path of the cuvette used.

This equation when plotted A vs c will give a straight line since A is a function of c

Thus we can determine ε by using the calibration curve in the Excel plot

y = 128459 x - 0.0042

the slope of this equation is the molar absorptivity since the length is one, so :

ε = 128459 L/mol·cm

For part (b)

A = 0.652 = 128459 L/mol·cm x c x 1 cm - 0.0042

⇒ 0.652 +0.0042 = 128459 c = 5.11 x 10⁻⁶ M

In absorbance spectrophotometry it is necessary to work with very dilute solutions to avoid deviations characteristic of high concentrations. One can always dilute the unknown and knowing exactly the dilution factor one can trace back the value for the a more concentrated solution.