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Alik
1 month ago
14

You are performing a simple distillation of roughly 50:50 liquid solution containing two components, hexane and nonane You place

15 ml of the mixture in a round bottom flask, and then collect the distillate sequentially as four three ml samples labeled S1, S2, S3, and S4. Pure hexane has a refractive index of 1.375 and pure nonane has a refractive index of 1.407. You measure a refractive index of 1.385 for one of the four samples. Assuming the refractive index varies linearly with mole fraction, estimate the mole fraction of hexane in this sample.
Chemistry
1 answer:
VMariaS [2.9K]1 month ago
6 0

Answer:

Explanation:

The refractive index is a one-dimensional measure that indicates the speed of light traveling through a substance. This property is utilized to determine the purity of liquids.

In this scenario, you begin with a 50:50 solution of hexane and nonane and conduct fractional distillation.

During distillation, the collection will be enriched with the more volatile component (hexane). Therefore, the final sample will contain a greater proportion of hexane compared to nonane.

Assuming the refractive index interrelates linearly with the mole fraction, the following values are obtained:

1.375 for 100% hexane

1.407 for 100% nonane

1.385 corresponds to???

mole fraction of hexane:

100 - (1.385 - 1.375) / (1.407 - 1.375) × 100 = 69 %

mole fraction of nonane:

100 - (1.407 - 1.385) / (1.407 - 1.375) × 100 = 31 %

The ratio of the distilled fractions is 69:31 (hexane:nonane)

This demonstrates that distillation serves as a method for separating liquid mixtures effectively.

I hope this information is useful!

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Sea water's density can be calculated as a function of the compressibility, B, where p = po exp[(p - Patm)/B]. Calculate the pre
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Answer:

A pressure of 137.14 MPa exists 10,000 m beneath the ocean surface.

At this same depth, the density measures 2039 kg/m3.

Explanation:

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\frac{dP}{dy}=\rho*g=\rho_0*g*e^{(P-P_0)/\beta\\\\

By rearranging

\frac{dP}{e^{(P-P_0)/\beta}}= \rho_0*g*dy\\\\\int\limits^{P}_{P_0} {e^{-(P-P_0)/\beta}}dP =\int\limits^y_0 {\rho_0*g*dy}\\\\(-\beta*e^{-(P-P_0)/\beta})-(\beta*e^0)=\rho_0*g*(y-0)\\\\-\beta*(e^{-(P-P_0)/\beta}-1)=\rho_0*g*y\\\\e^{-(P-P_0)/\beta}=1-\frac{\rho_0*g*y}{\beta}\\\\-\frac{P-P_0}{\beta} =ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\

This equation allows for computation of P at 10,000 m beneath the ocean's surface:

P-P_0=-\beta*ln(1-\frac{\rho_0*g*y}{\beta})\\\\P-P_0=-200MPa*ln(1-\frac{1027kg/m^3*9.81m/s^2*10,000m}{200MPa})\\\\P-P_0=-200MPa*ln(1-\frac{1027*9.81*10,000Pa}{200*10^6Pa})\\\\P-P_0=-200MPa*ln(1-0.5037)\\\\P-P_0=-200MPa*(-0.6857)=137.14MPa

The density found at a depth of 10,000 m in the ocean is

\rho=\rho_0*e^{(P-P_0)/\beta}\\\rho=1027kg/m^3*e^{(137.14/200)}=1027*e^{0.686}kg/m^3\\\rho=1027*1.985 kg/m^3\\\rho=2039\,kg/m^3

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2 months ago
At the boiling point, the density of the liquid is 809 g/l and that of the gas is 4.566 g/l. how many liters of liquid nitrogen
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Result: 1.68 L of liquid nitrogen is generated during the gas liquefaction process.

Clarification:

This process involves transforming gaseous nitrogen into its liquid form.

The two states possess distinct densities, thus occupying varying volumes; however, the mass remains constant.

Step 1: Calculate the mass of nitrogen gas

Let’s determine the mass of nitrogen gas associated with 297 L.

The density formula is:

Density = \frac{Mass}{Volume}

With a density of nitrogen gas at 4.566 g/L and a volume of 297 L, we can compute the mass of nitrogen gas as follows:

Using these values yields:

Mass = Density \times Volume

Mass = \frac{4.566g}{L} \times 297L

Mass = 1356g

The mass of nitrogen gas calculates to be 1356 g.

Step 2: Derive the volume of liquid nitrogen from the mass obtained

The mass for liquid nitrogen remains the same.

With the density of liquid nitrogen at 809 g/L, we can substitute this into our formula to find the volume of liquid.

Volume = \frac{Mass}{Density}

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