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Tems11 [23]
3 years ago
5

Paper chromatography is a technique used to separate molecules based upon their size and solubility in a particular solvent. If

pigments from a particular species of plant are extracted and subjected to paper chromatography, which of the following results is most likely?
Paper chromatography would separate the pigments into several bands that appear green or yellow/orange.
Chemistry
1 answer:
dalvyx [7]3 years ago
6 0

Answer:

Paper chromatography would separate the pigments into several bands that appear green or yellow/orange.

Explanation:

Chloroplasts have a mixture of pigments with different colors: intense green chlorophyll-a, green chlorophyll-b, yellow carotenes and orange-yellow xanthophylls, in different proportions. All these substances have a different degree of solubility in non-polar solvents, which allows their separation when a solution of them ascends by capillarity through a porous paper strip, vertically arranged on a film of an organic solvent, where the most soluble move faster, and the less soluble move less on the filter paper strip. Therefore, several bands of different colors will appear that will be more or less distant from the solution according to the greater or lesser solubility of the pigments.

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1. The molar mass of the unknown gas obtained is 0.096 g/mol

2. The pressure of the oxygen gas in the tank is 1.524 atm

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>1. How to determine the molar mass of the gas </h3>
  • Rate of unknown gas (R₁) = 11.1 mins
  • Rate of H₂ (R₂) = 2.42 mins
  • Molar mass of H₂ (M₂) = 2.02 g/mol
  • Molar mass of unknown gas (M₁) =?

R₁/R₂ = √(M₂/M₁)

11.1 / 2.42 = √(2.02 / M₁)

Square both side

(11.1 / 2.42)² = 2.02 / M₁

Cross multiply

(11.1 / 2.42)² × M₁ = 2.02

Divide both side by (11.1 / 2.42)²

M₁ = 2.02 / (11.1 / 2.42)²

M₁ = 0.096 g/mol

<h3>2. How to determine the pressure of O₂</h3>

From the question given above, the following data were obtained:

  • Volume (V) = 438 L
  • Mass of O₂ = 0.885 kg = 885 g
  • Molar mass of O₂ = 32 g/mol
  • Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
  • Temperature (T) = 21 °C = 21 + 273 = 294 K
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Pressure (P) =?

The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Divide both side by V

P = nRT / V

P = (27.65625 × 0.0821 × 294) / 438

P = 1.524 atm

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

Learn more about ideal gas equation:

brainly.com/question/4147359

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