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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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6 0
2 years ago
Read 2 more answers
A 1.28-kg sample of water at 10.0 °C is in a calorimeter. You drop a piece of steel with a mass of 0.385 kg at 215 °C into it. A
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Answer:

T_{2}=16,97^{\circ}C

Explanation:

The specific heats of water and steel are  

Cp_{w}=4.186 \frac{KJ}{Kg^{\circ}C}

Cp_{s}=0.49 \frac{KJ}{Kg^{\circ}C}

Assuming that the water and steel are into an <em>adiabatic calorimeter</em> (there's no heat transferred to the enviroment), the temperature of both is identical when the system gets to the equilibrium T_{2}_{w}= T_{2}_{s}  

An energy balance can be written as

m_{w}\times Cp_{w}\times (T_{2}- T_{1})_{w}= -m_{s}\times Cp_{s}\times (T_{2}- T_{1})_{s}  

Replacing

1.28Kg\times 4.186\frac{KJ}{Kg^{\circ}C}\times (T_{2}-10^{\circ}C)= -0.385Kg\times 0.49 \frac{KJ}{Kg^{\circ}C} \times (T_{2}-215^{\circ}C)

Then, the temperature T_{2}=16,97^{\circ}C

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