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2 years ago

How many colors would you see if you were analyzing a pure substance through chromatography?

Chemistry

2 answers:

harina [27]2 years ago

4 0

Answer :

The correct answer is no color will observed if a pure substance is analyzed ,

Chromatography :

It is a technique commonly used to separate a mixture of chemical substances based on their affinity with stationary phase and mobile phase .

The mobile phase travels through the stationary phase in a definite direction . The component of mixture travels along with mobile according to their affinity , ionic interactions , diffusion , solubility on stationary phase .

The component has more affinity with mobile phase goes higher with mobile phase .

While component with stationary phase remain at lower position on stationary phase .

This is how separation of components takes place.

A pure substance has only one type of component in it . There is no mixture of components.

Hence when pure substance will be analyzed by chromatography technique, there will be no separation of components .

So there will no color be observed.

AfilCa [17]2 years ago

3 0

For separation of components of mixture chromatography technique is used. Parts of mixture are separated by using the different solubilities of the parts for a stationary phase (that does not move) compared to the mobile phase (that carries the parts). Due to capillary action, the solvent will move up with paper or plate in apparatus. The mobile phase will move with solvent and the stationary phase will remain adhere to the paper or plate. And thus, giving separate colors for the substances present in the mixture.

Hence, for pure substance, only one color would be observed through chromatography.

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2 years ago

An acetate buffer solution is prepared by combining 50. mL of 0.20 M acetic acid,

natima [27]

Answer:

The answer is "Option B".

Explanation:

$\to CH_3COOH + NaOH \longleftrightarrow CH_3COONa + H_2O\\\\\to CH_3COONa + NaOH\longleftrightarrow CH3COONa\\\\\therefore \ mol\ NaOH = (5 \ E-3\ L)\times(0.10 \ \frac{mol}{L}) = 5 \ E-4\ mol\\\\$

$\to mol\ CH_3COOH = (0.05 \ L)\times(0.20 \frac{mol}{L}) = 0.01 \ mol\\\\\to C \ CH_3COOH = \frac{(0.01 \ mol - 5 \ E-4\ mol) }{(0.105 \ L)}\\\\\to C \ CH_3COOH = 0.0905 \ M\\\\\therefore \ mol \ CH_3COONa = (0.05\ L )\times (0.20 \ \frac{mol}{L}) = 0.01 \ mol\\\\$

$\to C \ CH_3COONa = \frac{(0.01\ mol + 5 \ E-4\ mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\$

$\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\ E-6 - 1.75\ E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to [H_3O^+] = 1.5835\ E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7$

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2 years ago

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Anastaziya [24]

Answer:

• The Na atom is formed when the Na+ ion gains an electron.

• A Na+ ion has 10 electrons and a Na atom has 11 electrons.

• A sodium atom has one more electron shell than a sodium ion.

• A Na+ ion is formed when a Na atom loses an electron

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1 year ago