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emmasim [6.3K]
3 years ago
13

A(n) _____, used as part of the stationary phase, has an affinity for the solvent and the chemical components of the mixture. ad

sorbent eluent Rf factor chromatogram
Chemistry
1 answer:
tester [92]3 years ago
4 0

The answer is Adsorbent

The adsorbent has a high affinity for solvent and the chemical components of the mixture.

Chromatography is a method of separation in which the mixture of substances is introduced into a mobile phase (solvent). The separation occurs as the solvent interacts with an adsorbent(stationary phase).

The extent of separation of the components of the mixture depends on the extent of interaction between the mobile and the stationary phase . This interaction also determines the retention factor (Rf) of the separation.

For a definition of chromatography, see

brainly.com/question/19334271

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Green plants use light from the Sun to drive photosynthesis, a chemical reaction in which liquid water and carbon dioxide gas fo
Artyom0805 [142]

Answer:

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Explanation:

3 0
3 years ago
Does the excess reactant get used up completely in a reaction??
Alex787 [66]

Answer:

In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents. The reagent that is completely used up or reacted is called the limiting reagent, because its quantity limits the amount of products formed.

Explanation:

4 0
2 years ago
A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a
Mandarinka [93]
There are two ways to solve this problem. We can use the ICE method which is tedious and lengthy or use the Henderson–Hasselbalch equation. This equation relates pH and the concentration of the ions in the solution. It is expressed as

pH = pKa + log [A]/[HA] 

 where pKa = - log [Ka]
[A] is the concentration of the conjugate base
[HA] is the concentration of the acid

Given:
Ka = 1.8x10^-5
NaOH added = 0.015 mol
HC2H3O2 = 0.1 mol
NaC2H3O2 = 0.1 mol

Solution:
pKa = - log ( 1.8x10^-5) = 4.74

[A] = 0.015 mol + 0.100 mol = .115 moles
[HA] = .1 - 0.015 = 0.085 moles

pH = 4.74 + log (.115/0.085)
pH = 4.87
6 0
3 years ago
Imagine a small synthetic vesicle made from pure phospholipids enclosing an interior lumen containing 1 mM glucose and 1 mM sodi
Alexeev081 [22]

Answer:

C. H2O diffuses in.

Explanation:

<em>The phospholipids-made synthetic vesicle in this case will act like a semi-permeable membrane while the solution in the interior lumen will be hypertonic to the surrounding pure water. </em>

<em>Hence, water molecules will diffuse into the lumen through the semi-permeable membrane because of the osmotic gradient that exist between the internal and the surrounding solution of the vesicle.</em>

7 0
3 years ago
Which isomer of 1-tert-butyl-3-ethyl-5-methylcyclohexane below is thermodynamically the most stable?
Agata [3.3K]
<h2>Answer:</h2>

Option A is correct.

<h3>Explanation:</h3>

In the options given below the isomer given in the option A of 1-tert-butyl-5-methylcyclohexane is the most stable of all. The IUPAC name for this compound is 1-tert-butyl-3-ethyl-5-methylbenzene.

7 0
3 years ago
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