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bija089
1 month ago
10

Suppose that there are two components, A and B, in a mixture. The chemical affinity of A for the stationary phase is less than t

hat of B and the chemical affinity of A for the mobile phase is greater than that of B. If the mixture is subjected to paper chromatography, which substance will have traveled further at the completion of the experiment?Which substance will have the largest Rf value?
Chemistry
1 answer:
eduard [2.7K]1 month ago
8 0

Answer: Substance A will move farther by the end of the experiment and will also display a higher Rf value than substance B.

Explanation:

The tendency of a substance to adhere to either the stationary or mobile phase indicates its binding capacity. Component B sticks more to the stationary phase due to its stronger attraction, while component A prefers the mobile phase. Through differences in affinity, chromatography can separate components A and B. The stationary phase remains stationary, whereas the mobile phase is in motion. As a result, component A travels with the mobile phase because of its greater affinity, while component B remains attached to the stationary phase. Thus, substance A will cover a greater distance by the end of the experiment.

The Rf value (Retention Factor) is defined as the ratio of the distance a substance travels in the mobile phase (solute) to the distance the solvent front moves.

Rf = Distance traveled by substance / Distance traveled by solvent front

Since the solvent's front movement is always greater than that of the solute, Rf values are constrained to a range between 0 and 1. The longer a compound travels, the higher its Rf value will be. Therefore, in this scenario, substance A will have a larger Rf value than substance B.

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An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as subs
eduard [2782]

Answer:

Fatty acids with an even number of carbons, like palmitate, undergo complete β-oxidation in the liver mitochondria, resulting in CO₂, as acetyl-CoA, their end product, can enter the TCA cycle.

On the other hand, odd-number fatty acids such as undecanoic acid generate acetyl-CoA and propionyl-CoA during their final pass. To allow entry into the TCA cycle, propionyl-CoA must go through additional processes, including carboxylation.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is facilitated by propionyl-CoA carboxylase, a biotin-dependent enzyme that is inhibited by avidin. In contrast, the oxidation of palmitate does not require carboxylation.

Explanation:

Fatty acids with an even number of carbons, such as palmitate, are completely oxidized to CO₂ in the liver mitochondria due to the ability of their oxidation product, acetyl-CoA, to enter the TCA cycle where it is further oxidized to CO₂.

Undecanoic acid is classified as an odd-number fatty acid, consisting of 11 carbon atoms. The last stage of β-oxidation for odd-number fatty acids, like undecanoic acid, produces a five-carbon fatty acyl substrate that is oxidized and split into acetyl-CoA and propionyl-CoA. To enter the TCA cycle, propionyl-CoA needs additional reactions such as carboxylation. Since the oxidation occurs using a liver extract, CO₂ must be supplied externally for propionyl-CoA carboxylation, enabling the complete oxidation of undecanoic acid.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, which contains biotin. The function of biotin is to activate CO₂ before it is transferred to the propionate group. The addition of avidin obstructs the complete oxidation of undecanoic acid as it binds very tightly to biotin, thereby hindering the activation and transfer of CO₂ to propionate.

In contrast, palmitate oxidation does not require carboxylation, meaning that the presence of avidin doesn't influence its oxidation.

6 0
19 days ago
Commercial grade fuming nitric acid contains about 90.0% HNO3 by mass with a density of 1.50 g/mL, calculate the molarity of the
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x is greater than or equal to 56Step-by-step explanation: He requires at least 56 additional cans. Hence, x should be a minimum of 56.
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14 days ago
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Answer:

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Where

n is the number of moles of hydrogen

n is the mass of hydrogen

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1 month ago
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Diethyl ether (DTH) and Tetrahydrofuran (THF).

Clarification:

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Therefore, solvents like anhydrous diethyl ether or tetrahydrofuran (THF), as well as poly(tetramethylene ether) glycol (PTMG), are used in experimental procedures to limit the exposure of Grignard reagents to air and moisture.
These solvents are chosen because the oxygen they contain stabilizes the magnesium reagent.
THF is a stable compound.
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