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Anastasy [175]
3 years ago
9

Which element is used to paint the skin to pervent infection from cuts and scratches

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
4 0
Iodine is the element 


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3. Briefly discuss the results of the TLC. Was there evidence of unreacted p-nitrobenzaldehyde in either product
Tems11 [23]

TLC means Thin Layer Chromatography. It is a method that can best be described as "Affinity-Based" used in the separation of compounds that are in a mixture.

<h3>What is unreacted p-nitrobenzaldehyde?</h3>

Unreacted p-nitrobenzaldehyde is simply an organic aromatic compound that contains a nitro group para-substituted to an aldehyde. in this case, if it is unreacted, that means it is the same as before the chemical reation.

Note that the question is missing key information hence the general answer.

Learn more about TCL at:

brainly.com/question/10296715

4 0
2 years ago
Why can't methanol, CH3OH, be used as a solvent for sodium amide, NaNH2? Sodium amide is nonpolar and methanol is polar. Sodium
Elodia [21]

Answer: sodium amide undergoes an acid -base reaction

Explanation:

sodium amide is a ionic compound and basically exists as sodium cation and amide anion. Amide anion is highly basic in nature and hence as soon as  there is amide anion generated in the solution , Due to its very pronounced acidity it very quickly abstracts the slightly acidic proton available on methanol.

This leads to formation of ammonia and sodium methoxide.

Hence sodium amide reacts with methanol and abstracts its only acidic proton and form ammonia and sodium Methoxide.

Hence the 3rd statement is a corrects statement.

So we cannot use methanol for sodium amide because sodium amide itself would react with methanol and the inherent molecular natur of sodium amide would then change.

The 1st and 2nd statements both are incorrect because both the compounds methanol as well as sodium amide have dipole moments and hence are polar molecules.

The 4th statement is also incorrect as both the molecules have dipole moment and hence there would be ion-dipole forces operating between them.

The following reaction occurs:

NaNH₂+CH₃OH→NH₃+CH₃ONa

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