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Reil [10]
3 years ago
14

Referring back to question 6 above, in the TLC plate on the left the mobile solvent phase was ethyl acetate, in the plate on the

right it was 97% ethyl acetate and 3% methanol. Explain why the Rf values in the plate on the right are larger than those in the plate on the left. Your explanation must be in the terms of intermolecular forces, you are not allowed to only use the word "polarity".
Chemistry
1 answer:
iren2701 [21]3 years ago
3 0

Answer:

Polarity is a measure of molecule's electrical charge which means that molecule or it's groups can have a dipole moment and also make hydrogen bonds.

Explanation:

So in terms of polarity methanol is more polar then ethyl acetate due to it's higher dipole moment and that it is a stronger hydrogen bond acceptor and also can be a hydrogen bond donor. So if we add methanol to the elutioning system it becomes more polar because and our compound moves faster on tlc due to methanol providing some more hydrogen bonds to it and increasing dipole-dipole interactions.

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sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

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Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4  × \frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}

= 21284.56606

mass of CaCO3 =  2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

brainly.com/question/12498769

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