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Vitek1552 [10]
3 years ago
5

Which orbital can be modeled by a "peanut" shape?

Chemistry
1 answer:
kobusy [5.1K]3 years ago
4 0

Answer:

d - orbital

Explanation:

The d-orbital can modelled by a peanut shape.

A peanut shape closely resembles the shape of a double dumb-bell shape. The double dumb-bell shape is like that of a peanut.

  • The s-orbital has a spherical shape.
  • p - orbital is dumb-bell shaped
  • d - orbital is a double dumbell
  • f - orbital has a complex shape.

Therefore, since a peanut is like a double dumb-bell, it clearly models the d-orbital.

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The reactant that will be the best reactant for a nucleophilic aromatic substitution is NO₂- NO₂. The correct option is b.

<h3>What is nucleophilic aromatic substitution?</h3>

Nucleophilic aromatic substitution is a substitution process of nucleophile substance is substituted by halides in an aromatic ring. Aromatic compounds contain this type of substitution.

In option b, the compound is the one nitroxide group substituted by halogen, that is fluorine. The fluorine group is substituted in these given aromatic compounds.

Thus, the correct option is b, NO₂- NO₂.

To learn more about nucleophilic aromatic substitution, refer to the link:

brainly.com/question/28265482

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NO₂F

NO₂- NO₂-F

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CH₃-O-F

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1.00 L of a gas at STP is compressed to 473mL. What is the new pressure of gas?
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Hello!

1.00 L of a gas at STP is compressed to 473 mL. What is the new pressure of gas?

  • <u><em>We have the following data:</em></u>

Vo (initial volume) = 1.00 L  

V (final volume) = 473 mL → 0.473 L  

Po (initial pressure) = 1 atm (pressure exerted by the atmosphere - in STP)  

P (final pressure) = ? (in atm)

  • <u><em>We have an isothermal transformation, that is, its temperature remains constant, if the volume of the gas in the container decreases, so its pressure increases. Applying the data to the equation Boyle-Mariotte, we have:</em></u>

P_0*V_0 = P*V

1*1 = P*0.473

1 = 0.473\:P

0.473\:P = 1

P = \dfrac{1}{0.473}

\boxed{\boxed{P \approx 2.11\:atm}}\:\:\:\:\:\:\bf\green{\checkmark}

<u><em>Answer:  </em></u>

<u><em>The new pressure of the gas is 2.11 atm  </em></u>

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\bf\blue{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

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