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3 years ago

Assume that a compound is a cyclic, planar, completely conjugated ring. Which number of p electrons would make it aromatic?

Chemistry

2 answers:

DochEvi [55]3 years ago

8 0

Answer:

option b is correct

2 p electron makes aromatic

Explanation:

An aromatic compound which is cyclic, planar and has a complete conjugate ring must have (4n + 2)pi electrons(Huckel's rule)

Huckel's Standard (4n+2 rule): For a compound to be an aromatic, a particle must have a specific number of pi (electrons with pi bonds, or lone pairs inside p orbitals) inside a shut loop of parallel, adjoining p orbitals. The pi electron tally is characterized by the arrangement of numbers created from 4n+2 where n = zero or any positive whole number (i..e, n = 0, 1, 2, and so forth.). The most widely recognized case in six pi electrons (n = 1) which is found for example in benzene, pyrrole, furan, and pyridine.

where n is the number of pi electrons

where n = 0

(4n +2) pi electrons = 2pi electrons

attached is an example of aromatic which is cyclic, planar and a complete conjugate ring

den301095 [7]3 years ago

8 0

Answer:

2 p electrons.

Explanation:

For any compound to be considered an aromatic compound it must be cyclic,flat, conjugated and it must obey Huckel's rule that states an aromatic compound must have 4n + 2 pi electrons in it's p orbitals for it to be an aromatic compound.

n can represent an integer from 2, 6,10, 14,........

The lone pair is actually in a pure 2p orbital perpendicular to the ring, which means they count as π electrons.

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Use the concentration to obtain the moles. I am assuming you mean to write capital M. because little m means molality.

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2 years ago

What is the theoretical yield if 35.5g of Al reacts 39.0g of Cl2

atroni [7]

Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

$Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}$

$Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}$

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

$Mole of AlCl_3 = Mole of Al * Mole ratio$

$Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}$

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = $133.34 \frac{g}{mol}$

Mass of AlCl3 can be calculated using mole formula as:

$1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}$

Multiplying both side by $133.34 \frac{g}{mol}$

$1.32 mole * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34 \frac{g}{mol}$

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

$Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}$

Mole of Cl₂ = 0.55 mol

b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

$Mole of AlCl_3 = mole of Cl_2 * mole ratio$

$Mole of AlCl_3 = 0.55 mole * \frac{2}{3}$

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

$0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}$

Multiplying both side by

$133.34 \frac{g}{mol}$

$0.367 mol of AlCl_3 * 133.34 \frac{g}{mol} = \frac{mass(g)}{133.34\frac{g}{mol}} * 133.34 \frac{g}{mol}$

Mass of AlCl₃ = 48.93 g

Hence mass of AlCl₃ produced by Cl₂ = 48.93 g

Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

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3 years ago

Read 2 more answers

This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press

Solnce55 [7]

Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

[A]₀ is the initial concentration of A

k is the rate constant, and

t is the time

We also know that for a first order reaction

k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 / 35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀ = - kt

thus:

(pPH₃ )t = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

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