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1 year ago

What descriptive term is applied to the type of diene represented by 2,4-hexadiene?.

Chemistry

1 answer:

Varvara68 [4.7K]1 year ago

6 0

The descriptive term applied to the type of diene represented by 2,4-hexadiene is conjugated diene.

Dienes are compounds which contains two double bonds. These dienes can be non conjugated or conjugated.

Conjugated diene are those compound which have two double bonds joined by a single σ bond. Conjugated dienes can also be called 1,3-diene. To know if diene is conjugated or non conjugated, sp³ hybridization is to b checked and the number of double bonds and single sigma bond is checked.

Conjugated dienes are found in many different molecules. 2,4-hexadiene is a conjugated diene with two carbon-carbon double bonds that are separated by one sigma bond.

The stabilization of dienes by conjugation is better than the aromatic stabilization. Conjugated dienes are more stable than non conjugated or cumulative diene because it has higher electron density of molecules delocalized.

To learn more about conjugated dienes,

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In a particular experiment, 2.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.

Neporo4naja [7]

Answer:

4.18 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Given: For Li

Given mass = 2.50 g

Molar mass of Li = 6.94 g/mol

Moles of Li = 2.50 g / 6.94 g/mol = 0.3602 moles

Given: For $N_2$

Given mass = 2.50 g

Molar mass of $N_2$ = 28.02 g/mol

Moles of $N_2$ = 2.50 g / 28.02 g/mol = 0.08924 moles

According to the given reaction:

$6Li+N_2\rightarrow 2Li_3N$

6 moles of Li react with 1 mole of $N_2$

1 mole of Li react with 1/6 mole of $N_2$

0.3602 mole of Li react with $\frac {1}{6}\times 0.3602$ mole of $N_2$

Moles of $N_2$ that will react = 0.06 moles

Available moles of $N_2$ = 0.08924 moles

$N_2$ is in large excess. (0.08924 > 0.06)

Limiting reagent is the one which is present in small amount. Thus,

Li is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of Li gives 2 mole of $Li_3N$

1 mole of Li gives 2/6 mole of $Li_3N$

0.3602 mole of Li react with $\frac {2}{6}\times 0.3602$ mole of $Li_3N$

Moles of $Li_3N$ = 0.12

Molar mass of $Li_3N$ = 34.83 g/mol

Mass of $Li_3N$ = Moles × Molar mass = 0.12 × 34.83 g = 4.18 g

Theoretical yield = 4.18 g

5 0

3 years ago

Based on the article "Will the real atomic model please stand up?,” describe one major change that occurred in the development o

nydimaria [60]

I don't know this article, but I do know some major changes: first, the change from the plum pudding model (no nucleus, just electrons) to the gold foil experiment, which had Rutherford shoot alpha particles at a sheet of gold only to find them rebounding, proving the existence of a positively charged mass, i.e a nucleus, in the atom. However, this changed again when Bohr realized that the negatively charged electrons should be attracted to the positively charged center, so that there must be something else inside the nucleus.

3 0

3 years ago

Read 2 more answers

What is the transfer of electrons in Al + Cl = AlCl3

otez555 [7]

Answer:

3 e⁻ transfer has occurred.

Explanation

This is a redox reaction.

Oxidation (loss of electrons or increase in the oxidation state of entity)

Reduction (gain of electrons or decrease in the oxidation state of the entity)
An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
[Ne]= (1s²) (2s² 2p⁶)

A combination of both the reactions( Half-reactions) leads to a redox reaction.

Let us look at initial configurations of Al and Cl

[Al]= 1s² 2s² 2p⁶ 3s² 3p¹

[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵

Hence, Al can lose 3 electrons to achieve octet config.

and, Cl can gain 1e to achieve nearest noble gas config. [Ar]

This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.

Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃

Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)

Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.

3 0

3 years ago

what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?

IgorLugansk [536]

Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

5 0

3 years ago

If I have a molar concentration of Na2S2O3 of .02M and a volume of .001L. As well as a Molar Concentration of KI .3M and H2O2 of

Aleonysh [2.5K]

You have to use everything that is given since you have to know which is the limiting reactant. We find the limiting reactant by calculating the number of moles of each reactant and compare the number of moles. The limiting reactant would be the one that is consumed fully by the reaction.

4 0

3 years ago