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saul85 [17]
2 years ago
14

Choose the statements about benzene that are correct.

Chemistry
1 answer:
FromTheMoon [43]2 years ago
3 0

Answer:

All of the C-C bonds are actually identical,

The structure of the molecule is closed like a ring structure,

Explanation:

1) all benzene hydrocarbons are ChHp-2 ⇔ this is <u>false</u> the formula for benzene is C6H6 ( CnHn).

2)All C-C bonds share 2 pairs of electrons ⇔ <u>False </u>

There are not enough electrons to form double bonds on all the carbon atoms

3) Three C-C bonds share 2 pairs of electrons  ⇔ False

Since there are not enough electrons to form double bonds on all the carbon atoms, although the electrons do strengthen all of the bonds on the ring equally.

4) All of the C-C bonds are actually identical ⇔ <u>correct </u>

⇒The actual structure of benzene is a resonance hybrid, with six delocalised π-electrons.  (4n + 2 = π)

⇒Due to this structure with delocalised π-electrons, the structure of benzene is very stable and it's reactivity quite low.

⇒ All the C-C bonds of bezene have sp2-sp2 overlap on both sides, what makes them to form 3 π-bonds

5) The structure of the molecule is closed like a ring structure ⇔ <u>correct</u>

Since there are not enough electrons to form double bonds on all the carbon atoms, although the electrons do strengthen all of the bonds on the ring equally. This results in symmetric molecular orbital. The delocalisation of the electrons is also known as aromaticity or ring structure, what gives benzene a great stability.

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garri49 [273]

Answer:

Mg would blow off. AI would be affective to copper but not to MG

Explanation:

7 0
2 years ago
Consider the chemical reaction: N2 3H2 yields 2NH3. If the concentration of the reactant H2 was increased from 1.0 x 10-2 M to 2
Brilliant_brown [7]

The equilibrium constant of a reaction is defined as:

"The ratio between equilibrium concentrations of products powered to their reaction quotient and  equilibrium concentration of reactants powered to thier reaction quotient".

The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.

To solve this question we need this additional information:

<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>

<em>[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M</em>

<em />

Thus, for the reaction:

N₂ + 3H₂ ⇄ 2NH₃

The equilibrium constant, K, of this reaction, is defined as:

K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}

And Q, is:

Q = \frac{[NH_3]^2}{[N_2][H_2]^3}

Where actual concentrations are:

[NH₃] = 1.0x10⁻⁴M

[N₂] = 4.0M

[H₂] = 2.5x10⁻¹M

Replacing:

Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}

<h3>Q = 1.6x10⁻⁷</h3>

As Q < K,

<h3>The chemical system will shift to the right in order to produce more NH₃</h3>

Learn more about chemical equililbrium in:

brainly.com/question/24301138

8 0
2 years ago
Ron and Hermione begin with 1.50 g of the hydrate copper(II)sulfate ∙ x-hydrate (CuSO4 ∙ xH2O), where x is an integer. Part of t
Gwar [14]

Answer

5

Explanation:

We can go about this using the percentage compositions.

First, we calculate the percentage composition of the copper sulphate. This is obtainable by using the mass.

0.96/1.5 * 100 = 64%

Hence the percentage by mass of the water present is 36%

The molar mass of the anhydrous sulphate is 64 + 32 +4(16) = 160g/mol

The molar mass of the water is 2(1) + 16 = 18g/mol

Not forgetting that it is in multiples of x, the total molar mass of the water is 18x moles

The total mass of the copper sulphate hydrate is 160+ 18x

Now how do we get x? Like it is said earlier, the percentage composition is constant.

Hence, 64/100 * (160 + 18x) = 160

16000 = 64(160 + 18x)

16000 = 10,240 + 1152x

16,000 - 10,240 = 1152x

1152x = 5760

x = 5760/1152

x = 5

7 0
2 years ago
The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C u
GenaCL600 [577]

Answer:

the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

7 0
2 years ago
The reaction pcl3(g)+cl2(g)←−→pcl5(g) has kp=0.0870 at 300 ∘c. a flask is charged with 0.50atmpcl3, 0.50atmcl2, and 0.20atmpcl5
bearhunter [10]
The question only asks regarding the direction of the equilibrium reaction. The general expression of Kp is:

Kp = [PCl₅]/[PCl₃][Cl₂]

The higher the value of K (greater than 1), the more spontaneous the reaction (favors the product side). Otherwise, it favors the reactant side. Since Kp = 0.087 which is less than 1, the direction favors the forward reaction towards the product side.
6 0
2 years ago
Read 2 more answers
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