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Aleks [24]
3 years ago
5

What kind of structure do carbon atoms form in an aromatic compound?

Chemistry
1 answer:
Amanda [17]3 years ago
6 0
Carbon atoms form aromatic rings. When there are 3 double bonds and 6 carbons this is called a benzene ring.
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Compare and contrast the compositions of binary ionic and binary molecular compounds
ICE Princess25 [194]
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How many electrons can the third energy level hold?
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4 years ago
What is the rate law for the reaction 2A + 2B + 2C --&gt; products
-Dominant- [34]

Answer:

R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

The reaction rate R could be expressed as  

R = k ([A]^m)*([B]^n)*([C]^p)                      (1)

where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

R = ka ([A]^m)                   (2)

Where ka=(k[B]^n)*([C]^p);    

R = kb ([B]^n)                    (3)

Where kb=(k[A]^m)*([C]^p); and  

R = kc ([C]^p)                     (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.  

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A]  0.273   0.819   0.273   0.273

Initial [B]  0.763   0.763   1.526   0.763

Initial [C]  0.400   0.400    0.400   0.800

Rate           3.0       9.0       12.0       6.0

Run#  1  2 3  4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

3 = (3)^m = 3^1  -> m = 1

Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):  

R3/R1=kb[B3]^n/kb[B1]^n=([B3]/[B1])^n

12/3= (1.526/0.763)^n

4= 2^n, -> n = 2

And finally for the determination of p we have using runs 4 and 1:

R4/R1=kc[C4]^p/kc[C1]^p=([C4]/[C1])^p

6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

R = k [A]*([B]^2)*[C]

The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

k = R /( [A]*([B]^2)*[C]) = 3/0.273*(0.763^2)*0.4 =

47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

7 0
4 years ago
At 218°C, Keq=1.2 x 10^-4 for the equilibrium NH4SH(s) ⇋ NH3(g) + H2S(g). Calculate the equilibrium concentrations of NH3 and H2
maksim [4K]

At 218 °C, solid NH₄SH decomposes to form 0.011 M NH₃ and H₂S, as given by its equilibrium constant.

<h3>What is the equilibrium constant?</h3>

The equilibrium constant (Keq) is the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, all raised to their stoichiometric coefficients.

Only gases and aqueous species are included.

  • Step 1. Make an ICE chart.

        NH₄SH(s) ⇋ NH₃(g) + H₂S(g)

I                                 0           0

C                               +x          +x

E                                x            x

  • Step 2. Write the equilibrium constant.

Keq = 1.2 × 10⁻⁴ = [NH₃] [H₂S] = x²

x = 0.011 M

At 218 °C, solid NH₄SH decomposes to form 0.011 M NH₃ and H₂S, as given by its equilibrium constant.

Learn more about equilibrium here: brainly.com/question/5081082

#SPJ1

3 0
2 years ago
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