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lidiya [134]
4 years ago
15

Bromination occurs on alkene functional groups, but NOT on alkenes found within aromatic functional group, such as the phenyl ri

ng of cinnamic acid. Chemists would describe this difference in reactivity as an example of _____

Chemistry
1 answer:
liq [111]4 years ago
8 0

Answer:

When cinnamic acid react with bromine ,addition reaction rapidly occur on alkene functional group to form dibromo product

Explanation:

Phenyl ring is an aromatic hydrocarbon ,when aromatic hydrocarbons react with Cl2,Br2 or KMnO4 no reaction occur ,where as unsaturated hydrocarbon  like alkene react .Aromatic hydrocarbon with these reagents undenr different conditions undergoes subtituition reaction.They react with bromine in presence of lewis acid catalyst ferric bromide.

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A fusion example is the combining of hydrogen isotopes. <br> a. True<br> b. False
solmaris [256]
This statement is true. The combining of hydrogen isotopes is indeed a fusion example. Fusion is the process of the combining two distinct entities and most fusion reactions combine hydrogen isotopes to come up with such product.
3 0
3 years ago
C6H6 miscible or immiscible
Licemer1 [7]
C6H6 is benzene, an organic compound. I would imagine it would be immiscible, since most organic compounds don’t dissolve in water very well.
5 0
3 years ago
Whenever quantities of two or more reactants are given in a stoichiometric problem, you must identify the __________. This is th
artcher [175]

Answer:

1. Limiting reactant.

2. Product.

3. Theoretical yield.

4. Reactants.

5. Actual yield

Explanation:

1. The limiting reactant. The limiting reactant is the reagent that is completely up in the reaction.

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3. Theoretical yield. This is the result obtained from the stoichiometry calculations.

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5 0
3 years ago
Consider the following elementary reaction:
Marat540 [252]

Answer:

[NO]=\frac{k_{-1}}{k_1} [N_2O_2]

Explanation:

Hello!

In this case, since the reaction may be assumed in chemical equilibrium, we can write up the rate law as shown below:

r=-k_1[NO]+k_{-1}[N_2O_2]

However, since the rate of reaction at equilibrium is zero, due to the fact that the concentrations remains the same, we can write:

0=-k_1[NO]+k_{-1}[N_2O_2]

Which can be also written as:

k_1[NO]=k_{-1}[N_2O_2]

Then, we solve for the concentration of NO to obtain:

[NO]=\frac{k_{-1}}{k_1} [N_2O_2]

Best regards!

8 0
3 years ago
How do you label the delta E, on an energy diagram
vladimir1956 [14]
Delta energy on labelled diagram is attached below

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