Hi,
Homogeneous mixture 2
Heterogeneous 3
Mixture 1
Solution 5
Compound 4
Answer:
See explanation below
Explanation:
In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.
At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.
Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.
In the picture below, you have the mechanism for all the 4 products.
Hope this helps
Answer:
Explanation:
In the solution of AB , they are split to give ions as follows
AB ⇄ A⁺ + B⁻
Product of concentration of A⁺ and B⁻ in saturated solution of AB is constant .
This is called Ksp
Ksp = [A⁺] [ B⁻]
If product of concentration of A⁺ and B⁻ exceeds Ksp , the equilibrium shifts to the left side and excess ions come out of solution in the form of precipitate. So second option is the answer.
Answer:
Carbon dioxide and hydrogen monoxide
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