a. True.
There is always an equilibrium of the type
NH₃⁺CHRCOOH ⇌ NH₃⁺CHRCOO⁻ ⇌ NH₂CHRCOO⁻
The compound is <em>always in an ionized form</em>.
There are no unionized NH₂CHRCOOH molecules in the solution.
Answer:
2341, last option is the correct choice.
Explanation:
Boiling points of the given compounds are given as:
Best Regards!
<span>You have to use a Newman projection to make sure that the H on C#2 is anti-coplanar with the Br on C#1. (Those are the two things that are going to be eliminated to make the alkene.)
My Newman projection looks like this when it's in the right configuration:
Front carbon (C#2) has ethyl group straight up, H down/left, and CH3 down/right
Back carbon (C#1) has H straight down, Ph up/left, and Br up/right.
Then when you eliminate the H from C#2 and the Br from C#1, you will have Ph and the ethyl group on the same side of the molecule, and you'll have the remaining H and CH3 on the same side of the molecule.
This is going to give you (Z)-2-methyl-1-phenyl-1-butene.</span>
Answer:
The symbol is the right answer.
Explanation:
The “ Symbol” is the correct answer because chemist uses the letters of the alphabet to denote the element. For instance, the element oxygen is denoted by the letter of the alphabet “O”, the hydrogen is denoted by the letter of alphabet “H”, Boron is denoted by the letter of alphabet “B”, etc. Here these are the examples that use one letter but there are other elements that use more than 1 letter as the symbol. For example, the Chlorine is represented by the Cl.
