If you add salt to water then it will boil faster than water without salt
There are two hydrogen atoms in one molecule of water.
The answer is B) 2
Answer:
Explanation:
The metabolic pathway by which energy can be obtained from a fatty acid is called <u>"beta-oxidation"</u>. In this route, acetyl-Coa is produced by removing <u>2 carbons</u> from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the <u>number of rounds</u> that will take place for an <u>18-carbon fatty</u> acid using the following equation:
Where "n" is the <u>number of carbons</u>, in this case "18", so:
We also have to calculate the amount of Acetyl-Coa produced:
Now, we have to keep in mind that in each round in the beta-oxidation we will have the <u>production of 1 
and 1 
</u>. So, if we have 8 rounds we will have 8 and 8 .
Finally, for the total calculation of ATP. We have to remember the <u>yield for each compound</u>:
-) 
-) 
-) 
Now we can do the total calculation:
We have to <u>subtract</u> "2 ATP" molecules that correspond to the <u>activation</u> of the fatty acid, so:
In total, we will have 128 ATP.
I hope it helps!
First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation.
