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.
Answer: He spends 29.17 hours per week
Explanation:
1 class = 50mins
7 classes = 7 x 50mins = 350mins
For 5 days in a week, we have
5 x 350mins = 1750mins
Converting to hours, we have
1750 /60 = 29.17 hours
Answer:
I think it is three times I'm not sure
The answer is 'equal'. Hydroxide ions are OH- and Hydrogen ions are H+. Have you noticed they're opposite charges? Positive + negative = neutral. That's all there is to it :)
First of all, as you seen the gases are noble which means that will not react with each other and in this case each gas create individual pressure.
P
= total pressure
P
= pressure of neon
P
= pressure of argon
P
= pressure of helium {which is required}
P = P + P + P
1.25 = 0.68 + 0.35 + P
P = 1.25 - [0.68 + 0.35] = 0.22 atm
