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.
I believe the answer is background radiatin
Answer:
It depends on the objects mass, the gravitational pull when up or down slopes, and the height of the reference point
Answer:<span>d. 145 minutes
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Half-life is the time needed for a radioactive to decay half of its weight. The formula to find the half-life would be:
Nt= N0 (1/2)^ t/h
Nt= the final mass
N0= the initial mass
t= time passed
h= half-life
If 25.0% of the compound decomposes that means the final mass would be 75% of initial mass. Then the half-live for the compound would be:
Nt= N0 (1/2)^ t/h
75%= 100% * (1/2)^ (60min/h)
3/4= 1/2^(60min/h)
log2 3/4 = log2 1/2^(60min/h)
0.41503749928 = -60min/h
h= -60 min / 0.41503749928= 144.6min
Cells divide in asexual reproduction!
I hope this helps
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