This problem is easily solvable because radioactivity equations are common and well-established. The pseudo-first reaction is written below:
A = A₀(1/2)^(t/h)
where
A is the final amount
A₀ is the original amount
t is the time
h is the half life
5,000 = A₀(1/2)^(24,000/6,000)
Solving for A₀,
<em>A₀ = 80,000 atoms</em>
An atom having 52 protons and 54 electrons would have an atomic number of 52 and a net charge of -2. This element would be 52 Te 2-, or choice A.
Answer:
hydrogen helium lithium berrylium boron carbon nitrogen
You will feel way better because in our class we learned that lukewarm baths are good for fevers and warm washcloths :)
Answer:
an increase in 1-butene was observed when t-butoxide was used
Explanation:
When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.
Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.
The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.
The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.
