Answer:
is the rate constant for this reaction.
It will take
to concentration to reach 12.5% of its original value.
Explanation:
A decomposition reaction follows first order kinetics:
Half life of the reaction = ![t_{1/2}=578 years](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D578%20years)
Rate constant of the reaction = k
For first order reaction, half life and rate constant are linked with an expression :
![k=\frac{0.693}{t_{1/2}}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.693%7D%7Bt_%7B1%2F2%7D%7D%20)
![k=\frac{0.693}{578 years}=0.001199 year^{-1}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B0.693%7D%7B578%20years%7D%3D0.001199%20year%5E%7B-1%7D)
is the rate constant for this reaction.
Initial concentration of reactant =
= x
Final concentration of reactant after time t =
= 12.5% of x = 0.125x
The integrated law of first order reaction :
![[A]=[A_o]\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA_o%5D%5Ctimes%20e%5E%7B-kt%7D)
![0.125x=x\times e^{-0.001199 year^{-1}\times t}](https://tex.z-dn.net/?f=0.125x%3Dx%5Ctimes%20e%5E%7B-0.001199%20year%5E%7B-1%7D%5Ctimes%20t%7D)
t = 1,734.31 years =![1.73\times 10^3 years](https://tex.z-dn.net/?f=1.73%5Ctimes%2010%5E3%20years)
It will take
to concentration to reach 12.5% of its original value.
Answer:
See explanation
Explanation:
In order to do this, we need to use 3 reagents to get the final product.
The first one, and logic is the halogenation of the alkene. Doing this, with Br2/CCl4, we'll get an alkane with two bromines, one in carbon 2 and the other in carbon 3.
Then, the next step is to eliminate one bromine of the reactant. The best way to do this, is using sodium ethoxide in ethanol. This is because sodium ethoxide is a relatively strong base, and it will promove the product of elimination in major proportions rather than the sustitution product. If we use NaOH is a really strong base, and it will form another product.
When the sodium ethoxide react, it will form a double bond between carbon 1 and 2 (The carbon where one bromine was with the methyl, changes priority and it's now carbon 3).
The final step, is now use acid medium, such H3O+/H2O or H2SO4/H2O. You can use any of them. This will form an carbocation in carbon 2 (it's a secondary carbocation, so it's more stable that in carbon 1), and then, the water molecule will add to this carbon to form the alcohol.
See the attached picture for the mechanism of this.
You can't usually just use a single spectrum line to confirm the identity of an element because there are cases that the emission line id not clearly defined. When the emission line is very weak compared to surrounding noise, in which case the more datapoints you have to build up confidence for the existence of a particular emission spectra, the better.