Answer is: A) Sr (strontium).
The reactivity series<span> is a series of metals from highest to lowest reactivity.</span><span> Metal higher in the reactivity series will displace another.
</span>Strontium is only higher in this group from magnesium. Strontium is stronger reducing agent than magnesium, gives electrons easier.
Answer:
The catalyzed reaction will take 2.85 seconds to occur.
Explanation:
The activation energy of a reaction is given by:
For the reaction without catalyst we have:
(1)
And for the reaction with the catalyst:
(2)
Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:
Since the reaction rate is related to the time as follow:
![k = \frac{\Delta [R]}{t}](https://tex.z-dn.net/?f=%20k%20%3D%20%5Cfrac%7B%5CDelta%20%5BR%5D%7D%7Bt%7D%20)
And assuming that the initial concentrations ([R]) are the same, we have:
![\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bk_%7B1%7D%7D%7Bk_%7B2%7D%7D%20%3D%20%5Cfrac%7B%5CDelta%20%5BR%5D%2Ft_%7B1%7D%7D%7B%5CDelta%20%5BR%5D%2Ft_%7B2%7D%7D%20)
Therefore, the catalyzed reaction will take 2.85 seconds to occur.
I hope it helps you!
Answer:The chemical energy
Explanation:because The chemical energy in the food you eat is changed into another kind of chemical energy that your body can use. Your body then uses that energy to give you the kinetic energy that you use in everything you do.
It is a physical change because only the states as being changes, not the actual bonds in the compound.
The question is incomplete, the complete question is;
Why is a terminal alkyne favored when sodium amide (NaNH2) is used in an elimination reaction with 2,3-dichlorohexane? product. A) The terminal alkyne is more stable than the internal alkyne and is naturally the favored B) The terminal alkyne is not favored in this reaction. C) The resonance favors the formation of the terminal rather than internal alkyne. D) The strong base deprotonates the terminal alkyne and removes it from the equilibrium.
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Answer:
E) The positions of the Cl atoms induce the net formation of the terminal alkyne.
Explanation:
In this reaction, sterric hindrance plays a very important role. We know that sodamide is a strong base, it tends to attack at the most accessible position.
The first deprotonation yields an alkene. The strong base attacks at the terminal position again and yields the terminal alkyne. Thus the structure of the dihalide makes the terminal hydrogen atoms most accessible to the base. Hence the answer.
