Answer:
See explanation
Explanation:
The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.
Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.
Electronic Configuration of elements in a period is same because If you see the electronic Configuration of elements in a period you will notice that the valence shell electrons for all elements are present in the same Shell. For example, in first period consisting of Hydrogen and Helium, both the elements' valence electrons are present in the same Shell.
Electronic Configuration of Hydrogen,
1s^1
Electronic Configuration of Helium,
1s^2
Both elements' valance electrons are present in the 1st shell
(This is just a small example to understand the concept because other periods are long but the first period is short that's why I gave the example of the first period)
B. White Dwarf.
<h3>Explanation</h3>
The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.
As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:
Chandrasekhar Limit: 
.
A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its 
, which is much smaller than the Chandrasekhar Limit.
As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.
The momentum of the body is 12 m/s.
<h3>What is momentum?</h3>
Momentum is the product of the mass of a body and velocity. According to Newton's second law, the rate of change of momentum is equal to the impressed force.
The details required to answer the first two questions are missing hence we can't answer those questions. For the last question;
p = mv
m = mass
v = velocity
p = momentum
v = p/m
v = 72,000 kgm/s/6,000 kg
v = 12 m/s
Learn more about momentum: brainly.com/question/904448
