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sashaice [31]
4 years ago
11

Ernest Rutherford's gold-foil experiment showed which of the following?

Chemistry
1 answer:
loris [4]4 years ago
4 0
Ernest Rutherford's gold-foil experiment showed the density of atoms.
<span>The experiment proved that most of an atom is empty space with a very small positively charged nucleus in the middle. 
So, from the given statements he following is true: 
</span><span>Ernest Rutherford's gold-foil experiment showed </span>the existence of a dense, positively charged center in an atom. 
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Answer:

C. 1-ethyl, 3-methylcyclohexane

(Photo for proof at the bottom.)

Explanation:

The 1-ethyl is because you start numbering from the longest branch, towards the next closest branch. Prefix "eth-" means two, there are 2 carbons in the longest branch. 3-methyl is because the next branch is at number 3, and prefix "meth-" means 1, there is 1 carbon in that chain. "Cyclo" in cyclohexane means the skeletal model is shaped like a ring, and the "hexane" means there are 6 carbons in the ring. Prefix "hex" means 6.

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See explanation

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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.

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