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3 years ago

Why doesn’t a broken wineglass spontaneously fix itself, from a thermodynamic point of view?

2 answers:

8 0

The answer is the option<span> 3.) Fixing a wineglass requires a large decrease in entropy.

Entropy is a termodynamical variable and it is defined as the degree of disorder of the system.

As per the second law of thermodynamics the spontaneous changes only occur when the entropy increases.

</span>Fixing the broken glasses by itself meant to decrease the entropy spontaneusly which violate the principle or second law of thermodynamics. So, it is not possible.

ale4655 [162]3 years ago

5 0

Fixing a wineglass will require a large decrease in entropy.

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Match each element to the number of electrons in its valence shell

blondinia [14]

Answer: Be= 2, C =4, Li = 1 and B=3

Explanation:

The valence shell can be define as the outermost shell of an atom that contains the valence electrons.

Beryllium (Be), electronic configuration; 1s2 2s2, = 2 electrons in its valence shell.

Carbon (C), electronic configuration; 1s2 2s2 2p2, = 4 electrons in its valence shell.

Lithium (Li), electronic configuration; 1s2 2s1 = 1 electron in its valence shell.

Boron (B) , electronic configuration; 1s2 2s2 2p1 = 3 electron in its valence shell.

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2 years ago

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tresset_1 [31]

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2 years ago

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yulyashka [42]

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2 years ago

Complete the mechanism for the base-catalyzed opening of the epoxide below by adding any missing atoms, bonds, charges, nonbondi

worty [1.4K]

Complete Question:

check the first image for complete part of the question

Answer and Explanation:

Epoxide is a three membered ring made up of two carbon atoms and one oxygen atom. Epoxides are cyclic ethers. Due to its ring size, it is highly strained and very reactive. Epoxide ring opening takes place with respect to addition of acid and base.

Ring opening of epoxide with acid:

In the presence of base, the nucleophile attacks the epoxide ring at more substituted site and inverse stereochemistry takes place.(check file 2 attached)

Ring opening of epoxide with base:

The backside attack of nucleophile takes place in less substituted site and then it undergoes protonation to form a product.

(check file 2 attached)

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