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3 years ago

1.) Why does carbon form covalent bonds?

Chemistry

2 answers:

juin [17]3 years ago

7 0

For #1, I'd say it's "It will usually bond to multiple atoms which can provide a total of 4 additional electrons."

2. Ionic (I'm quite certain because anions/cations (-1 & +1) are Ionic from what I recall, if that's true it's Ionic.
3. "comparison of the associated families to which the elements belong" and
"the difference in electronegativities" are what I would choose, as I mentioned in a comment earlier.

If I'm wrong let me know, but I am at least 80% sure that these responses are correct from what I remember in Chemistry.

Oksi-84 [34.3K]3 years ago

7 0

Explanation:

1). A bond, that is, formed by sharing of electrons between the combining atoms is known as a covalent bond.

For example, atomic number of carbon atom is 6 with electronic distribution 2, 4.

So, in order to completely fill the octet carbon atom will combine with atoms that can provide 4 additional electrons to a carbon atom.

Hence, carbon form covalent bonds because it will usually bond to multiple atoms which can provide a total of 4 additional electrons.

2). When an atom loses one electron then it acquires a +1 charge. Whereas when an atom tends to gain one electron then it acquires a -1 charge.

An ionic bond is formed when there is transfer of electrons from one atom to another.

For example, fluorine on gaining one electron will form a $F^{-1}$ ion and sodium on losing an electron will form a $Na^{+1}$ ion.

Therefore, we can conclude that ionic bond will be formed for atoms that have a +1 or -1 charge.

3). When there is great difference in electronegativity of combining elements then it also results in the formation of an ionic bond. This is because difference in electronegativity causes partial positive charge on cation and partial negative charge on anion.

Also, families of the combining elements help in determining the difference in electronegativity.

Thus, we can conclude that the relative ionic or covalent character of chemical bonds may be estimated by:

comparison of the associated families to which the elements belong .

the difference in electronegativities.

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3 years ago

(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2

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Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

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