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Irina-Kira [14]
3 years ago
6

Carbon can bond to itself to form which of the following?

Chemistry
1 answer:
svetoff [14.1K]3 years ago
4 0

B

Carbon has 4 valence electrons that can be involved in the formation of a covalent bond. This is why it can form various types of bonds with itself or other elements. Depending on the number of valence electrons involved in the binding, the bond can be a single, double or triple bond.

Explanation:

Remember the atoms sharing electrons in a covalent bond are aimed at achieving stable electron configuration. Carbon (2.4) being in the middle of the 2nd period seeks to be either 2 or 2.8.

This is why carbon based structures can vary so much due to the large variability in which the carbon atoms can bond. Remember carbon is the same element that forms graphite, diamond, and coal.

Learn More:

For more on the carbon atom check out;

brainly.com/question/13255170

brainly.com/question/12882572

#LearnWithBrainly

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Answer:

C. Soil

Explanation:

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3 years ago
A. Electrons have a charge of_____
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Draw the major product for the reaction of 1-butyne with water in the presence of catalytic TfOH (i.e., CF3SO3H). Then answer th
Mademuasel [1]

Answer:

2-Butanone

Explanation:

From the given information:

The presence of mercury as an acid catalyst brings about the addition of water to the triple bond which yields enol. Then, according to Markownikov's rule and after tautomerism has occurred, we have a methyl ketone ( 2- Butanone) as the product.

The answer regarding the transformation is addition and hydration.

4 0
3 years ago
Determine the [OH−] of a solution that is 0.115 M in CO32−. For carbonic acid (H2CO3), Ka1=4.3×10−7 and Ka2=5.6×10−11.
lianna [129]

Answer:

[OH⁻] = 4.3 x 10⁻¹¹M in OH⁻ ions.

Explanation:

Assuming the source of the carbonate ion is from a Group IA carbonate salt (e.g.; Na₂CO₃), then 0.115M Na₂CO₃(aq) => 2(0.115)M Na⁺(aq) + 0.115M CO₃²⁻(aq). The 0.115M CO₃²⁻ then reacts with water to give 0.115M carbonic acid; H₂CO₃(aq) in equilibrium with H⁺(aq) and HCO₃⁻(aq) as the 1st ionization step.

Analysis:

            H₂CO₃(aq)     ⇄     H⁺(aq)    +    HCO₃⁻(aq); Ka(1) = 4.3 x 10⁻⁷

C(i)          0.115M                      0                  0

ΔC              -x                        +x                  +x

C(eq)    0.115M - x                   x                    x

            ≅ 0.115M

Ka(1) = [H⁺(aq)][HCO₃⁻(aq)]/[H₂CO₃(aq)] = [(x)(x)/(0.115)]M = [x²/0.115]M

= 4.3 x 10⁻⁷  => x = [H⁺(aq)]₁ = SqrRt(4.3 x 10⁻⁷ · 0.115)M = 2.32 x 10⁻⁴M in H⁺ ions.

In general, it is assumed that all of the hydronium ion comes from the 1st ionization step as adding 10⁻¹¹ to 10⁻⁷ would be an insignificant change in H⁺ ion concentration. Therefore, using 2.32 x 10⁻⁴M in H⁺ ion  concentration, the hydroxide ion concentration is then calculated from

[H⁺][OH⁻] = Kw => [OH⁻] = (1 x 10⁻¹⁴/2.32 x 10⁻⁴)M = 4.3 x 10⁻¹¹M in OH⁻ ions.

________________________________________________________

NOTE: The 2.32 x 10⁻⁴M  value for [H⁺] is reasonable for carbonic acid solution with pH ≅ 3.5 - 4.0.

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Ilia_Sergeevich [38]

Answer:

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Explanation:

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