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garik1379 [7]
4 years ago
12

which functional group is more electronegative, amine C-NH, or alcohol C-OH, in assigning carbon environment?

Chemistry
1 answer:
Studentka2010 [4]4 years ago
6 0

Answer:

The correct answer is: alcohol (C-OH) functional group

Explanation:

Electronegativity is described as the ability or the tendency of an element to attract the electron density or shared bonding electrons towards itself.  

The electronegativity value of oxygen atom, nitrogen atom and, carbon atom is 3.44, 3.04, and 2.55.

From this data we can conclude that oxygen is more electronegative than nitrogen. Also, the electronegativity difference of oxygen and carbon (0.89) is greater than the electronegativity difference of nitrogen and carbon (0.49).

Therefore, <u>the</u><u> electronegativity </u><u>of the oxygen containing </u><u>alcohol</u><u> functional group (C-OH) will be greater than the electronegativity of the nitrogen (C-NH) containing amine functional group.</u>

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Answer:

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I hope this answer helps

5 0
3 years ago
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"How much NH4Cl, when present in 2.00 liters of 0.200 M ammonia, will give a solution with pH = 8.20? For NH3, Kb = 1.8 x 10-5"
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Answer:

245.66g of NH₄Cl is the mass we need to add to obtain the desire pH

Explanation:

The mixture of NH3/NH4Cl produce a buffer. We can find the pH of a buffer using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where [A⁻] is the molar concentration of the base, NH₃, and [HA] molar concentration of the acid, NH₄⁺. This molar concentration can be taken as the moles of each chemical</em>

<em />

First, we need to find pKa of NH₃ using Kb. Then, the moles of NH₃ and finally replace these values in H-H equation to solve moles of NH₄Cl we need to obtain the desire pH.

  • <em>pKa NH₃/NH₄⁺</em>

pKb = - log Kb

pKb = -log 1.8x10⁻⁵ = <em>4.74</em>

pKa = 14 - pKb

pKa = 14 - 4.74

pKa = 9.26

  • <em>Moles NH₃</em>

<em>2.00L ₓ (0.200mol NH₃ / L) = 0.400 moles NH₃</em>

  • <em>H-H equation:</em>

pH = pKa + log [NH₃] / [NH₄Cl]

8.20 = 9.26 + log [0.400 moles] / [NH₄Cl]

-1.06 =  log [0.400 moles] / [NH₄Cl]

0.0087 =  [0.400 moles] / [NH₄Cl]

[NH₄Cl] = 0.400 moles / 0.0087

[NH₄Cl] = 4.59 moles of NH₄Cl we need to add to original solution to obtain a pH of 8.20. In grams (Using molar mass NH₄Cl=53.491g/mol):

4.59 moles NH₄Cl ₓ (53.491g / mol) =

<h3>245.66g of NH₄Cl is the mass we need to add to obtain the desire pH</h3>

<em />

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A dilute solution is prepared by transferring 40.00 ml of a 0.3433 m stock solution to a 750.0 ml volumetric flask and diluting
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     (0.3433 M)(40 mL) = M₂(750 mL)
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