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Norma-Jean [14]
3 years ago
8

Place each charge form of alanine under the pH condition where it would be the predominant form. The pKa values for the carboxyl

group and amino group of alanine are approximately 2.3 and 9.7, respectively.Note: If you answer any part of this question incorrectly, a single red X will appear indicating that one or more of the items were placed incorrectly.(A) pH < 1 for H3C-C(H)(NH3+)-COO-(B) pH = pl for H3C-C(H)(NH2)-COO- (C) pH > 11 for H3C-C(H)(NH2)-COOH(D) Does not occurs in any significant pH (for H3C-C(H)(NH3+)-COOH)
Chemistry
1 answer:
damaskus [11]3 years ago
8 0

Answer:

(A) pH < 1 the predominant form is the cation: H3C-C(H)(NH3+)-COOH  

(B) pH = pl the predominant form is the zwitterion H3C-C(H)(NH3+)-COO-

(C) pH > 11 the predominant form is the anion: H3C-C(H)(NH2)-COO-

(D) Does not occurs in any significant pH: H3C-C(H)(NH2)-COOH

Explanation:

Amino acids are bifunctional because they have an amine group and a carboxyl group. The amine group is a weak base and the carboxyl group is a weak acid, but the pKa of both groups will depend on the whole structure of the amino acid. Also, every amino acid has an isoelectric point (pI), which means the pH were the predominant form of the amino acid is the zwitterion. The structure of the alanine (CH3CH2NH2COOH) shows it has the carboxyl group at C1 with a pKa1 of 2.3  and the amino group at C2 whit the pKa2 of 9.7. The isoelectric poin (pI) of Alanine is 6. Consequently, the protonation of the molecule will depend on the pH of the solution. There are three possibilities:

1) If the pH is under the pKa of the carboxyl group (2.3) the predominant form will be with the amino group protonated, forming a cation (CH3CH(NH3+)COOH).

2) If the pH is between pKa1 (2.3) and pKa2 (9.7) the predominant form will be the zwitterion (CH3CH(NH3+)(COO-)).

3) If the pH is upper the pKa2 of the amino group (9.7) the predominant form will be with the carboxyl group deprotonated, forming an anion (CH3CHNH2(COO-)).

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What volume of 1.00 m hcl in liters is needed to react completely (with nothing left over) with 0.750 l of 0.100 m na2co3?
kotykmax [81]
The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol 
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then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol 
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3 0
3 years ago
PLS ANSWER THANK YOUU
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Answer:

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What is thePercent composition of dichlorine heptoxide?
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A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o
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Answer:

The concentration the student should write down in her lab is 2.2 mol/L

Explanation:

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S: 32.065 u

O: 15.999 u

Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.

Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.

For mole(s) of Na2S2O3 = (mass taken)/(molar mass)

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Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)

= 0.05029 L.

To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:

= (moles of sodium thiosulfate)/(volume of solution in L)

= (0.1090 mole)/(0.05029 L)

= 2.1674 mol/L

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