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Semmy [17]
3 years ago
13

The pka of the α-carboxyl group of glutamine is 2.17, and the pka of its α-amino group is 9.13. calculate the average net charge

on glutamine if it is in a solution that has a ph of 8.20. if the charge is positive, do not enter a \" \" sign.
Chemistry
2 answers:
IgorLugansk [536]3 years ago
4 0
I think u add the numbers and get the total and divide by the set of numbers to get the average.
Serjik [45]3 years ago
3 0

Answer:

-0.1051

Explanation:

First, we need to find the degree of dissociation (α) of each of the groups using the Handerson-Halsebach equation:

pH = pKa + log (α/1-α)

log (α/1-α) =  pH - pKa

log (1-α/α) = pKa - pH

1-α/α =10^{pKa - pH}

1-α = 10^{pKa - pH}*α

α = \frac{1}{10^{pKa-pH} +1}

So, fo the α-carboxyl group:

α = \frac{1}{10^{-6.03} +1}

α = 1.0000

And for the α-amino group

α = \frac{1}{10^{0.93} +1}

α = 0.1051

In the ionization, the α-carboxyl group goes from neutral to a negative ion with charge -1, and the α-amino group goes from a positive ion with charge +1 to a neutral compound. So, the net charge is how many ions are presented multiplied by the charge.

For the first one, the value of α is the value of ions, but for the second, α represents the neutral, so the ions are 1-α:

-1*(1.000) + (+1)*(1-0.1051) = -0.1051

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And your unbalanced and balanced equations are correct.

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3 years ago
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egoroff_w [7]

Answer:

Volume occupied by oxygen gas at 15 degree centigrade​ is equal to 316.5 centimeter cube

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Volume occupied by oxygen gas at 15 degree centigrade​ is equal to 316.5 centimeter cube

5 0
2 years ago
Question 15 How many grams of NaCl are required to make 500.0 mL of a 1.500 M solution? 58.40 g 175.3 g 14.60 g 43.83 g
ExtremeBDS [4]
Hi!

To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g

To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:

gNaCl=500mLsol* \frac{1L}{1000 mL}* \frac{1,500 mol NaCl}{1Lsol}* \frac{58,4428 g NaCl}{1 mol NaCl} \\ =43,83gNaCl

Have a nice day!
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