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baherus [9]
2 years ago
12

The pK1, pK2, and pKR of the amino acid lysine are 2.2, 9.1, and 10.5, respectively. The pK1, pK2, and pKR of the amino acid arg

inine are 1.8, 9.0, and 12.5, respectively. A student at SDSU wants to use ion exchange chromatography to separate lysine from arginine. What pH is likely to work best for this separation?
Chemistry
1 answer:
almond37 [142]2 years ago
5 0

Answer:

pH 9,8 is likely to work best for this separation

Explanation:

Ion exchange chromatography is a chemical process where molecules are separated by affinity to an ion exchange resin. To separate different aminoacids you must use the isoelectric point (That is the pH where the aminoacid will be in its neutral form).

For lysine, PI is:

pH = \frac{1}{2} (9,1+10,5) = 9,8

For arginine:

pH = \frac{1}{2} (9,0+12,5) = 10,75

At pH = 9,8 lysine will be in its neutral form and will not be retain in the column but arginine will be in +1 charge being retained by the ion exchange resin.

Thus, <em>pH 9,8 is likely to work best for this separation</em>

<em></em>

I hope it helps!

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Which of the following species contains manganese with the highest oxidation number?
ioda

In NaMnO₄, Mn has the highest oxidation number.

The question is incomplete, the complete question is;

Which of the following species contains manganese with the highest oxidation number?

A) Mn

B) MnF₂

C) Mn₃(PO₄)₂

D) MnCl₄

E) NaMnO₄

In order to ascertain the specie that contains manganese with the highest oxidation number, we must calculate the oxidation number of manganese in each of the species one after the other.

1) For Mn, the oxidation number of Mn is zero because the atom is uncombined.

2) For MnF₂;

Mn has an oxidation number of +2

3) For Mn₃(PO₄)₂

Mn has an oxidation number of +2

4) For MnCl₄

Mn has an oxidation number of +4

5) For NaMnO₄

Mn has an oxidation number of +7

Hence in NaMnO₄, Mn has the highest oxidation number.

Learn more: brainly.com/question/10079361

7 0
1 year ago
HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
2Na2O2 + 2CO2 → 2Na2CO3 + O2
ahrayia [7]

the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment

theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations

we need to first find the theoretical yield

2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂

molar ratio of Na₂O₂ to Na₂CO₃ is 2:2

number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed

number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol

therefore number of Na₂CO₃ moles formed is - 0.10 mol

mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g

therefore theoretical yield is 10.6 g

percent yield = actual yield / theoretical yield  x 100%

81.0  % = actual yield / 10.6 g x 100 %

actual yield = 10.6 x 0.81

actual yield = 8.59 g

therefore actual yield is 8.59 g

7 0
2 years ago
Read 2 more answers
Describe how you would prepare exactly 100 mL of 0.109 M picolinate buffer, pH 5.61. Possible starting materials are pure picoli
Pepsi [2]

Answer:

1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL

Explanation:

<em>The pKa of the picolinic acid is 5.4.</em>

Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:

pH = pKa + log [Picolinate] / [Picolinic]

<em>Where [] could be taken as moles of each species</em>

<em />

5.61 = 5.4 + log [Picolinate] / [Picolinic]

0.21 = log [Picolinate] / [Picolinic]

1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>

<em></em>

Now, both picolinate and picolinic acid will be:

0.100L * (0.109mol / L) =

0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>

<em></em>

First, as we will start with picolinic acid, we need add:

0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid

Now, replacing (2) in (1):

1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]

1.62181 [Picolinic] = 0.0109 moles - [Picolinic]

2.62181 [Picolinic] = 0.0109 moles

[Picolinic] = 4.157x10⁻³ moles

And:

[Picolinate] = 0.0109 - 4.157x10⁻³ moles =

<h3>6.743x10⁻³ moles</h3><h3 />

To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:

Picolinic acid + NaOH → Picolinate + Water

<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>

<em />

6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:

6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =

<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />

Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:

<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>
3 0
2 years ago
As photosynthesis occurs in chloroplasts, O2 is produced from _____ via a series of reactions associated with _____. View Availa
Monica [59]

Answer:

H₂O, Photosystem II

Explanation:

  • Photosynthesis is the process that enables autotrophs such as green plants and algae to generate food using water, carbon dioxide and energy from the sun.
  • It occurs in two phases, that is, the light-dependent phase and the light-independent phase.
  • During photosystem II energy from the sun is used to break-down water molecules to yield oxygen and hydrogen ions. Oxygen is released away to the atmosphere while hydrogen ions are used in the next phase to generate ATP molecules.

6 0
2 years ago
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