Suppose that you are given a mixture of proteins with their properties provided in the following chart.
Isoelectric point PI Molecular weight
Protein A 4.1 80
Protein B 9.0 81
Protein C 8.8 37
Protein D 3.9 172
Choose one combination of techniques that can be used to isolate Protein B from Proteins A, C, and D.
a. gel filtration chromatography and ultracentrifugation
b. dialysis and ultracentrifugation
c. dialysis and ion exchange chromatography
d. ion exchange chromatography and gel filtration chromatography

Answers

Answer 1

Answer:

The correct option is d. ion exchange chromatography and gel filtration chromatography

Explanation:

The mixture contains proteins with different isoelectric points and molecular weights. Ion exchange chromatography separates the proteins with different electric charges, which depend on the isoelectric point. At certain pH of the aqueous medium, the proteins will have a certain electric charge and they will be retained differently in the stationary phase. Proteins A and D have similar isoelectric points (around 4.0) and maybe they cannot be efficiently separated by this method; the same occurs with protein B and C (with an isoelectric point around 9.0). To separate proteins A from D and protein B from C, it could be used gel filtration chromatography. This separation method is based on differences in molecular weights. So, at the end of the process, we will obtain the proteins A, B, C, D separated one from each other.


Related Questions

chemistry
Definition in your own words. I will check if you got it from online.

Word:
Malleable
(malleability)

Answers

mallebable- a material that is able to be hammered or pressed permanently without breaking .

As the food burned,
energy was
nergy. Thus, a
thermal
transformed intos
chemical
form of Select
nuclear
$ converted to a
form of
Select
energy.
Check

Answers

Answer:

I don't get it is it even a question?

Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. 4 Hg(l) + 2 O2(g) LaTeX: \rightarrow → 4 HgO(s) Determine the value of LaTeX: \Delta ΔH°rxn for the synthesis, given that

Answers

Joseph Priestly is frequently credited with the discovery of oxygen, and was reported to have produced molecular oxygen from the decomposition reaction of mercury(II) oxide, which is the reverse of the synthesis of HgO depicted in the following equation. [tex]4Hg(l)+2O_2(g)\rightarrow 4 HgO(s) [/tex]Determine the value of [tex]\Delta ΔH°rxn[/tex] for the synthesis, given that [tex]\Delta H_f^0[/tex] for HgO is -90.7 kJ/mol.

Answer: The enthalpy change for this reaction is, -362.8 kJ

Explanation:

The balanced chemical reaction is,

[tex]4Hg(l)+2O_2(g)\rightarrow 4HgO(s)[/tex]

The expression for enthalpy change is,

[tex]\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

[tex]\Delta H=[(n_{HgO}\times \Delta H_{HgO})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{Hg}\times \Delta H_{Hg})][/tex]

where,

n = number of moles

[tex]\Delta H_{O_2}=0[/tex] (as heat of formation of substances in their standard state is zero

[tex]\Delta H_{Hg}=0[/tex] (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

[tex]\Delta H=[(4\times -90.7)]-[(2\times 0)+(4\times 0)][/tex]

[tex]\Delta H=-362.8kJ[/tex]

Therefore, the enthalpy change for this reaction is, -362.8 kJ

Every morning, Jeremiah uses a blender to make a smoothie for breakfast. Which of the
following shows the energy transformation that the blender demonstrates?
А
Electrical - light
B
Chemical → mechanical
с
Mechanical → chemical
D
Electrical mechanical

Answers

Answer:

D

Explanation:

Answer:

D: Electrical -> mechanical

Explanation:

Hope this helps!

Classify the following compounds as binary ionic, ternary ionic, or molecular.

Answers

Answer:bro u already have an answer why are you asking?

Explanation:

an unknown substance has a mass of 57.4 g and occupies a volume of 34.3 ml. what is the density in g/ml?

Answers

Answer:

1.6734 g\ml..hope it helps

a titanium bicycle frame displays 0.250 L of water and has a mass of 1.21kg. what is the density of the titanium on g/cm3?

Answers

Answer:

$4.49 g/cm3

Explanation:

Density=Mass/Volume

plzz put brainiest

You want to clean a 500-ml flask that has been used to store a 0.9M solution. Each time the flask is emptied, 1.00 ml of solution adheres to the walls, and thus remains in the flask. For each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. What is the minimum number of such rinses necessary to reduce the residual concentration of 0.00001 M or below

Answers

Answer:

In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M

Explanation:

In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10

In the first rinse the concentration must be of 0.9M  10 = 0.09M

2nd = 0.009M

3rd = 0.0009M

4th = 0.00009M

5th = 0.000009M →

In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M

A sample of saturated clay was placed in a container and weighed. The weight was 6N. The clay in its container was placed in an oven for 24 hours at 105° C. The weight reduced to a constant value of 5N. The weight of the container is 1N. If G-2.7, determine the:
(a) water content;
(b) void ratio;
(c) bulk unit weight;
(d) dry unit weight;
(e) effective unit weight.

Answers

Answer is given below

Explanation:

given data

weight = 6N

temp = 105° C

weight reduced = 5N

solution

weight of container is 1N

SO W = (6-1) = 5

And Wd = 5 - 1 = 4

so

moisture content is

moisture content = [tex]\frac{W-Wd}{Wd} \times 100[/tex]       .......1

moisture content = [tex]\frac{5-4}{4} \times 100[/tex]

moisture content = 25%

and

as we know density of soil soild = 2700 kg/m³

density of water = 1000 kg/m³

and sp gravity of soil =  [tex]\frac{2700}{1000}[/tex]  = 2.7

so

now we get here bulk unit weight

bulk unit wt = [tex]Yw \times [\frac{G+e}{1+e}][/tex]     ..........2

bulk unit wt = [tex]9.01 \times [\frac{2.7 + 0.675}{1+0.675}][/tex]

bulk unit wt = 19.766 KN/m³

and

so dry unit wt  will be

dry unit wt = [tex]\frac{Ysat}{1+w}[/tex]           ..............3

dry unit wt = [tex]\frac{19.766}{1+0.25}[/tex]

dry unit wt = 15.813 kN/m³

is C5H10 ionic or covalent?

Answers

Covalent because it is 5 and 10 so there even numbers I think
covalent. there is 5 c-c bonds 2 hydrogen atoms attach to each. total # of bonds is 15

How many significant figures are in 3.20x10^2 g?

Answers

Answer:

3

Explanation:

For numbers with decimals, count the number after the decimal.

identify which element is oxidized and which element is reduced.
PLZ HELPP...​

Answers

In the first three equations, Magnesium, sodium and aluminum are oxidized because they loose electrons. Sulfur, oxygen and chlorine are reduced because they gain electrons. In the last equation, magnesium is reduced because it gains hydrogen, and hydrogen is oxidized because it loses electrons.

How many moles of hydrogen gas are present in 65.0 liters at STP?
1456 moles
1.45 moles
3.00 moles
2.90 moles

Answers

Answer:

2.9moles of hydrogen gas

Explanation:

convert liters to dm³

since 1liter= 1dm³

thus, 65.0liters = 65.0dm³

number of moles = volume given/22.4dm³

= 65.0/22.4

=2.9moles

How many moles of water can be formed from 0.57 moles of hydrogen gas?

Answers

Answer:

0.57 water

Explanation:

To solve this problem, we need to write the reaction expression first.

The reactants are oxygen gas and hydrogen gas.

They react to give a product of water

       2H₂    +    O₂   →   2 H₂O  

Given that;

Number of moles of hydrogen gas = 0.57moles

From the balanced reaction expression;

       2 moles of hydrogen gas produces 2 moles of water

   So;

    0.57mole of hydrogen gas will also produce 0.57 water

What produces the magnetic force of an electromagnet?

O magnetic fields passing through the device

O static charged particles on the wire

O movement of charged particles through the wire

O positive and negative charges repelling each other

Answers

Answer:

movement of charged particles through the wire .

Explanation:

When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .

In each row, checkbox under the compound that can reasonably be expected to be more acidic in aqueous solution, e.g have the larger
Ka
H₂ SO₃ H₃ SO ₄
H₃ PO₄ H₃ PO₃
HCH₃ SO₂ HCH₃CO₂

Answers

Explanation:

H2SO3 is more acid than H2TeO3. Since S is more electronegative than Te is. In H2SO3, thus, dissociation of H+ would be smoother.

So, H2SO3's got high Ka.

HCH3SO2 is more acid than HCH3CO2. Since S is more electronegative than C. So, HCH3SO2 is a high Ka.

HClO2 is more acid than HClO. Since in HClO2, after the donation of H+ ion, the negative charge is set by two oxygen atoms, while in HClO, only one oxygen atom stabilizes the negative charge.

So, HClO2 is a high Ka

Two volumes of nitric oxide react with one volume of oxygen gas to form two volumes of a reddish-brown gas. Deduce the formula of this gas and sketch particle representations of its molecules.

Answers

Answer:

Explanation:

Nitric oxide is the gas NO, it reacts with oxygen as shown below;

2NO(g) + O2(g) -----> 2NO2(g)

Now the gas formed is the gas NO2 which is known to be reddish brown in colour.

A diagrammatic representation of this reaction is shown in the image attached to this answer.

Image credit: Chemlibretext

definition of solubility
(science)

Answers

Answer:

th relative ability of a solute to devolve into a solvent

An ionic compound has a generic formula of QR2.

Which elements could the Q and R represent?

Once you choose an answer, check it by plugging those elements into the QR2 formula to see if it looks right.

Q= Sodium R= Oxygen

Q= Magnesium R= Chlorine

Q= Oxygen R= Sodium

Q= Chlorine R= Magnesium

Answers

Answer:

Q= Magnesium R= Chlorine

Explanation:

The element Q should be magnesium and R is chlorine.

An ionic compound is a compound that is formed by the combination of a metal and non-metal. Such bonds forms when there is a transfer of electrons from the metals to the non-metals. This leaves a net positive charge on the metal and a negative charge on the non-metal.

The electrostatic attraction leads to the formation of the bond.

 To solve this problem, the hypothetical compound is QR₂

       Mg                        Cl

     2 8 2                    2 8 7

So, Mg transfers 2 electrons to two atoms of chlorine.

 This leads to the formation of the compound MgCl₂

A series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume. What is the molar concentration of NaCl in solutions A, B and C

Answers

Answer:

Solution A: 0.00400M

Solution B: 0.00400M

Solution C: 4.00x10⁻⁵M

Explanation:

Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:

250mL / 10mL = 25 times.

That means molar concentration of sln A is:

0.100M / 25 = 0.00400M

Solution B is obtained diluting 25mL to 100mL:

100mL / 25mL = 4 times

0.00400M / 4 times = 0.00100M

And solution C is obtained diluting the solution C from 20mL to 500mL:

500mL / 20mL = 25 times

Solution C:

0.00100M / 25 times = 4.00x10⁻⁵M

The formula for serial dilution can be used to obtain the molarity of solution A, B , C.

For solution A

M1V1 = M2V2

M2 = 0.100 M ×  10 mL/250-mL

M2 = 0.004 M

For solution B

M1V1 = M2V2

M2 = 0.004 M × 25 mL/100-mL

M2 = 0.001 M

For solution C

M1V1 = M2V2

M2 = 0.001 M × 20 mL/500-mL

M2 = 0.00004 M

Learn more about serial dilution: https://brainly.com/question/2167827

water is a unique material in that the density of the solid is lower than the density of the liquid (which is why ice forms at the top of a pond and why ice floats in our drinks). if the density for ice at 0C is .917g/mL and the density for water at 0C is .999 g/mL, what is the calculated free space (as %) for each of these materials. you will need to estimate the volume of water as the sum of 2 H atoms and 1 O atom with radii 37 and 66 pm respectively. note that you will also have to assume a quantity of water to perform this exercise

Answers

Answer:

% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

% Free space in ice  = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

Explanation:

As given ,

Density for ice at 0⁰C = 0.917 g/ml

Density for water at 0⁰C = 0.999 g/ml

Radii of H atoms = 37 pm

Radii of O atoms = 66 pm

Now,

Consider 1 ml of water = 1 cm²

As , we know that mass of water in 1 cm² = 0.999 g

Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]

Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²

Now,

Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 5.48×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%

Now,

Consider 1 ml of ice  = 1 cm²

S.I unit of ice = 1×[tex]10^{-6}[/tex] m²

As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g

Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]

Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012

Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]

Now,

Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²

                                              = 1.17×[tex]10^{-9}[/tex] m²

⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²

Now,

Free space = 1×[tex]10^{-6}[/tex]  - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²

% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%

what is the electron configuration for Na?

Answers

1s22s22p63s1. That's the ans

Calculate the number of oxygen atoms in a 50.0g sample of scheelite CaWO4

Answers

Answer:

0.696 atoms of oxygen

Explanation:

We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:

Mass of CaWO₄ = 50 g

Molar mass of CaWO₄ = 40 + 184 + (4×16)

= 40 + 184 + 64

= 288 g/mol

Mole of CaWO₄ =?

Mole = mass / Molar mass

Mole of CaWO₄ = 50 / 288

Mole of CaWO₄ = 0.174 mole

Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:

1 mole of CaWO₄ contains 4 atoms of oxygen.

Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.

Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.

Molecule A undergoes isomerization to molecule B in acetone. Using curved arrows, showing key intermediates and any formal charges, propose a detailed mechanism for this isomerization. Provide a brief explanation why this isomerization occurs.

Answers

Answer:

hello your question is incomplete attached below is the complete question

answer :

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

Explanation:

Reason for the mechanism

we use Isomerization because conjugated allylic carbocation  is more stable when compared to a Non-conjugated Allylic carbocation

attached below is the detailed mechanism

16. Using the average atomic masses given in the inside front cover of this book, calculate the indicated quantities.

d. the number of moles of cobalt represented by 5.99 x 1021 cobalt atoms e. the mass of 4.23 mol of cobalt

f. the number of cobalt atoms in 4.23 mol of cobalt

g. the number of cobalt atoms in 4.23 g of cobalt

Answers

Answer:

d. 9.95 × 10⁻³ mol

e. 249 g

f. 2.55 × 10²⁴ atoms

g. 4.32 × 10²² atoms

Explanation:

d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol

e. the mass of 4.23 mol of cobalt

The molar mass of cobalt is 58.93 g/mol.

4.23 mol × 58.93 g/mol = 249 g

f. the number of cobalt atoms in 4.23 mol of cobalt

We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.

4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms

g. the number of cobalt atoms in 4.23 g of cobalt

First, we will calculate the moles of cobalt using the molar mass of cobalt.

4.23 g × 1 mol/58.93 g = 0.0718 mol

Then, we will calculate the number of cobalt atoms using Avogadro's number.

0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms

When preparing for work in the fume hood, be sure to gather all necessary tools, glassware, and chemicals _________ to minimize the number of times the hood sash is raised and lowered. Work as much as possible in the _________ of the work surface to keep the area tidy and promote air flow. If you need to step away from the experiment to obtain another item, _________ the sash during this time.

Answers

Answer:

In advance

middle

lower

Explanation:

These are the safety precautions needed when carrying out duties in the fume hood.

When planning and preparing to work in a fume hood (a locally designed area to reduce exposure to hazardous fumes). It is advisable to make all equipment readily available at your disposal in advance to reduce and minimize the raising and lowering of the hood sash at intervals.

It is also pertinent to understand that working in the middle of the work surface helps to promote the movement of air and keeps the area neat and tidy.

However, if any case where there is a need to get a new tool or equipment during the process of working in a fume hood, it is advisable to lower the sash at that point in time.

3.4 x 10-25 kg = ? microounces

Answers

Answer: 1.2 x 10^-17 microounces

Explanation:

Ounce = 28.5G microounce = 28.5*10^-6g

3.4*10^-25 kg = 3.4*10^-22 g = (3.4/2.85)*10^(-22+5) = 1.2*10-17

Which of the following is the Arrhenius equation?

Answers

Answer:

A is the answer the reason why is because I got it right when I was doing my test

A species of desert plant produces flowers that only bloom at night. How does this enhance the survival of the species?

A. This allows the plants to conserve water and not bloom during the heat of the day.

B. This species relies on nocturnal animals like moths for pollination and reproduction

C. This species relies on moonlight for photosynthesis.

D.This allows flowers to stay closed durng the day when herbivores are more likely to eat them.

Od

Answers

Answer:

A. This allows the plants to conserve water and not bloom during the heat of the day

Explanation:

Most desert plants only bloom at night because they take advantage of animals like moths and insects that fly at night for pollination and reproduction.

Because these plants are in the desert and do not get enough water except from short occasional rainfalls, they conserve water and not bloom during the heat of the day. They bloom at night when the temperature is low and this enhances their water conservation and survival.

Suppose a 500.mL flask is filled with 0.40mol of N2 and 1.0mol of NO. The following reaction becomes possible:
N2g+O2g ->2NOg
The equilibrium constant K for this reaction is 5.93 at the temperature of the flask. Calculate the equilibrium molarity of N2. Round your answer to two decimal places.

Answers

Answer:

[N₂] = 1.1M

Explanation:

Based on the chemical reaction:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Equilibrium constant, K, is defined as:

K = 5.93 = [NO]² / [N₂] [O₂]

Where [] are equilibrium concentrations of each specie

As initial concentrations are:

N₂ = 0.40mol / 0.500L = 0.8M

NO = 1mol / 0.500L = 2M

The equilbrium concentrations are:

[NO] = 2M - 2X

[N₂] = 0.8M +X

[O₂] = X

Replacing:

5.93 = [2 - 2X]² / [0.8+X] [X]

5.93 = 4 - 8X + 4X² / 0.8X + X²

4.744X + 5.93X² = 4 - 8X + 4X²

1.93X² + 12.744X - 4 = 0

Solving for X:

X = -6.9M → False solution. There are no negative concentrations

X = 0.3M. Real solution.

[N₂] in equilibrium is:

[N₂] = 0.8M +0.3M

[N₂] = 1.1M

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