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3 years ago

¿para que se utilizan los métodos de fraccionamiento de fases?

1 answer:

5 0

Answer: What are phase fractionation methods used for? well They apply to the prefractionation of the sample to obtain less complex protein mixtures for an easier analysis; they are also used as a means to evidence specific proteins or protein classes otherwise impossible to detect.

Explanation: in spanish /

Se aplican al prefraccionamiento de la muestra para obtener mezclas de proteínas menos complejas para un análisis más fácil; también se utilizan como un medio para evidenciar proteínas específicas o clases de proteínas que de otro modo serían imposibles de detectar.

You might be interested in

This transition metal is in group 12 and has less than 34 protons

slava [35]

Zinc (Zn) has less than 34 protons, 30 to be exact, and is a transition metal in Group 12. Note: it is also called a "post-transition metal."

4 0

2 years ago

A sealed 1.0L flask is filled with 0.500 mols of I_2 and 0.500 mols of Br_2. When the container achieves equilibrium the equilib

DochEvi [55]

Answer:

[IBr] = 0.049 M.

Explanation:

Hello there!

In this case, according to the balanced chemical reaction:

$I_2+Br_2\rightarrow 2IBr$

It is possible to set up the following equilibrium expression:

$K=\frac{[IBr]^2}{[I_2][Br_2]} =0.0110$

Whereas the the initial concentrations of both iodine and bromine are 0.50 M; and in terms of $x$ (reaction extent) would be:

$0.0110=\frac{(2x)^2}{(0.50-x)^2}$

Which can be solved for $x$ to obtain two possible results:

$x_1=-0.0277M\\\\x_2=0.0245M$

Whereas the correct result is 0.0245 M since negative results does not make any sense. Thus, the concentration of the product turns out:

$[IBr]=2x=2*0.0249M=0.049M$

Regards!

7 0

2 years ago

Rank the compounds in each set in order of increasing acid strength.

Viktor [21]

Answer:

See explanation

Explanation:

For this question, we have to remember the effect of an atom with high electronegativity as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an inductive effect. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be weaker and the compound will be more acid (because is easier to produce the hydronium ion $H^+$ ).

With this in mind, for A in the last compound, we have 2 Br atoms near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have more acidity. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.

In B, the difference between the molecules is the position of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a higher inductive effect and more acidity.

See figure 1

I hope it helps!

5 0

3 years ago

If a substance melts at 20oC, what is its freezing point?

shutvik [7]

Answer:

It is difficult, if not impossible, to heat a solid above its melting point because the heat that ... in a solid are packed in a regular structure that is characteristic of that particular substance.

<h3>#carryONlearning </h3>

4 0

3 years ago

Why is there not a constant molar volume for solids and

iragen [17]

Answer:

all of the above

Explanation:

i got the answer right on cK-12

7 0

2 years ago