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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.

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What is the [OH-] of solution with pH of 3.4?

arsen [322]

Answer:

(A) 4×0.0001M

Explanation:

[OH-] = 10^-3.4 which is equivalent to 4×0.0001M

5 0

3 years ago

Read 2 more answers

Which one of the following is a pure substance? 1.milk2. Concrete3. Wood 4. Salt water5. Elemental copper

Sauron [17]

Elemental copper would be a pure substance.

5 0

3 years ago

I NEED HELP PLEASE!!

VikaD [51]

Answer:

The amount of heat absorbed is 5.183889 kJ .

Explanation:

In conversion of water to ice it rejects some heat while in conversion of ice to water it absorbs heat which is called latent heat which is given as 6.02 kJ/mol.

The amount of ice given is 15.5 g.

Converting it to moles as the latent heat is given in per moles:

$\frac{given\\weight \\ (in\\grams)}{molecular\\weight\\(in\\grams)}$

Molecular mass of Hydrogen (H) and Oxygen (O) is 1 u and 16 u respectively.

Molecular mass of water is 18 g ( $H_{2} O$ ⇒ 2*1+16=18 ).

mole = 15.5/18 ≈ 0.8611 moles

Therefore the amount of heat absorbed by 15.5 g of ice ( 0.8611 moles) = Latent heat * moles

Heat absorbed = 6.02*0.8611

= 6.02*(15.5/18)

≈ 5.183889 kJ

5 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

Answer: The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

Explanation:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

Someone please help! I will give Brainliest to who answers the best! Write an explanation of how mass determines (or affects) th

xxTIMURxx [149]

Gravity is the force of motion pulling down objects to the ground. If there was no gravity, everyone would walking as if they were on the moon.

Mass is what gravity needs. If an object has a little amount of mass, gravity will be able to easily bring it to the ground.

If an object has a very huge amount of mass, gravity will still be able to bring it to the ground but it will be hard.

For example: An airplane has a HUGE amount of mass. Gravity pulls it down but the airplane needs to be steering up in order for it to be straight. Gravity is applied on the airplane when it is landing.

BUT..... if a table is in the way of an object it depends if it will fall down to the ground or stay on the table.

If an object has little mass and a table is in the way of gravity pulling it down to the ground, the object will stay on the table. Like a plate of food on a table.

If an object has a very big amount of mass and a table is in the way of gravity pulling it to the ground, the object will break the object and make it's make to the ground. That is mostly why most of the time people have very strong tables/ anything to hold a heavy object.

Another example is if you're lifting weights and you have little amount of mass, you're most likely to get the little sized weight. It depend on you mass.

Here are some pictures I included here as well of Mass and gravity.

Glad to help! :) :D

8 0

3 years ago