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1 year ago

Before the development of electrophoresis to separate macromolecules, ____________ was used.

1 answer:

max2010maxim [7]1 year ago

4 0

Before the development of electrophoresis to separate macromolecules, high-speed centrifugation was used to isolate DNA.

A laboratory procedure called electrophoresis is used to divide DNA, RNA, or protein molecules according to their size and electrical charge. The molecules are moved by an electric current through a gel or other matrix. The technology of electrophoresis is crucial for the separation and examination of nucleic acids. At the lab bench, cloned DNA fragments are frequently isolated and worked with using nucleic acid electrophoresis.

High-speed centrifugation employs centrifugal force to separate particles with various densities or masses suspended in a liquid. High-speed rotation of the solution inside the tube causes each particle's angular momentum to experience centrifugal forces inversely proportionate to its mass.

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Describe briefly the changes from the control values in the MEAN values for volume, specific gravity and NaCl after drinking pur

Yuri [45]

Answer:

As specific gravity is defined as weight per unit volume,so urine is the less dense fluid,so urine will be in hyperosmotic category.

8 0

3 years ago

A certain reaction at equilibrium has more moles of gaseous products than of gaseous reactants.(b) Write a statement about the r

mojhsa [17]

To be equal as $K_p$ , $K_c$ needs to be divided by $(RT)^{\Delta n}$

<h3>Briefly explained</h3>

When the amount of gaseous reactant is greater than the amount of gaseous product, where $n_{products} < n_{reactants}$ ,

then = $K_p < K_c$

It's because $\Delta n$ is negative, which places (RT) in the denominator. This is how the equation will now appear.

$K_p = K_c(RT)^{-\Delta n} = \frac{ K_c}{(RT)^{\Delta n}}$

Here you can observe the value of $K_c$ is greater than $K_p$ . To be equal as $K_p$ , $K_c$ needs to be divided by $(RT)^{\Delta n}$

<h3>What is Δngas?</h3>

The number of moles of gas that move from the reactant side to the product side is denoted by the symbol ∆n or delta n in this equation.

Once more, n represents the growth in the number of gaseous molecules the equilibrium equation can represent. When there are exactly the same number of gaseous molecules in the system, n = 0, Kp = Kc, and both equilibrium constants are dimensionless.

<h3>Definition of equilibrium</h3>

When a chemical reaction does not completely transform all reactants into products, equilibrium occurs. Many chemical processes eventually reach a state of balance or dynamic equilibrium where both reactants and products are present.

Learn more about equilibrium

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6 0

1 year ago

How much aluminum oxide are produced when 46.5g of Al react with 165.37g of MnO?

solong [7]

Aluminum oxide produced : = 79.152 g

<h3>Further explanation</h3>

Given

46.5g of Al

165.37g of MnO

Required

Aluminum oxide produced

Solution

Reaction

2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)

mol Al(Ar = 27 g/mol) :

mol = mass : Ar

mol = 46.5 : 27

mol = 1.722

mol MnO(Ar=71 g/mol) :

mol = 165.37 : 71

mol = 2.329

mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776

MnO as a limiting reactant(smaller ratio)

So mol Al₂O₃ based on MnO as a limiting reactant

From equation , mol Al₂O₃ :

= 1/3 x mol MnO

= 1/3 x 2.329

= 0.776

Mass Al₂O₃ (MW=102 g/mol) :

= 0.776 x 102

= 79.152 g

7 0

3 years ago

If calcium carbonate (CaCO3) decomposes, what would the product of the reaction be?

NemiM [27]

It was C) CaO on edg

8 0

2 years ago

Read 2 more answers

4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60

Bingel [31]

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

t = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Half-life of Co-60 = 5.3 years

Input the value :

$\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}$

8 0

2 years ago