Commercially prepared cloning vectors such as pUC18 are designed to contain several useful features. An example of one of these features is ________.

Answers

Answer 1

Vectors may be plasmids. An example of one of several useful features of commercially prepared cloning vectors is MULTIPLE CLONING SITES.

The pUC18 vector is a widely used standardized cloning vector for replication in Escherichia coli.

A multiple cloning site can be defined as a short DNA fragment observed in genetically engineered plasmids.

These DNA fragments (multiple cloning sites) contain twenty (20) or more sites where restriction enzymes can cut in order to generate recombinant DNA molecules.

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Related Questions

A chemical reaction involves the reactants A and B, and the product C.
A+B→C
Trial 1 2 3
[A](mol/L) 0.10 0.10 0.20
[B](mol/L) 0.10 0.20 0.10
−Δ[A]Δt(molL s) 3.08×10−9 2.464×10−8 1.232×10−8

Answers

Explanation:

a+b =ab

answer

ab added all

is give answer

How many grams of sodium acetate NaCH3COO, , must be dissolved to prepare 200. mL of a 0.216 M aqueous solution of the salt? g

Answers

Answer:

Explanation:

You have to find the mass in the mole before the molarity and then in the mole.

A mixture is made of 40 ml of salt water to 200 ml of solution. What percent of the solution is salt water?

Answers

Answer:

16.7%

Explanation:

40 ml of salt water + 200 ml of solution = 240 ml

40/240 = 4/24 = 1/6=16.7%

The anion O3− does not obey the octet rule. Draw its Lewis structure and state the type of octet-rule exception. Indicate the values of nonzero formal charges and include lone pair electrons.

Answers

One of the oxygen atoms in the anion O3− is hypervalent and the formal charge on this oxygen atom is -1.

Ozone is a triatomic molecule. The anion formed from ozone is called the ozonide anion. This anion is also triatomic. The resonance structures of the ozonide anion are shown in the image attached to this answer.

We can see that one of the oxygen atoms in the ozonide ion is hypervalent because it contains ten instead of eight electrons. This hypervalent oxygen atom has a formal charge of -1 while the two other oxygen atoms has a formal charge of zero.

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Using the Kf value of 1.2×109 calculate the concentration of Ni2+(aq) and Ni(NH3)62+ that are present at equilibrium after dissolving 1.47 gNiCl2 in 100.0 mL of NH3(aq) solution such that the equilibrium concentration of NH3 is equal to 0.20 M.
Express your answers in moles per liter to two significant figures separated by a comma.

Answers

This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]

Afterwards, we set up an equilibrium expression for this chemical reaction:

[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]

Which can be written in terms of the reaction extent, [tex]x[/tex]:

[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]

Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:

[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]

[tex][Ni^{2+}]=0M[/tex]

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what is the expected hybridization of the central atom tetrahedral

Answers

Answer:

Tetrahedral molecules are normally spy hybridized.

Explanation:

So the neclues it not the lower horse

6) 0.5 moles of gas is kept at 2.0 L of volume and 0.75 atm of pressure. What is the temperature of the gas in K?

Answers

Answer:

310K

Explanation:

Rearrange PV=nRT to get T=PV/nR

T=(2.0L)(0.75atm)/(0.5mol)(0.08206)

=36.5 or 37

add 273 for K to get

310

pls help 8th grade k12

Answers

Look it up online you will find it

PLEASE HELPPPPPPPPPPPPPPPP

Answers

In this problem, you’re given a mass (m) and a force (F) applied to an object and you’re asked to calculate the acceleration (a). If we plug this into the provided equation:

F = ma
(7.3 N) = (3.2 kg)a
a = 2.3 m/s/s

The acceleration of the object is 2.3 m/s/s.

Hope this helps!

Is brass a solution:

Answers

Answer:

Brass is an alloy, and either a "solid solution".

Alloys in general may be solid solutions or they simply be mixtures

Explanation:

Hope it Helps you!!

) A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at 25 °C, and the density of water at 25 °C is 0.9970 g>mL)


At the left, grains of the mineral calcite float on the surface of the liquid bromoform (d = 2.890 g/mL) At the right, the grains sink to the bottom of liquid Chloroform (d = 1.444 g/mL). By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined

Answers

Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]

Thus, the density of the calcite sample will be:

[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

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Regards!

. Nhiệt độ ban đầu của 344 g một mẫu sắt là 18,2oC. Nếu mẫu sắt này hấp thụ 2,25 kJ nhiệt lượng thì nhiệt độ cuối của mẫu sắt này là bao nhiêu? (CFe = 0,449 J/g.K).

Answers

Answer:

i can't understand the question

Compound A is neutral and Compound B is acidic. Both are water-insoluble solids. A and B are dissolved in dichloromethane (DCM) and extracted with aqueous base. The layers are then separated. What must be done to obtain the compound in the aqueous layer

Answers

In order to extract the compound in the aqueous layer, a strong acid must be added to the system.

Liquid - Liquid extraction is a common method for obtaining substances that can partition between two layers. In this case, compound A is neutral and compound B is acidic.

When the both compounds are dissolved in dichloromethane and extracted using an aqueous base, the acid substance will form a salt in the aqueous layer. In order to extract the compound in the aqueous layer, a strong acid must be added to the system.

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Escriba un poema usando 4 de las siguientes palabras: lluvia, nieve, lluvia, granizo, tornado, huracanes, frente cálido, frente frío, alta presión, baja presión, termómetro, barómetro, anemómetro, pluviómetro, veleta o higrómetro.
plis ayúdenme porfa

Answers

Answer:

i dont speak mexican

Explanation:

2. Alex pulls on the handle of a claw hammer with a force of 15 N. If
the hammer has a mechanical advantage of 5.2, how much force
is exerted on the nail in the claw?

Answers

Answer:

78n

Explanation:

The output force exerted on the nail in the claw is equal to 78 N which has a mechanical advantage of 5.2.

What is the mechanical advantage?

The mechanical advantage can be demonstrated as the ratio of the output force to the Input force. The mechanical advantage of any machine can be expressed in the form of the ratio of the forces utilized to do the work.

The ratio of the resistance to the effort is said to be the actual mechanical advantage which will be less. The efficiency of a machine can be evaluated by equating the ratio of the output to its input.

Given, the input force = 15 N

The mechanical advantage of the hammer = 5.2

Mechanical advantage = Output force/ Input force

5.2 = Output/15

Output force = 15 ×5.2 = 78 N

Therefore, the force is exerted on the nail in the claw is equal to 78 N.

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#SPJ2

The box shows the chemical formula for bleach.

Answers

Bleach contains sodium chlorine and oxygen

HELP ME OUT PLEASE!!!!!!!!!

Compare and contrast model A with model C. How are they alike: How are they different?

A) They are composed of different elements in the same ratio.

B) They are composed of the same elements in different ratios.

C) They are composed of different elements in different ratios

D) The are composed of the same elements but the bond types differ​

Answers

Answer:

Model C has more double the oxygen than A. I think it is B

Explanation:

B I had got this on my chemistry test before and it was right

GIVING BRAINLY AND 20 POINTS

A sound wave in air has the wavelength of 1.36 m. Calculate its frequency? Assume the speed is 340 m/s.

Answers

Answer: the answer is 0.074Hz

Explanation:

Given, (In air)

Velocity V=340m/s

Frequency f=20,000Hz

Wavelegth λ=?

V=f.λ

λ=

F

V

=

20,000

340

=0.017Hz

Also, Given (in Water)

Velocity, V=1480m/s

Frequencyf=20,000Hz

wavelength, λ=?

V=F.λ

λ=

F

V

=

20,000

1480

=​0.074Hz

Answer:

frequency

Explanation:

frequency is velocity/ wavelength

340/1.36

250

A student performs a titration of 51.0 mL of a phosphoric
acid (H PO) solution of unknown concentration with a
standardized 1.25 M NaOH solution. The titration requires
26.2 mL of base to reach the third equivalence point. What is
the concentration of the H3PO4
solution?

Answers

From the information available in the question, the concentration of the acid is 0.21 M.

H3PO4(aq) + 3NaOH(aq) -----> Na3PO4(aq) + 3H2O(l)

Volume of acid(VA) = 51.0 mL

Concentration of acid (CA) = ?

Volume of base (VB) = 26.2 mL

Concentration of base (CB) = 1.25 M

Number of moles of acid (NA) = 1

Number of moles of base (NB) = 3

Using the formula;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA =  CBVBNA/VANB

CA =  1.25 M × 26.2 mL × 1/51.0 mL × 3

CA = 0.21 M

The concentration of the acid is 0.21 M.

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________ are neutral particles found in the nucleus of an atom.

Answers

Neutrons is your answer my friend
The answer is Neutrons

a material that is not a mixture; has the same properties all the way through

Answers

Answer:

Explanation:

The material that is not a mixture; it has the same properties all the way through is called a substance. Thus the material that is not a mixture; it has the same properties all the way through is called a substance.

ALL THE BEST :)

Grace wanted to find out the best conditions for growing lettuce plants.
She took 4 trays and planted 8 lettuce plants in each.
The results of her investigation are shown below.

How many days did the investigation last?
Use the table to help you.

Answers

Explanation:

the investigation lasts for 7 days !

hope this helps you.

A mixture of 3.25 moles of oxygen gas and 2.75 moles of nitrogen gas exert a total pressure of 22.4 atm. What is the partial pressure of oxygen

Answers

The partial pressure of oxygen is equal to 10.2 atm.

Given the following data:

Number of moles of oxygen = 3.25 moles.Number of moles of nitrogen = 2.75 moles.Total pressure = 22.4 atm.

To determine the partial pressure of oxygen:

First of all, we would find the total number of moles of the elements:

[tex]n=3.25+2.75[/tex]

n = 6 moles

Next, we would determine the mole fraction of the oxygen by using this formula:

[tex]Molefraction \;of \;a \;substance =\frac{No.\; of \; moles \;of \;substance}{Total \;no. \;of \; moles \;of \;substances}[/tex]

Substituting the values, we have:

[tex]Molefraction \;of \;a \;substance =\frac{2.75}{6} \\\\Molefraction \;of \;a \;substance =0.4583[/tex]

For oxygen:

[tex]Partial \;pressure = Molefraction \times Total\;pressure\\\\Partial \;pressure = 0.4583 \times 22.4[/tex]

Partial pressure of oxygen = 10.2 atm.

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Answer:

12.1

Explanation:

3.25 + 2.75= 6 moles total

then take: 3.25/6 to get 0.542

multiply 0.542 by 22.4 to get 12.1

the other answer solved for nitrogen, not oxygen.

A metal (FW 341.1 g/mol) crystallizes into a body-centered cubic unit cell and has a radius of 1.74 Angstrom. What is the density of this metal in g/cm3

Answers

This  problem provides the molar mass and radius of a metal that has an BCC unit cell and the density is required.

Firstly, we consider the formula that relates molar mass and also includes the Avogadro's number and the volume of the unit cell:

[tex]\rho =\frac{Z*M}{V*N_A}[/tex]

Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since BCC is able to hold 2 atoms and M and NA are given, we calculate the volume of the atom in the unit cell given the radius in meters:

[tex]V=a^3=(\frac{4R}{\sqrt{3} } )^3=(\frac{4*1.74x10^{-10}m}{\sqrt{3} } )^3=6.49x10^{-29}m^3[/tex]

 

And finally the required density in g/cm³:

[tex]\rho =\frac{2*341.1g/mol}{6.49x10^{-29}m^3\frac{m^3}{atom} *6.022x10^{23}\frac{atom}{mol} } =17455257.8g/m^3\\\\\rho=17.5g/cm^3[/tex]

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Determine the mass in grams of 5.15 × 10²¹ atoms of chromium. (The mass of one mole of chromium is 52.00 g.)

Answers

The mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Cr

But:

1 mole of Cr = 52 g

Thus, we can say that:

6.02×10²³ atoms = 52 g of Cr

With the above information, we can obtain the mass of 5.15 × 10²¹ atoms of chromium. This can be obtained as follow:

6.02×10²³ atoms = 52 g of Cr

Therefore,

5.15×10²¹ atoms = (5.15×10²¹ × 52) / 6.02×10²³

5.15×10²¹ atoms = 0.44 g of Cr

Thus, the mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

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in a certain reaction, Fe and O2 combine to form iron (iii) oxide. 14.7 moles of Fe and 13.0 moles of O2 are placed in a container and the reaction proceeds iwth 100% yield. which is the excess reactant

Answers

Answer:

Fe is the excess reactant

Reaction yields are the amount of the reactant and the products of a chemical reaction. In the reaction between iron and oxygen, iron is the excess agent.

What is excess reactant?

The reactant present in an extra quantity than the other reactant in a chemical reaction which reacts with the limiting reactant is called an excess reactant.

In a reaction mixture, the excess reactant is present even when the limiting agent is completely consumed.

The chemical reaction between iron and oxygen is shown as,

[tex]\rm 4 Fe(s) + 3 O_{2}(g) \rightarrow 2 Fe_{2}O_{3}[/tex]

From the reaction, it can be said that oxygen is a limiting reagent that limits the formation of iron (iii) oxide.

Therefore, iron is an excess reactant.

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The sum of the number of proteins and neutrons in an atoms nucleus is its __________ ___________.

Answers

Answer:

Mass Number

Explanation:

In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.

Please vote brainliest!

Which of the following best describes the scientific exploration of the atom?

Answers

where do we pick or where is it

Each element can be indentified by the number of _______ found in its nucleus, which also equals the elements _______ _______.

Answers

Defination or recotubation

If 5 g of sodium chloride saturates 12.5 g of water at 10 °C, what mass of sodium chloride would saturate 50 g of water at constant temperature? ​

Answers

Explanation:

since 5g saturates 12.5g of water at 10°c

so......x would saturate 50g of water at 0°c

then you can cross multiply

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