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Olin [163]
1 year ago
8

What is the relationship between nucleic acids and nucleotides?

Chemistry
1 answer:
maw [93]1 year ago
7 0

The nucleotides are monomers of nucleic acid.

The nucleic acids are the macro molecules. They are two types- DNA and RNA. The nucleic acid, DNA consists of nucleotide which are monomers of the DNA. The nucleotides have three components such as pentose sugar, nitrogen bases, and phosphate.

They are the backbone of DNA. DNA is the polymer of nucleotide, and this DNA is coming under nucleic acids. The nucleotides form DNA, and DNA forms the nucleic acid. They are interlinked with each other.

Nucleic acids are the center for storing all the information. The deoxyribose nucleic acid or DNA is the center for storing all the genetic information.

The genetic material is inherited from the parents to the children. The characters are transferred from one generation to another.

The characters such as hair color, hair type, eye color are some of the genetic information that is carried from one generation to another via nucleic acids.

Nucleotide is the basic unit of nucleic acid (DNA and RNA) which store the genetic code of the cell.

It consists of sugar base (ribose or deoxyribose), nitrogenous base (A, T, G, C and U), and phosphate group.

Nucleotides (specially ATP) provides the energy for most of the cellular processes.

Therefore, with the above statements we got to know that both nucleic acid and nucleotides are genetically modified.

To know more about nucleic acid, refer: brainly.com/question/29794338

#SPJ4

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arsenic

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The equilibrium constant in terms of concentration that is, K_c=3.6243\times 10^{3} .

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PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)

The relation of K_c\& K_p is given by:

K_p=K_c(RT)^{\Delta n_g}

K_p= Equilibrium constant in terms of partial pressure.=98.1

K_c= Equilibrium constant in terms of concentration  =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

\Delta n_g = Difference between gaseous moles on product side and reactant side=n_{g,p}-n_{g.r}=1-2=-1

98.1=K_c(RT)^{-1}

98.1 =\frac{K_c}{RT}

K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}

The equilibrium constant in terms of concentration that is, K_c=3.6243\times 10^{3} .

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what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?
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Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

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