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2 years ago

The structure of DNA resembles a twisted ladder. Determine the structural components that form the rungs of the ladder.

2 answers:

8 0

Answer: Nitrogenous bases guanine(G), adenine(A), cytosine(C) and thymine(T) forms the rungs of the ladder in twisted ladder structure of DNA.

Explanation:

Rungs of the ladder are normally horizontal small logs placed one after the other with small gaps between them on the two vertically standing logs of the ladder.

Since, DNA resembles the twisted ladder structure the rungs of the ladder structure of DNA are the organic nitrogenous basic molecules which are associated to their respective base with the help of hydrogen bonding.

In DNA, there are four bases guanine(G), adenine(A), cytosine(C) and thymine(T). These bases through hydrogen bonding are associated with there respective bases :adenine is bonded to thymine and guanine is bonded cytosine.

(A=T): Two hydrogen bonds between the bases

(C≡G): Three hydrogen bonds between the bases.

Mamont248 [21]2 years ago

6 0

The structural components that form the rungs of the ladder are the nitrogeneous bases of the DNA molecule. These bases are adenine, guanine, cytosine and thymine. Hope this answers the question. Have a nice day.

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Be sure to answer all parts. Nitric oxide, NO·, is a radical thought to cause ozone destruction by a mechanism similar to that o

SIZIF [17.4K]

Answer:

See image attached and explanation

Explanation:

The stratospheric ozone layer is very important in absorbing high-energy ultraviolet radiation that is harmful to living systems on earth. The concentration of ozone in the stratosphere is determined by both thermal and photochemical pathways for its decomposition. Nitric oxide, NO, is a trace constituent in the stratosphere that reacts with ozone to form nitrogen dioxide, NO2, and the diatomic oxygen molecule. The nitrogen-oxygen bond in NO2 is relatively weak. When an NO2 molecule encounters an oxygen atom, it transfers an oxygen, forming O2 and NO. The chemical reactions involved are formations of NO2 following by reaction of NO2 with atomic oxygen for form NO and O2. The sum of both reactions show that the overall reaction is simply the reaction of ozone with atomic oxygen to form two molecules of molecular oxygen. Hence, NO only serves as a catalyst, it does not undergo a permanent change itself.

6 0

3 years ago

A student observes two strips of magnesium, mg ribbon that are each 3cm long. One strip of magnesium

ankoles [38]

Therefore option c , i.e. The substances in both test tubes are reactive only at high temperatures. is the only statement which is NOT supported by the student's observations.

<h3>What is the reaction between Magnesium and Hydrogen ?</h3>

Magnesium reacts with hydrochloric acid to produce hydrogen gas

Mg (s) + 2 HCl (aq) → MgCl₂ (aq) + H₂ (g)

In this reaction, the magnesium and acid are gradually used up , which can be seen in the test tube 2 .

A chemical reaction is taking place in Test tube 2 ,

Hydrogen gas is released in test tube 2 ,

Energy is released in the reaction involving hydrochloric acid and we can see in test tube 2 the reaction is going on

therefore option C i.e. The substances in both test tubes are reactive only at high temperatures. is the only statement which is NOT supported by the student's observation.

To know more about the chemical reaction between Magnesium and Hydrogen and this test.

brainly.com/question/19062002

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6 0

2 years ago

Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be

AlexFokin [52]

Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

$\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles$

$\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles$

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

$\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles$

The concentration of benzoic acid is:

$C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M$

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺

0.14 - x x x

$Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}$

$Ka = \frac{x*x}{0.14 - x}$

$6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0$ (2)

By solving equation (2) for x we have:

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

$pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52$

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻ (3)

The number of moles of C₆H₅CO₂⁻ is:

$\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles$

The volume of NaOH added is:

$V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L$

The concentration of C₆H₅CO₂⁻ is:

$C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M$

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻

0.14 - x x x

$Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}$

$(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0$

$(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0$

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

$pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33$

$pH = 14 - pOH = 14 - 5.33 = 8.67$

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

$\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles$

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:

$\eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles$

The concentration of NaOH is:

$C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M$

Therefore, the pH is given by this excess of NaOH:

$pOH = -log([OH^{-}]) = -log(0.067) = 1.17$

$pH = 14 - pOH = 12.83$

I hope it helps you!

4 0

3 years ago

What can be formed by atoms of same element (H) or atoms of different 20 points<br> elements (CO)? *

NARA [144]

Answer:

Molecules

Explanation:

If you had more than one atom chemically bonded together, then regardless of the types of atoms that are bonded, you're going to have a molecule regardless.

4 0

3 years ago

What is the limiting reactant when 8.4 moles of lithium react with 4.6 moles of oxygen gas?

serious [3.7K]

Answer: Lithium

Explanation: The balanced chemical equation is:

$4Li+O_2\rightarrow 2Li_2O$

It can be seen, 4 moles of lithium combines with 1 mole of oxygen gas to produce 2 moles of lithium oxide.

Thus 8.4 moles of lithium combines with= $\frac{1}{4}\times {8.4}=2.1moles$ of oxygen gas to produce 4.2 moles of lithium oxide.

As, Lithium limits the formation of product, it is the limiting reagent and Oxygen gas is present in excess, it is called the excess reagent. (4.6-2.1)=2.5 moles of oxygen gas are present in excess.

5 0

3 years ago

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