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Alexus [3.1K]
3 years ago
12

Are the bases on the interior or the exterior of the double helix? Are they randomly arranged or neatly stacked?

Chemistry
2 answers:
alina1380 [7]3 years ago
5 0
Each helix corresponds to a nucleotide chain, and the two chains are joined throughout their length by adenine-thymine or guanine-cytosine pairs. These base pairs are STACKED one above the other with their planes perpendicular to the axis of the two spirals. This places the hydrophobic base pairs inside the structure and allows the hydrophilic sugar and phosphate groups to contact water on the exterior. The whole helix will just fit inside a cylinder 2000 pm in diameter.
Lemur [1.5K]3 years ago
3 0

the double helix is hydrogen bonded through the bases only so the bases are inside the helix only

as adenine combines with thymine and guanine with cytosine

phosphate are in the exterior of it

sugar groups constitute the double helix.

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The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c
crimeas [40]

Answer : The rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

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According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

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\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 785.0K = ?

Ea = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 785.0K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]

K_2=1.45\times 10^{-2}s^{-1}

Therefore, the rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

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3 years ago
2 Points<br> Which of the following is a thermal conductor?
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8 0
2 years ago
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Brackett Series (n = 4)

Explanation:

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If a bus traveled 20 miles north on a straight path, and then traveled 20 miles back to the starting point, what would the veloc
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Velocity is calculated by the formula:

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