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Elena L [17]
3 years ago
7

What compound is composed of deoxyribose sugars and nitrogenous bases

Chemistry
1 answer:
klasskru [66]3 years ago
7 0

Answer:

Deoxyribonucleic acid (DNA) is a long, double-stranded, helical molecule composed of building blocks called deoxyribonucleotides. A deoxyribonucleotide is composed of 3 parts: a molecule of the 5-carbon sugar deoxyribose, a nitrogenous base, and a phosphate group.

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Which of the following factors can be changed for a gas without changing the mass of the gas? Select all that apply.
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The different molecules that make up the air in a room have on average the same kinetic energy. How does the speed of the differ
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The formula for kinetic energy is KE=1/2(mv²).  Since both mass and velocity are multiplied by each other, particle with a larger mass needs to be moving slower than a particle with less mass if both have the same kinetic energy. You can think of it as 2KE/m=v² or 2KE/v²=m, If you increase the mass the velocity needs to decrease to keep the same KE value.

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3 years ago
<
victus00 [196]

Answer:

This isotope has 59 electrons giving it a charge of -2.

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6 0
2 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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