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Lelu [443]
3 years ago
5

What is the enthalpy for reaction 1 reversed? reaction 1 reversed: co2→co + 12o2?

Chemistry
1 answer:
Serhud [2]3 years ago
7 0
<span>Answer: CO2→CO + 12O2 is not a balanced reaction, I think you meant to post " CO2→CO + 1/2O2" The simple answer is that the enthalphy for reaction 1 reversed is the same enthalpy as for the reaction as written but with the opposite sign. If A + B = AB with 30 kJ released then AB = A + B with 30 kJ absorbed. We would write it as A + B = AB heat of reaction = -30 kJ AB = A + B heat of reaction = +30 kJ</span>
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The backbone of nucleic acids consists of
V125BC [204]

Answer:

c. a phosphodiester bond between the 3' and 5' hydroxyl groups of neighboring sugars

Explanation:

Phosphodiester bond is the bond which is formed between the hydroxyl group of one nucleotide to the phosphate group of the another nucleotide. These are ester bonds. These bonds are central to all the life which is in existence on Earth. These bonds forms the backbone of the strands of the nucleic acid.

The bond is formed by the linkage of 3' carbon atom of one of the sugar unit to the 5' carbon atom of the another succeeding sugar unit.

<u>Hence, the answer is:- c. a phosphodiester bond between the 3' and 5' hydroxyl groups of neighboring sugars</u>

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Answer:

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3 years ago
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hoa [83]
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vazorg [7]

<u>Answer:</u> The correct statement is low temperature only, because entropy decreases during freezing.

<u>Explanation:</u>

The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

\Delta G=\Delta H-T\Delta S

Where,

\Delta G = change in Gibb's free energy

\Delta H = change in enthalpy

T = temperature

\Delta S = change in entropy

It is given that freezing of methane is taking place, which means that entropy is decreasing and Delta S is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the \Delta H is also negative.

For a reaction to be spontaneous, \Delta G must be negative.

-ve=-ve-[T(-ve)]\\\\-ve=-ve+T

From above equations, it is visible that \Delta G will be negative only when the temperature will be low.

Hence, the correct statement is low temperature only, because entropy decreases during freezing.

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