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Musya8 [376]
3 years ago
9

Does a molecule want a geometry that has high energy of low energy

Chemistry
2 answers:
Alex3 years ago
4 0
High energy hope this helps
navik [9.2K]3 years ago
3 0
That depends on what you mean by "high energy molecules" If you are referring to molecules that release a lot of energy upon a chemical reaction, then butane with the formula C4H10 is a random example. Butane releases a good chunk of energy upon combustion, methane and other C-H molecules are similar, but those are just examples, there are TONS. 

<span>You could also be referring to molecules which give a lot of energy to the body during organic processes. These are sugars and carbohydrates. An example would be glucose, with the formula C6H12O6. </span>
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Explanation:

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3 years ago
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27 Which energy conversion must occur in an operating electrolytic cell?
Vitek1552 [10]
<u> electrical energy to chemical energy</u>
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3 years ago
N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol
docker41 [41]

Answer:

A

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:


\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)

Therefore, from the chemical equation, we have that:


\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} +  \Delta H^\circ_f \text{ H$_2$O}  \right]   -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}

In conclusion, our answer is A.

5 0
2 years ago
John Fang has 5 apples, Jonny ate 3, how many are left!
Doss [256]
Answer:5-3=2

As 5apple eaten 3apple so left 2apple
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3 years ago
32P can be used to make any nucleotide (A, C, G, or T) radioactive. Which of the following explains why this is true?
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B. The answer is: All nucleotides have a phosphorus atom that can be replaced with 32P.

Nucleotides contain a nitrogenous base, a five-carbon sugar, and, at least, one phosphate group. Exactly that phosphate group in the nucleotide has the phosphorus atom. Therefore, the phosphorus atom in the nucleotide can be replaced with radioactive phosphorus-32 (32P). 
3 0
2 years ago
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