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Bezzdna [24]
3 years ago
13

In the following pair, determine whether the two represent resonance contributors of a single species or depict different substa

nces. If two structures are not resonance contributors, explain why. Select the single best answer.
N-N ≡ N: and N=N=N:
Chemistry
1 answer:
Aleksandr-060686 [28]3 years ago
5 0

Answer:

They are resonance contributors

Explanation:

Resonance structures are structures that differ only in the distribution or placement of electrons.

Considering the two structures, we can easily see that the two species have the same total number of bonds and electrons differing only in the distribution of these electrons.

Hence, they are resonance contributors.

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Which colorless and odorless gas, produced by radioactive decay of Uranium-238, is considered to be a cancer-causing agent?
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After a covalent bond has stabilized an atom, the atom will have
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Please review the attachment
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Answer: The correct answer is -297 kJ.

Explanation:

To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:

2SO3 —> O2 + 2SO2 (196 kJ)

Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.

Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:

SO3 —>1/2O2 + SO2 (98 kJ)

S + 3/2O2 —> SO3 (-395 kJ)

Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.

S + O2 —> SO2

Now, we must add the enthalpies together to get our final answer.

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3 years ago
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What is the product if an atom of Po-209<br> undergoes alpha decay?
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<h3>Answer:</h3>

Lead-205 (Pb-205)

<h3>Explanation:</h3>

<u>We are given;</u>

  • An atom of Po-209

We are supposed to identify its product after an alpha decay;

  • Polonium-209 has a mass number of 209 and an atomic number of 84.
  • When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
  • Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
  • The product from this decay is Pb-205, because Pb-205 has a mass number of 205 and atomic number 82.
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  • Note; An alpha particle is represented by a helium nucleus, ⁴₂He.

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