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Dimas [21]
3 years ago
12

The double bonds in an aromatic ring are not fixed in position as in an alkene. The electrons are considered to be ____ and are

delocalized around the ring.
Chemistry
2 answers:
Dima020 [189]3 years ago
7 0

Answer: The double bonds in an aromatic ring are not fixed in position as in an alkene. The electrons are considered to be in conjugation and are delocalized around the ring.

Explanation:

Chemical compounds that have conjugated planar ring systems where there are delocalized pi-electron clouds are present in place of individual alternate and single bonds.

This means that electrons present in a double bond of alkene are in conjugation.

Therefore, we can conclude that the double bonds in an aromatic ring are not fixed in position as in an alkene. The electrons are considered to be in conjugation and are delocalized around the ring.

anastassius [24]3 years ago
7 0

Answer:

In resonance

Explanation:

Got it right on Acellus.

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If 42.7 of 0.208 M hydrochloric acid are needed to completely neutralize a solution of calcium hydroxide, how many grams of calc
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0.329 g

Explanation:

In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.

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Ca(OH)_2(aq)+2 HCl (aq)\rightarrow CaCl_2 (aq)+2 H_2O (l)

Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:

c_{HCl}=\frac{n_{HCl}}{V_{HCl}}

Rearranging for moles of HCl, n:

n_{HCl}=c_{HCl}V_{HCl}

Based on stoichiometry of the balanced chemical equation, notice that 1 mole of calcium hydroxide reacts with 2 moles of HCl, meaning:

n_{Ca(OH)_2}=\frac{1}{2} n_{HCl}=\frac{1}{2}c_{HCl}V_{HCl}

Now that we have the expression for moles, we may also express moles of calcium hydroxide as the ratio between its mass and molar mass:

n_{Ca(OH)_2}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}

Using the last two equations, we see that:

\frac{1}{2}c_{HCl}V_{HCl}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}\\\therefore m_{Ca(OH)_2}=\frac{1}{2}c_{HCl}V_{HCl}M_{Ca(OH)_2}

Substitute the given data, as well as the molar mass of calcium hydroxide:

m_{Ca(OH)_2}=\frac{1}{2}\cdot0.208 M\cdot0.0427 L\cdot74.093 g/mol=0.329 g

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