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crimeas [40]
3 years ago
15

Select the reason why two molecules of benzaldehyde cannot react via the aldol condensation.

Chemistry
1 answer:
Lesechka [4]3 years ago
4 0
Benzaldehyde or C6H5CHO would not undergo the aldol condensation because it does not contain an alpha-hydrogen in its structure. Aldol condensation is a type of reaction that happens between an enolate and an aldehyde or ketone leading to a alkene that has a planar structure. The lack of an alpha-hydrogen would not allow for it to undergo such process since it cannot enolize. Benzaldehyde undergoes a nucleophilic reaction known as Claisen-Schmidt condensation. It has somehow same mechanism of the aldol reaction however, the nucleophilic attack on the carbonyl happens even without the alpha-hydrogen but with an enolate that is from a ketone.
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Answer:

This problem is providing a chemical equation between two hypothetical elements, X and Y and asks for the molesof X that are needed to 

produce 21.00 moles of D in excess Y. After the following work, the answer turns out to be 15.75 mol X:Mole ratios:In chemistry, one the most crucial branches is stoichiometry, which allows us to perform calculations with grams, moles and particles (atoms, molecules and ions). It is based on the concept of mole ratios, whereby the moles of a specific substance can be converted to moles of another one, say product to reactant, reactant to reactant, reactant to product and product to product.

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Explanation:

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3 years ago
How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide
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Answer:

\boxed{\text{2.00 mol}}

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r:      58.32

          Mg(OH)₂ + … ⟶ … + 2HOH

m/g:       58.3

(a) Moles of Mg(OH)₂

\text{Moles of Mg(OH)$_{2}$} =\text{58.3 g Mg(OH)$_{2}$} \times \dfrac{\text{1 mol Mg(OH)$_{2}$}}{\text{58.32 g Mg(OH)$_{2}$}}\\\\=\text{0.9997 mol Mg(OH)$_{2}$}

(b) Moles of H₂O

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\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}

The reaction will form \boxed{\textbf{2.00 mol}} of water.

6 0
3 years ago
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