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faltersainse [42]
3 years ago
5

When 2-phenylacetaldehyde is added dropwise to benzaldehyde in the presence of sodium hydroxide they undergo a crossed aldol rea

ction to form 3-hydroxy-2,3-diphenylpropanal?

Chemistry
1 answer:
egoroff_w [7]3 years ago
7 0
Carbonyl Compounds containing alpha proton readily donates alpha acidic proton when treated with strong Base. In given statement <span>2-phenylacetaldehyde is converted into Enolate and Enolates acts as a nucleophile and attacks carbonyl group of Benzaldehyde. The mechanism of given reaction is as follow, the movement of electrons is shown by RED arrows,</span>

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Answer:

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Explanation:

5 0
3 years ago
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Find the concentraion of [OH-] in a .003 M solution of HBr
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ahrayia [7]

Answer:

D

Explanation:

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3 years ago
When the reaction h3po4(aq) ca(oh)2(aq)→ occurs what is the formula for the salt formed?
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7 0
3 years ago
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Calculate the solubility product constant, Ksp, of lead(II) chloride, PbCl2, which has a
V125BC [204]

Answer:

0.0159m

Explanation:

9 M

Explanation:

Lead(II) chloride,  

PbCl

2

, is an insoluble ionic compound, which means that it does not dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution.

Instead of dissociating completely, an equilibrium rection governed by the solubility product constant,  

K

sp

, will be established between the solid lead(II) chloride and the dissolved ions.

PbCl

2(s]

⇌

Pb

2

+

(aq]

+

2

Cl

−

(aq]

Now, the molar solubility of the compound,  

s

, represents the number of moles of lead(II) chloride that will dissolve in aqueous solution at a particular temperature.

Notice that every mole of lead(II) chloride will produce  

1

mole of lead(II) cations and  

2

moles of chloride anions. Use an ICE table to find the molar solubility of the solid

 

PbCl

2(s]

 

⇌

 

Pb

2

+

(aq]

 

+

 

2

Cl

−

(aq]

I

 

 

 

−

 

 

 

 

 

0

 

 

 

 

 

0

C

 

 

x

−

 

 

 

 

(+s)

 

 

 

 

(

+

2

s

)

E

 

 

x

−

 

 

 

 

 

s

 

 

 

 

 

2

s

By definition, the solubility product constant will be equal to

K

sp

=

[

Pb

2

+

]

⋅

[

Cl

−

]

2

K

sp

=

s

⋅

(

2

s

)

2

=

s

3

This means that the molar solubility of lead(II) chloride will be

4

s

3

=

1.6

⋅

10

−

5

⇒

s

=  √ 1.6 4 ⋅ 10 − 5  = 0.0159 M

8 0
3 years ago
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