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harkovskaia [24]
3 years ago
5

Butanone draw both enolates formed when the ketone is treated with base. include charges. draw the oxyanion species; do not draw

carbanion resonance forms.

Chemistry
1 answer:
hjlf3 years ago
8 0
This is an example of deprotonation of Asymmetric Ketones. In this case two products are possible.

Thermodynamic Product:
                                        It gives highly substituted enolate, has greater transition energy to cross and has lower product energy.

Kinetic Product:
                         It gives less substituted enolate, the proton is accessible and easily removed and has less transition energy, and higher product energy compared to thermodynamic product.

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Balance the following chemical equation:<br> Mg(OH)₂ ​+ HCI → MgCI₂ ​+ H₂ ​O
geniusboy [140]

Answer:

Mg(OH)₂ ​+  2HCI → MgCI₂ ​+ 2 H₂ ​O

Explanation:

6 0
2 years ago
Compound that contains a terminal carbonyl?
malfutka [58]

Many compunds have a terminal carbonyl

Aldehyde, Ketone, Carboxylic acid, Amide, Imide, Acid anhydride are the first that come to my mind.

7 0
3 years ago
Which type of precipitation occurs when the water vapor in a cloud freezes to form ice crystals?
Verizon [17]

Answer:snow i think

Explanation:

6 0
3 years ago
Read 2 more answers
Hydrazine, N2H4, may react with oxygen to form nitrogen gas and water.
Sindrei [870]

Answer:

The percent yield of the reaction is 35 %

Explanation:

In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.

Let's verify the moles that were used in the reaction.

2.05 g . 1mol/ 32 g = 0.0640 mol

In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.

Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).

1atm . 0.550L = n . 0.082 . 295K

(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles

Percent yield of reaction = (Real yield / Theoretical yield) . 100

(0.0225 / 0.0640) . 100 = 35%

3 0
2 years ago
Read 2 more answers
What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7
mariarad [96]

 The  empirical  formula    of hydrocarbon is  CH2

The  molecular formula  of the  hydrocarbon is  C6H12


    <u><em>Explanation</em></u>

Hydrocarbon  is  made up  of carbon and hydrogen


<h3><u><em> </em></u>Empirical formula  calculation</h3>

 Step 1:  find  the  moles   CO2  and  H2O

moles =mass/molar mass

moles   of CO2 =  6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol =  0.15  moles

Step 2: Find the moles  ratio  of Co2:H2O  by diving  each mole by smallest mole(0.15)

that  is  for  CO2 = 0.15/0.15  =1

              For H2O = 0.15/0.15 =1

therefore  the mole ratio  of Co2 : H2O = 1:1  which  implies that 1 mole of Co2  and 1  mole of H2O is  formed  during combustion reaction.


From the  the law of mass conservation the number  of atoms in reactant side  must  be equal to  number of  atoms  in product side

therefore  since  there 1 atom  of C  in product side there  must be 1 atom of C  in reactant  side.

In addition  there is 2 H atom in product  side  which should be the  same  in reactant side.  

From information above the empirical formula is therefore = CH2


Molecular formula  calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n =  84 g/mol

n = 6

multiply the  each subscript  in CH2  by  6

 Therefore the molecular formula = C6H12




5 0
3 years ago
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