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pshichka [43]
3 years ago
6

Identify the intermediate leads to the major product for the reaction of 2-methyl-2-butene with hydrogen bromide in the presence

of peroxide.
Chemistry
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:

II

Explanation:

We must have a good idea of the fact that there are two mechanisms that come into play when we are discussing about the addition of hydrogen halides to alkenes. The first is the ionic mechanism and the second is the radical mechanism.

The ionic mechanism is accounted for by the Markovnikov rule while the radical mechanism occurs in the presence of peroxides and is generally referred to as anti Markovnikov  addition.

The intermediate in anti Markovnikov  addition involves the most stable radical, in this case, it is a tertiary radical as shown in the images attached. The most stable radical is II hence it leads to the major product shown in the other image.

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Any reaction that absorbs 150 kcal of energy can be classified as.
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The Answer should be Endothermic reaction

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Reactions in which energy is absorbed are called Endothermic reactions.

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6 Which element requires the least amount of
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Which element requires the least amount of

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<h3>Answer-</h3><h3>Na</h3>
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Is C3HNO an organic molecule
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Yes.

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I
ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:  

Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\

Equation:  

3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\

Calculating the amount of MnCO_3

= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\=  13.1 \ kg

For point b:

Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\  of\  Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\

=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg

Equation:  

3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\

Calculating the amount of MnO_2:  

= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\=  6516 \ g \\\\=6.52 \ kg\\\\

7 0
3 years ago
Calculate the work (kJ) done during a reaction in which the internal volume expands from 20 L to 43 L against an outside pressur
Alenkinab [10]

Answer:

\large \boxed{\text{-10.0 kJ}}

Explanation:

1. Calculate the work

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2. Convert litre-atmospheres to joules

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The negative sign indicates that the work was done against the surroundings.

4 0
3 years ago
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