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natta225 [31]
3 years ago
14

Show the mechanism for the following reaction: cyclohexene bromine yields a dibromocyclohexane Unlike previous problem, this pro

blem does require illustration of stereochemistry. Draw wedge-and-dash bond stereochemical structures – including H atoms on ALL chirality center (product and intermeidate both)– and include charges, electrons, and curved arrows. Details count. Draw one enantiomer only for any racemates. NOTE: A bromonium ion bromine should have TWO lone pairs, not three.

Chemistry
1 answer:
tester [92]3 years ago
4 0

Answer:

Check the explanation

Explanation:

Kindly check the attached image below for the step by step explanation to the question above.

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larisa86 [58]
Your finger will get burned
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3 years ago
Convert 1.50μm2 to square meters
IrinaVladis [17]
1 micro meter = 10^{-6}  meters
1 μm ^2 = 1 μm*1μm = 10^{-6} *10^{-6} =  10^{-12} meters

1.5 μm^2 = 1.5 * 10^{-12} meters

μm^2 is a unit for surface. First you want to convert μm to meters which is unit for length and if you multiply units for length you get unit for surface.
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Why is the enthalpy of formation of oxygen zero?
madreJ [45]
All elements in their standard states (oxygen<span> gas, solid carbon in the form of graphite, etc.) have a standard </span>enthalpy of formation<span> of </span>zero<span>, as there is no change involved in their </span>formation<span>.</span>
4 0
3 years ago
At a constant temperature, 600 mL of an ideal gas is at a pressure of 760 mmHg . If the volume is decreased to 300 mL, the press
belka [17]

Answer:

1520mmHg

Explanation:

Data obtained from the question include:

V1 (initial volume) = 600 mL

P1 (initial pressure) = 760 mmHg

V2 (final volume) = 300 mL

P2 (final pressure) =.?

Using the Boyle's law equation P1V1 = P2V2, the final pressure of the gas can easily be obtained as shown below:

P1V1 = P2V2

760 x 600 = P2 x 300

Divide both side by 300

P2 = (760 x 600) /300

P2 = 1520mmHg

The final pressure of the gas is 1520mmHg

7 0
3 years ago
On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to t
Ne4ueva [31]

The given question is incomplete. The complete question is as follows.

On the basis of periodic trends, choose the larger atom from each pair (if possible): match the elements in the left column to the appropriate blanks in the sentences on the right. make certain each sentence is complete before submitting your answer.

          Sn, F, Ge, not predictable, Cr

Of Ge or Po, the larger atom is .........

Of F or Se, the larger atom is ..........

Of Sn or I, the larger atom is .........

Of Cr or W, the larger atom is ........

Explanation:

When we move down a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.

Ge is a group 14 element which lies in period 4 and Po is a group 16 element that lies in period 6. As polonium is larger in size as compared to germanium.

Fluorine is a group 17 element and lies in period 2. Selenium is a group 16 element lies in 4. Therefore, selenium is larger in size as compared to fluorine.

Sn is a group 14 element that lies in period 5 and I is a group 17 element that lies in period 5. Hence, I is a larger atom.

Cr is a d-block element which lies in period 4 and W is also a d-block element which lies in period 6. Hence, W is larger in size than Cr.

Thus, we can conclude that given blanks are matched as follows.

  • Of Ge or Po, the larger atom is Po.
  • Of F or Se, the larger atom is Se.
  • Of Sn or I, the larger atom is I.
  • Of Cr or W, the larger atom is W.
7 0
3 years ago
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