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AVprozaik [17]
3 years ago
14

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1o) = carbon bonded to just 1 other ca

rbon group Secondary carbon (2o) = carbon bonded to 2 other carbon groups Tertiary carbon (3o) = carbon bonded to 3 other carbon groups Quaternary carbon (4o) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?

Chemistry
1 answer:
Alla [95]3 years ago
3 0

The question is incomplete, complete question is :

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?

Structure is given in an image?

Answer:

There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

Explanation:

Total numbers of carbon = 10

Number of primary carbons that is carbon joined to just single carbon atom = 6

Number of secondary carbons that is carbon joined to two carbon atoms = 1

Number of tertiary carbons that is carbon joined to three carbon atoms = 2

Number of quartenary carbons that is carbon joined to four carbon atoms = 1

So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

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Which of the following frequencies corresponds to light with the longest wavelength? A) 3.00 times 10^13 s^-1 B) 4.12 times 10^5
Vsevolod [243]

Answer:

1) B. 4.12 times 10^5 s^-1

2) B. frequency-v

3) C. 1.18 times 10^15 s^-1

4) B. 4.39 times 10^-19 J

5) B. Energy is absorbed

6) C. 3

7) 1s^2  2s^2  2p^6  3s^2  3p^6  4s^2  3d^10  4p^6  5s^2  4d^10  5p^6     6s^2  

Explanation:

1)

The wavelength is inversely proportional to the frequency. Thus, the smallest frequency shall correspond to the longest wavelength. Thus, the correct answer is <u>B. 4.12 times 10^5 s^-1</u>

2)

<u>B. frequency-v</u> are wrongly paired because frequency is represented by f. Thus, correct pair will be frequency-f.

3)

The relationship between wavelength (λ) and frequency (f) is:

c = fλ

f = c/λ

where, c = speed of light = 3 x 10^8 m/s

f = (3x10^8 m/s) / (254 x 10^-9 m)

f = 1.18 x 10^15 s^-1

Thus, the correct option is <u>C. 1.18 times 10^15 s^-1.</u>

4)

The energy of photon is given as:

E = hc/λ

where, c = speed of light = 3 x 10^8 m/s

            h = Plank's Constant = 6.625 x 10^-34 J.s

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s) / (453 x 10^-9 m)

E = 4.39 x 10^-19 J

Thus, the correct option is <u>B. 4.39 times 10^-19 J</u>.

5)

Since, the higher energy levels away from nucleus have higher energies. So, in order to move an electron from lower to higher or distant energy level, it must absorb energy from external source. So, the correct option is <u>B. Energy is absorbed</u>.

6)

The P-Sub-level has three orbitals, each having two electrons. Thus, P-Sub-Level accommodates total of 6 electrons. The correct option is <u>C. 3</u>

7)

The electronic configuration of barium atom is:

<u>1s^2  2s^2  2p^6  3s^2  3p^6  4s^2  3d^10  4p^6  5s^2  4d^10  5p^6  6s^2  </u>

8 0
3 years ago
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