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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°.

You might be interested in
Phosphorus-32 is radioactive and has a half life of 14.3 days. Calculate the activity of a 3.5mg sample of phosphorus-32. Give y
Andreyy89

Answer:

The activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.

Explanation:

The activity of P-32 can be calculated with the following equation:

A = \lambda N   (1)

Where:

N: is the number of atoms of P-32

λ: is the decay constant

We can find the number of atoms of P-32 as follows:

N = \frac{N_{A}*m}{M}  (2)

Where:

N_{A}: is the Avogadro's number = 6.022x10²³ atoms/mol

m: is the mass of P-32 = 3.5x10⁻³ g

M: is the molar mass of the radionuclide (P-32) = 32 g/mol    

Now, the decay constant is given by:

\lambda = \frac{ln(2)}{t_{1/2}}   (3)

Where:

{t_{1/2}}: is the half-life of P-32 = 14.3 days

Finally, we can find the activity of P-32 by entering equations (2) and (3) into (1):

A = \lambda N = \frac{ln(2)}{t_{1/2}}*\frac{N_{A}*m}{M} = \frac{ln(2)}{14.3 d*\frac{24 h}{1 d}*\frac{3600 s}{1 h}}*\frac{6.022 \cdot 10^{23} mol^{-1}*3.5 \cdot 10^{-3} g}{32 g/mol} = 3.7 \cdot 10^{13} dis/s      

Since a becquerel (Bq) is defined as a disintegration (dis) per second, the activity in Bq is:

A = 3.7 \cdot 10^{13} Bq

And, since a Curie (Ci) is 3.7x10¹⁰ Bq, the activity in Ci is:

A = 3.7 \cdot 10^{13} Bq*\frac{1 Ci}{3.7 \cdot 10^{10} Bq} = 1.0 \cdot 10^{3} Ci

Therefore, the activity of P-32 is 3.7x10¹³ becquerels = 1.0x10³ curies.  

               

I hope it helps you!

8 0
3 years ago
Please answer this for me shehsh
Levart [38]

Answer:

ih2706

Explanation:

7 0
3 years ago
Please help with this chemistry question!
marusya05 [52]

Hey there!

Compounds with ionic bonds have higher melting points because of the forces needed to break through the strong forces of attraction holding it together.

Compounds with covalent bonds have lower melting points because less energy is needed to break the weaker forces of attraction.  

So, your answer is C. Compound 1 is ionic, and compound 2 is molecular.

Hope this helps!

4 0
4 years ago
A gas has a volume of 25 ml, at a pressure of 1 atmosphere (1 atm). Increase the volume to 125 ml and the temperature remains co
vekshin1

<h2>0.2 atm </h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2 \\

Since we're finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

We have

P_2  =  \frac{25 \times 1}{125}  =  \frac{25}{125}   = \frac{1}{5}  = 0.2 \\

We have the final answer as

<h3>0.2 atm</h3>

Hope this helps you

8 0
3 years ago
Calculate the number of C atoms in 0.183 mol of C.
Alex Ar [27]

Answer: 1.10x10²³ atoms of C

110202600000000000000000 atoms C

Explanation:The solution process is shown below.

0.183 mole C x 6.022x10²³ atoms C / 1 mole C

= 1.10x10²³ atoms C

or 110202600000000000000000 atoms C

4 0
3 years ago
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