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arlik [135]
2 years ago
8

In the following structure, carbons (I),(2),(3) and (4) are classified respectively as

Chemistry
1 answer:
love history [14]2 years ago
7 0

<u>Answer:</u>

Carbon (i) : quaternary carbon

Carbon (ii) : secondary carbon

Carbon (iii) : tertiary carbon

Carbon (iv) : secondary carbon

<u>Explanation:</u>

Carbons can be classified into 4 categories:

(1) Primary carbon (1^o): These are the atoms where the carbon atom is attached to one other carbon atom.

(2) Secondary carbon (2^o): These are the atoms where the carbon atom is attached to two other carbon atoms.

(3) Tertiary carbon (3^o): These are the atoms where the carbon atom is attached to three other carbon atoms.

(4) Quaternary carbon (4^o): These are the atoms where the carbon atom is attached to four other carbon atoms.

In the given structure:

Carbon (i) is attached to 4 further carbon atoms and hence, it is a quaternary carbon.

Carbon (ii) is attached to 2 further carbon atoms and hence, it is a secondary carbon.

Carbon (iii) is attached to 3 further carbon atoms and hence, it is a tertiary carbon.

Carbon (iv) is attached to 2 further carbon atoms and hence, it is a secondary carbon.

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With circulation, the heart provides your body with:
Ivanshal [37]
<h3>Answer:  D) all of the above</h3>

Explanation:

The lungs pump oxygen in and carbon dioxide out, which goes through the blood stream to help with the cell's energy needs.

Nutrients pass through the blood stream as well. The nutrients start with the digestive system (mouth, esophagus, stomach, small intestine) before going into the blood stream. The nutrients are building blocks to help make new cells when old ones die off.

When those cells die off, the body sheds them like dead skin, but internal dead cells are passed off as waste. This waste and other byproducts the body doesn't need passes through the blood stream as well.

In short, the blood stream is basically the highway to help get desired materials (eg: oxygen and nutrients) and get rid of waste (eg: carbon dioxide and other unwanted byproducts or dead cell material)

So that's why the answer includes A, B and C.

8 0
3 years ago
A molecule of carbon dioxide is made up of 1 atom of carbon and 2 atoms of oxygen. Which of the following represents 2 carbon di
Degger [83]
Here’s a CO2 molecule :)

3 0
2 years ago
Read 2 more answers
when carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4g of carbon were burned in the presence of
SpyIntel [72]
Answer: C(s) + O2(g) --> CO2(g)12g (C) .... 50.8g (O2)................. initial amounts0g(C) .........18.8g(O2) ................. amounts when reaction completeThat means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 12 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.12g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) =44 g of CO2
7 0
2 years ago
What volume is needed to make a 2.45 M solution of KCl using 0.50 mol of KCl
Ivan

Answer:

Explanation:

91.4

grams

Explanation:

C

=

m

o

l

v

o

l

u

m

e

2.45

M

=

m

o

l

0.5

L

2.45

M

⋅

0.5

L

=

m

o

l

m

o

l

=

1.225

Convert no. of moles to grams using the atomic mass of K + Cl

1.225

m

o

l

⋅

(

39.1

+

35.5

)

g

m

o

l

1.225

m

o

l

⋅

74.6

g

m

o

l

=

1.225

⋅

74.6

g

=

91.4

g

4 0
2 years ago
A 5.024 mg sample of an unknown organic molecule containing carbon, hydrogen, and nitrogen only was burned and yielded 13.90 mg
Dafna1 [17]

Answer:

C8H17N

Explanation:

Mass of the unknown compound = 5.024 mg

Mass of CO2 = 13.90 mg

Mass of H2O = 6.048 mg

Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:

For carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C = 12/44 x 13.90 = 3.791 mg

For hydrogen, H:

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H = 2/18 x 6.048 = 0.672 mg

For nitrogen, N:

Mass N = mass of unknown – (mass of C + mass of H)

Mass of N = 5.024 – (3.791 + 0.672)

Mass of N = 0.561 mg

Now, we can obtain the empirical formula for the compound as follow:

C = 3.791 mg

H = 0.672 mg

N = 0.561 mg

Divide each by their molar mass

C = 3.791 / 12 = 0.316

H = 0.672 / 1 = 0.672

N = 0.561 / 14 = 0.040

Divide by the smallest

C = 0.316 / 0.04 = 8

H = 0.672 / 0.04 = 17

N = 0.040 / 0.04 = 1

Therefore, the empirical formula for the compound is C8H17N

8 0
3 years ago
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