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natali 33 [55]
3 years ago
5

Hexane is a chain organic compound that has the formula C6H14. What type of organic compound is hexane? Alcohol Amine Ether Hydr

ocarbon
Chemistry
1 answer:
Maslowich3 years ago
7 0

Answer:  C_6H_{14} is a hydrocarbon.

Explanation:

Alcohols are the compounds which contain -OH as the functional group. Example: CH_3OH

Amine are the compounds which contain -NH_2 as the functional group. Example: CH_3NH_2

Ether are the compounds which contain ROR as the functional group. Example: CH_3OCH_3

Hydrocarbons are defined as the compounds in which bonds are present between carbon and carbon atoms. Hydrocarbons with single bond between carbon and carbon atoms are called as alkanes.The general formula for alkanes is C_nH_{2n+2} Example: C_6H_{14} which is named as hexane.

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If I add 375 of water to 125 mL of a 0.90 M NaOH solution, what will the molarity of the diluted solution be?
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Answer:

0.225M

Explanation:

Step 1:

Data obtained from the question. This includes:

Volume of the stock solution (V1) = 125mL

Molarity of the stock solution (M1) = 0.90 M

Volume of the diluted solution (V2) = 125 + 375 = 500mL

Molarity of the diluted solution (M2) =..?

Step 2:

Determination of the molarity of the diluted solution.

This is obtained by using the dilution formula as follow:

M1V1 = M2V2

0.9 x 125 = M2 x 500

Divide both side by 500

M2 = (0.9 x 125) /500

M2 = 0.225M

Therefore, the molarity of the diluted solution is 0.225M

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Which of the following is always true when a substance undergoes a physical change ?
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When a substance undergoes a physical change no new substance is formed.
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How many molecules of water are there in 1 gram of water​
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Answer:

6.022*1023

Explanation:

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Are electrons exchanged between ionic or covalent bonds
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After the end of a normal inspiration, the volume of air in the lungs is about 2.8 L. Normally quiet inspiration is driven by a
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Answer:

The Volume of the lungs that would produce 2 mmHg pressure decrease is

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Explanation:

From the question we are told that

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     The pressure difference for quit normal inspiration is P = 2 \ mmHg

      The temperature of air in the lungs T = 37^oC

      The pressure  after normal  expiration is at  T =  760 \ mmHg

     

From ideal gas law we have that

         PV= nRT

Now since  nRT is constant we have that

        P_1 V_1 = P_2 V_2

As the pressure decreased by 2 mmHg the volume becomes

        V_2 = \frac{P_1 V_1}{P_2}

        V_2 = \frac{2.8 * 760}{758}

        V_2 = 2.81 \ L

       

     

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