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aalyn [17]
3 years ago
10

You are investigating an organic compound. you discover the following: it contains no hydrogen atoms. what property do you expec

t it to exhibit ?
a. has a foul odor
b. odorless
c. flammable
d. high boiling point
e. low boiling point
Chemistry
2 answers:
Juli2301 [7.4K]3 years ago
5 0

the answer is C because it has no water there will be nothing to boil and odor has nothing to do with hydrogen

I hope this helps! :)

-Ayden

Mashutka [201]3 years ago
4 0

Answer:

<u>Organic Compounds:</u>

The question itself is very wrong, as we can not attribute the name Organic compound just to any sort of compound which does not have the basic constituents inside its structure, as the absence of element hydrogen means that the compound which is under our observation is not O.C at all but it will be some other type of compound which will have a way different properties and structure.

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14. 60. g of NaOH is dissolved in enough distilled water to make 300 mL of a stock solution. What volumes of this solution and d
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The question is incomplete, the complete question is attached below.

Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

Explanation : Given,

Mass of NaOH = 60 g

Volume of stock solution = 300 mL

Molar mass of NaOH = 40 g/mol

First we have to calculate the molarity of stock solution.

\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}

Now put all the given values in this formula, we get:

\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M

Now we have to determine the volume of stock solution and distilled water mixed.

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of stock solution.

M_2\text{ and }V_2 are the molarity and volume of diluted solution.

From data (A) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 20mL=1M\times V_2\\\\V_2=100mL

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (B) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 20mL=1M\times V_2\\\\V_2=100mL

Volume of stock solution = 20 mL

Volume of distilled water = 100 mL - 20 mL = 80 mL

From data (C) :

M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From data (D) :

M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?

Putting values in above equation, we get:

5M\times 60mL=1M\times V_2\\\\V_2=300mL

Volume of stock solution = 60 mL

Volume of distilled water = 300 mL - 60 mL = 240 mL

From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.

Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.

5 0
3 years ago
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