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Ksivusya [100]
9 months ago
10

What descriptive term is applied to the type of diene represented by 1,5-octadiene?

Chemistry
1 answer:
Mrac [35]9 months ago
3 0

The descriptive term that is applied to the type of diene represented by 1,5-octadiene is isolated diene. The correct option is C.

<h3>What is diene?</h3>

Diene is a compound that contains two or more double bonds, usually  carbon bonds, which are separated by a single bond. They are covalent compounds. Alkene units are surely present in these compounds, whose quantity is two.

1,5-octadiene is a polymer of diene, which are generally elastomers, and they made of vulcanized rubber. They are isolated diene.

Thus, the correct option is C) Isolated diene.

To learn more about diene, refer to the below link:

brainly.com/question/17425564

#SPJ4

The question is incomplete. The complete question is given below:

A) Conjugated diene.

B) Cumulated diene.

C) Isolated diene.

D) Alkynyl diene.

E) None of the above.

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171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in o
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The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?

a. 16mm Hg

b. 24mm Hg  

c. 20mm Hg  

d. 12mm Hg

Answer: The vapor pressure (VP) of this solution at 25^{o}C is closest to the value 24 mm Hg.

Explanation:

Given: Mass of sucrose = 171 g

Mass of water = 1 L = 1000 g

Vapor pressure of water = 24 mm Hg

As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{1000 g}{18.02 g/mol}\\= 55.49 mol

Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{171 g}{342 g/mol}\\= 0.5 mol

Total moles = 55.49 + 0.5 mol = 55.99 mol

Mole fraction of water is as follows.

Mole fraction = \frac{moles of water}{total moles}\\= \frac{55.49}{55.99}\\= 0.99

Formula used to calculate vapor pressure of the solution is as follows.

P_{i} = P^{o}_{i} \times \chi_{i}

where,

P_{i} = vapor pressure of component i over the solution

P^{o}_{i} = vapor pressure of pure component i

\chi_{i} = mole fraction of i

Substitute the values into above formula to calculate vapor pressure of water as follows.

P_{i} = P^{o}_{i} \times \chi_{i}\\= 24 mm Hg \times 0.99\\= 23.76 \\or 24 mm Hg\\

Thus, we can conclude that the vapor pressure (VP) of this solution at 25^{o}C is closest to the value 24 mm Hg.

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A sample of 42 mL of carbon dioxide gas was placed in a piston in order to maintain a constant 101 kPa of pressure.
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Answer:

The answer to your question is 33.4 ml

Explanation:

Data

volume 1 = V1 = 42 ml

temperature 1 = T1 = 20°C

temperature 2 = T2 = -60°C

Volume 2 = V2 = x

Process

1.- Convert celsius to kelvin

T1 = 20 + 273 = 293°K

T2 = -60 + 273 = 233°K

2.- Use the Charles' law to solve this problem

               \frac{V1}{T1} = \frac{V2}{T2}

Solve for V2

                V2 = \frac{V1T2}{T1}

3.- Substitution

               V2 = \frac{(42)(233)}{293}

4.- Simplification

                V2 = \frac{9786}{293}

5.- Result

                V2 = 33.4ml

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2 years ago
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