Mixtures can be separated by physical processes.

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

Mrac [35]

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.

3 0

1 year ago

At Christmas time, I like to burn scented candles that smell like sugar cookies. How does the candle in one location change the

LekaFEV [45]

Answer:

its called diffusion

Explanation:

so the air is a substance and then it mixes with another substance which is the scent of the candle then the particles mix together to make a mixture

3 0

2 years ago

Which reactant in each of the following pairs is more nucleophilic? Explain.a. -NH2 or NH3b. H20 or CH3CO2-c. BF3 or F-d. (CH3)3

Anestetic [448]

Answer:

I^- or Cl^-

Explanation:

A nucleophile is any reagent that donates an unshared pair of electrons to form a new covalent bond. Nucleophiles are mostly bases also. Nucleophilicity is a kinetic property, it refers to the rate at which a nucleophile replaces a leaving group from an alkyl halide.

Looking at the options, Cl^- and I^- are the best nucleophiles among the options provided in the question.

3 0

3 years ago

The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that als

olasank [31]

Answer:

A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula $M_1*V_1 = M_2*V_2$ . It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL) with water

Explanation:

Given data includes:

Tris= 10mM

pH = 8.0

NaCl = 150 mM

Imidazole = 300 mM

In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.

Stock Concentration Volume to be Final Concentration

added

1 M Tris 2.5 mL 10 mM

5 M NaCl 7.5 mL 150 mM

1 M Imidazole 75 mL 300 mM

$M_1*V_1 = M_2*V_2$ . is the formula that is used to determine the corresponding volume that is added for each stock concentration

The stock concentration of Tris ( 1 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.01 M *250mL\\V_1 = 2.5mL$

The stock concentration of NaCl (5 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.15 M *250mL\\V_1 = 7.5mL$

The stock concentration of Imidazole (1 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.03 M *250mL\\V_1 = 75mL$

Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.

5 0

3 years ago

A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane

son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

= 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

= 75 - 67.5

= 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

= 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

= 25 - 21.25

= 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

= 64.125 moles

$CO$ formed is = 0.05 x 67.5

= 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

= 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

= 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21}$

= 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

= 2 x 64.125

= 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

= 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3 = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

= 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b). The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.

6 0

3 years ago