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3 years ago

The first step in the free radical mechanism for the preparation of polyethylene is:

Chemistry

1 answer:

kherson [118]3 years ago

3 0

Answer:

The first step in the free radical mechanism for the preparation of polyethylene is:

b. heating an organic peroxide to break the O-O bond

Explanation:

The preparation of ethylene occur by free radical mechanism , which occurs in three steps :

1. Chain Initiation

2. Chain propagation

3. Chain termination.

The first step i.e chain initiation require a radical. The compound are generally peroxide(O-O linkage) that produce radical is :

Benzoyl Peroxide

This compound when heated (with U.V radiation) produce two species of radicals .

(see the attached image)

These RADICALS are highly unstable and very reactive in nature.

They readily react with ethene and initiate the polymerisation.

Thus,

The first step in the free radical mechanism for the preparation of polyethylene is:

b. heating an organic peroxide to break the O-O bond

The sequence of Polymerisations are:

b. heating an organic peroxide to break the O-O bond

d. propagation of the free radicals

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Carbons from acetyl CoA are transferred to the citric acid cycle. Which is the first round of the citric acid cycle that could p

Zielflug [23.3K]

Answer:

In the third step of the citric acid cycle, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released.

Explanation:

In the first step of citric acid cycle, acetylCoA combines with a four-carbon molecule, oxaloacetate, forming a six-carbon molecule, citrate.

In the second step, the citrate in the presence of enzyme anicotase is converted into isocitrate.

In the third step, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released leaving behind one five-carbon molecule called as α-ketoglutarate. During this step, NAD⁺ is reduced to form NADH.

This is first round of the citric acid cycle that could possibly release a carbon atom originating from this acetyl CoA.

On series of reaction, another carbon dioxide molecule also being relased and oxaloacetate is regenerated again.

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3 years ago

Enter your answer in the provided box. When mixed, solutions of barium chloride, BaCl2, and potassium chromate, K2CrO4, form a y

ira [324]

Answer:

35.42g

Explanation:

Step 1:

The balanced equation for the reaction

BaCl2(aq) + K2CrO4(aq) → BaCrO4(s) + 2KCl(aq)

Step 2:

Determination of the limiting reactant.

It is important to determine which of the reactant is limiting the reaction as the limiting reactant is used to determine the maximum yield of the reaction. The limiting reactant can be determined as follow:

From the balanced equation above,

1 mole of BaCl2 reacted with 1 mole of K2CrO4.

Therefore, 0.7 mole of BaCl2 will also react with 0.7 mol of K2CrO4.

From the above illustration, we can see that it requires a higher amount of K2CrO4 to react with 0.7 mol of BaCl2. This simply means that K2CrO4 is the limiting reactant.

Step 3:

Determination of the number of mole of BaCrO4 produced from the reaction.

The limiting reactant is used in this case.

From the balanced equation above,

1 mole of K2CrO4 produced 1 mole of BaCrO4.

Therefore, 0.14 mole of K2CrO4 will also produce 0.14 mole of BaCrO4.

Step 4:

Converting 0.14 mole of BaCrO4 to grams.

This is illustrated below:

Molar Mass of BaCrO4 = 137 + 52 + (16x4) = 137 + 52 + 64 = 253g/mol

Number of mole BaCrO4 = 0.14 mole

Mass of BaCrO4 =?

Mass = number of mole x molar Mass

Mass of BaCrO4 = 0.14 x 253

Mass of BaCrO4 = 35.42g

Therefore, 35.42g of BaCrO4 is produced from the reaction.

4 0

3 years ago

How do we seperate a mixture of water and sugar

Firlakuza [10]

Answer:

The easiest way to separate a mixture of sugar and water is to use distillation, a process that separates substances based on their different boiling points. 《☆☆☆☆☆》

Explanation:

Hope it helps JOIN 《Æ §QŮÅĐ》

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3 years ago

The natural abundances of elements in the human body, expressed as percent by mass, are oxygen (o), 65 percent; carbon (c), 18 p

Vlada [557]

It is given that the person weighs 62 kg = 62,000 g

Natural abundances in mass percent are:

O = 65%

C = 18%

H = 10%

N = 3.0%

Ca = 1.6%

P = 1.2%

Corresponding weights of the elements are:

O = 65/100 * 62000 g = 40.30 * 10^3 g

C = 18/100 * 62000 g = 11.16 * 10^3 g

H = 10/100 * 62000 g = 62.00 * 10^2 g

N = 3.0/100 * 62000 g = 18.60 * 10^2 g

Ca = 1.6/100 * 62000 g = 9.92 * 10^2 g

P = 1.2/100 * 62000 g = 7.44 * 10^2 g

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3 years ago

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From a laboratory process designed to separate water into hydrogen and oxygen gas, a student collected 20.0g of Hydrogen and 158

denis23 [38]

Answer:

it is water 2

Explanation:

3 0

1 year ago

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