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3 years ago

Explain how chemists overcome the problem of low yield In industry

1 answer:

8 0

High temperature and pressure produce the highest rate of reaction. However, this must be balanced with the high cost of the energy needed to maintain these conditions. Catalysts increase the rate of reaction without affecting the yield. This can help create processes which work well even at lower temperatures.

I hope this helps you.

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The __________ system is made up of the nose, pharynx, trachea, lungs, bronchi, and alveoli. *

Reil [10]

Respiratory system is made of that

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3 years ago

Read 2 more answers

If gas in a sealed container has a pressure of 50 kPa at 300 K, what will the pressure be if the temperature rises to 360 K?

VLD [36.1K]

The answer is 60 kpa, also did you try look it up because that is what i did but i did not copy and paste it.

hope this helped have a good day

6 0

3 years ago

Give the structure of the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane.

Fudgin [204]

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

what is free radical halogenation?

A substitution reaction in which a hydrogen atom is replaced with a halogen atom, via a free radical reaction mechanism. when this reaction is carrid out by bromine radical, it is called free radicle bromination. When bromine ( $Br{2}$ ) treated with light (hν) it comes to homolytic cleavage of the Br-Br bond and give rise to bromine radicles.

free-radical bromination of 2,2,4−trimethylpentane

Bromination of an alkane includes the substitution of a bromine atom for a hydrogen atom. The following stages will be taken by 2,2,4-trimethylpentane during this reaction:

Initiation reaction: The production of a bromine free radical requires the initiation of heat or light.

$Br - Br$ ⇒ $2Br$ ·

Propagation: This reaction relies heavily on hydrogen. This reaction is impossible if hydrogen is not present. Because tertiary free radicals are more stable than secondary and primary free radicals, they are favoured in this reaction.

Termination: The remaining free radical of bromide reacts with the tertiary free radical of 2,2,4-trimethylpentane to form 2-bromo-2,4,4-trimethylpentane.

the principal organic product formed by free-radical bromination of 2,2,4−trimethylpentane is 2-bromo-2,4,4-trimethylpentane.

To know more about free radical halogenation, check out:

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7 0

1 year ago

Chemistry

madam [21]

Answer: 1. 3.914 × ^10-4 | 2. 4.781 × ^10-1

Explanation:

7 0

3 years ago

Based on the relative energy of the absorbed electromagnetic radiation, which absorber, a peptide bond or an aromatic side chain

alina1380 [7]

Aromatic side chain exhibits an electronic excited state that is closer in energy to the ground state.

In order to respond to this query, we must decide whether a peptide bond or an aromatic side chain is demonstrating an electronic exited state that is more closely related to the ground state in terms of energy.
When our energy is as low as possible, we are in the ground state.
What I want to point out is that if we can choose between the two options—peptide bond or aromatic side chain—without knowing the specific reasons, we can immediately rule out two potential answers.
Consider what we already know about energy, we have:

E = h x c/λ

That indicates that when we have more energy, a wavelength decreases. Lower energy corresponds to higher wavelength.

Aromatic side chains absorb between 250 and 290 nm, while peptide bonds do so between 190 and 250 nm.

According to our breakdown, we have an electron excited state that is more closely related to the ground state in terms of energy as wavelength increases.

Thus, Aromatic side chain exhibits an electronic excited state that is closer in energy to the ground state.

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4 0

2 years ago