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aleksklad [387]
1 year ago
11

Explain how impurities are selected for doping in group 14 semiconductors

Chemistry
1 answer:
aleksklad [387]1 year ago
6 0

Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

<h3>Meaning of Semiconductors</h3>

Semiconductors can be defined as any material that has the ability to exhibit some properties of a conductor and some properties of an insulator.

A semiconductor can be used as either a conductor or an insulator when worked upon.

In conclusion, Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

Learn more about semiconductors: brainly.com/question/1918629

#SPJ1

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g A gas is compressed in a cylinder from a volume of 20.0 L to 2.0 L by a constant pressure of 10.0 atm. Calculate the amount of
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Answer:

The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.

Explanation:

Thermodynamic work is called the transfer of energy between the system and the environment by methods that do not depend on the difference in temperatures between the two. When a system is compressed or expanded, a thermodynamic work is produced which is called pressure-volume work (p - v).

The pressure-volume work done by a system that compresses or expands at constant pressure is given by the expression:

W system= -p*∆V

Where:

  • W system: Work exchanged by the system with the environment. Its unit of measure in the International System is the joule (J)
  • p: Pressure. Its unit of measurement in the International System is the pascal (Pa)
  • ∆V: Volume variation (∆V = Vf - Vi). Its unit of measurement in the International System is cubic meter (m³)

In this case:

  • p= 10 atm= 1.013*10⁶ Pa (being 1 atm= 101325 Pa)
  • ΔV= 2 L- 20 L= -18 L= -0.018 m³ (being 1 L=0.001 m³)

Replacing:

W system= -1.013*10⁶ Pa* (-0.018 m³)

Solving:

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<u><em>The amount of work done on the system is 18234 J and the final positive sign means that this work corresponds to an increase in internal energy of the gas.</em></u>

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