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bearhunter [10]
3 years ago
12

What is the difference between polycrystalline and amorphous materials? How are the properties of a polycrystalline material aff

ected by the grain size? When might smaller grains be useful? When might larger grains be useful? Describe two applications of amorphous materials.
Chemistry
1 answer:
Ahat [919]3 years ago
5 0

Answer:

Explanation:

The difference between Polycrystalline  and Amorphous materials is given as:

Polycrystalline:

  • The atoms in the crystal lattice are arranged in an ordered manner.
  • The particles in the crystal posses a particular geometry
  • The crystal lattice have a specific temperature known as its Melting Point.

Amorphous:

  • There is no specific order in the arrangement of particles in the crystal.
  • They do not have any particular geometry.
  • There is no specific temperature but a range of temperature in which the crystal melts.

The properties of crystalline materials can be constrained by modifying the grain size at the hour of the amalgamation. The mechanical properties can be improved by choosing the grain size so that the quantity of disengagements and grain limits are expanded.  

Usually this should be possible by diminishing the grain size, yet it additionally relies on a ton of different elements relying on the application. The quality of the material is expanded when the grain size is decreased.

Usefulness of smaller grains:

At the point when the size of the grains is decreased to a degree of 100 nm to 1000 nm, we can say we had acquired smaller grain which can be called as ultra-fine grain materials.  

These can be utilized widely for the assembling of nanomaterial which are having a tremendous assortment of utilization and the new regions of use are expanding by step by step.

Usefulness of smaller grains:

Larger grains size is valuable in light dissipating applications, huge size grain has high perceivability to the light and it very well may be utilized in dispersing applications. Larger molecule size is utilized in specific responses to restrict the reactivity to a specific degree.

Applications of Amorphous Material:

  • The amorphous carbon is utilized for the production of Ta-C films which can be utilized for the applications in ultra-flimsy defensive coatings for attractive plates, in cells, batteries and sun powered cells, to keep up inactive layers in electronic gadgets, etc.
  • Amorphous silicon is utilized for the assembling of the Thin Film Transistor (TFT) which is in the end be utilized for computerized x-beam picture detecting, coordinated shading sensors, sensors for CMOS cameras, light-radiating diodes, and so on.
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A. Na₂S + FeBr₂ → 2NaBr + FeS.

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Iron sulfide (FeS) has low solubility in water.

Calcium sulfate (CaSO₄) has low solubility in water and it will form white precipitate.

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3 years ago
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The elements chlorine and iodine have similar chemical properties<br> because they
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Answer:

They are both halogens and have the same number of electrons on their outer shell.

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3 years ago
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3 years ago
Three isotopes of oxygen are oxygen-16 oxygen-17 and oxygen-18. Write the symbol for each, including the atomic number and mass
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All the isotopes of the same element have the same atomic number. They only vary the mass number.

So, all the isotopes of oxygen have atomic number 8.

The isotope oxygen-16 has mass number 16, so it is written with the symbol O preceded by the number 16 as a superscript and the number 8 as a subscript (the two numbers to the right of the chemical symbol).

The isotope oxygen-17 has mass number 17, so it is written with the symbol O preceded by the number 17 as a superscript and the number 8 as a subscript.

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4 years ago
How many millimeters of 1.25 M Copper(2) sulfate, CuSO4, will contain 55.75g CuSo4
Shkiper50 [21]

Answer:

280 mL

Explanation:

Given data:

Mass of CuSO₄ = 55.75 g

Molarity of CuSO₄ = 1.25 M

Volume of CuSO₄ = ?

Solution:

First of all we will calculate the number of moles.

Number of moles:

Number of moles = mass / molar mass

Number of moles = 55.75 g/ 159.6 g/mol

Number of moles = 0.35 mol

Volume of CuSO₄:

Molarity = number of moles / volume in litter

1.25 M = 0.35 mol / volume in litter

Volume in litter = 0.35 mol / 1.25 M

Volume in litter = 0.28 L

Volume in milliliter:

0.28 L × 1000 mL/ 1 L = 280 mL

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