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2 years ago

the main part of the microscope that is used to see and denlarge specimens what do you call this part?

Chemistry

1 answer:

Crazy boy [7]2 years ago

7 0

Answer:

I believe you are asking about the objective lens or eyepiece.

Explanation:

Objective lenses are used to magnify objects enough for you to see them in great detail

Eyepiece/Ocular is what you look through at the top of the microscope.

A tube connects the eyepiece lens to objective lenses, which enhance the magnification power of the eyepiece lens. Both the eyepiece and the objective lens contribute to the magnification of the object.

The eyepiece lens usually magnifies 10x, and a typical objective lens magnifies 40x.

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A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 234 picometers and its

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Answer:

$\delta=101.13 g/cm^3$

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3 years ago

If the molecular weight of a semiconductor is 27.9 grams/mole and the diamond lattice constant is 0.503 nm, what is the density

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Explanation:

The given data is as follows.

Mass = 27.9 g/mol

As we know that according to Avogadro's number there are $6.023 \times 10^{26}$ atom present in 1 mole. Therefore, weight of 1 atom will be as follows.

1 atoms weight = $\frac{38}{6.023 \times 10^{26}}$

In a diamond cubic cell, the number of atoms are 8. So, n = 8 for diamond cubic cell.

Therefore, total weight of atoms in a unit cell will be as follows.

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= $37.06 \times 10^{-26}$

Now, we will calculate the volume of a lattice with lattice constant 'a' (cubic diamond) as follows.

= $a^{3}$

= $(0.503 \times 10^{-9})^{3}$

= $0.127 \times 10^{-27} m^{3}$

Formula to calculate density of diamond cell is as follows.

Density = $\frac{mass}{volume}$

= $\frac{37.06 \times 10^{-26}}{0.127 \times 10^{-27} m^{3}}$

= 2918.1 $g/m^{3}$

or, = 0.0029 g/cc (as 1 $m^{3} = 10^{6} cm^{3}$ )

Thus, we can conclude that density of given semiconductor in grams/cc is 0.0029 g/cc.

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