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Gelneren [198K]
3 years ago
5

Describe the process used to measure out a specific mass of a solid

Chemistry
1 answer:
Svetlanka [38]3 years ago
7 0
Although the process varies slightly from one material to another, the general process is as follows:

1) Choose an appropriate container for the solid. This may be a petri dish or a beaker in which you want to prepare the solution of the solid or any other lab equipment.

2) Place the container on a mass balance, then turn the balance on. The mass balance will automatically zero-out the mass of the container, so that any mass that you add on the container will be the mass of the solid. Alternatively, you may first measure the mass of the empty container alone.

3) Add the solid using a lab spatula. The solid should be added more slowly when the reading on the scale comes close to the desired value.

4) Remove the container from the mass balance after the desired amount of solid has been added.
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A.

Explanation:

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Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

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