LAB: predicting products

The density of solid Ag is 10.5 g/cm3. How many atoms are present per cubic centimeter of Ag?

Sveta_85 [38]

Answer:

Unit cells which are present per cubic centimeter of Ag = $=1.45\times 10^{22}\ unit\ cells /cm^3$

Volume = $68.9\times 10^{-24}\ cm^3$

Edge length = $4.1\times 10^{-8}\ cm$

Explanation:

(a)

Given that:-

The density of the solid Ag = 10.5 g/cm³

Molar mass of silver = 107.8682 g/mol

So, Moles present per cm³ of Ag = $\frac{10.5\ g/cm^3}{107.8682\ g/mol}$ =0.0973 mol/cm³

Also, 1 mole = $6.023\times 10^{23}$ atoms.

So,

Atoms present per cm³ of Ag = $0.0973\ mol/cm^3\times 6.023\times 10^{23}\ atoms/mol=5.8\times 10^{22}\ atoms/cm^3$

Thus, answer = $5.8\times 10^{22}\ atoms/cm^3$

In FCC, the number of atoms in the unit cell = 4 unit cells

So,

Unit cells which are present per cubic centimeter of Ag = $\frac{5.8\times 10^{22}\ atoms/cm^3}{4}=1.45\times 10^{22}\ unit\ cells /cm^3$

Unit cells which are present per cubic centimeter of Ag = $=1.45\times 10^{22}\ unit\ cells /cm^3$

(b)

The reciprocal of the unit cell/cm³ is the volume of the unit cell.

So, $Volume=\frac{1}{1.45\times 10^{22}\ unit\ cells /cm^3}=68.9\times 10^{-24}\ cm^3$

Volume = $68.9\times 10^{-24}\ cm^3$

(c)

Also, Volume = ${(Edge\ length)}^3$

Thus, edge length = ${Volume}^{\frac{1}{3}}$ = $\left(68.9\times \:\:10^{-24}\right)^{\frac{1}{3}}\ cm=4.1\times 10^{-8}\ cm$

Edge length = $4.1\times 10^{-8}\ cm$

5 0

4 years ago

Drag each item to indicate whether it is related to Aristotle's or Ptolemy's model of the solar system, or to both. Some items m

Romashka-Z-Leto [24]

Answer:

Aristotle's Model :

geocentric

planets fixed on nested spheres

Ptolemy's model:

geocentric

planets fixed on epicycles

accurately predicted the positions of the planets

Explanation:

Let's begin by the fact both models are geocentric and make a brief explanation of them:

Aristotle's Model :

Aristotle built his model based on observations and philosophical assumptions, not on measurements or calculations, then he mistakenly thought the Earth was the center of the universe.

So, according to this model, the universe was spherical and finite, with the Earth immobile at its center, composed of the four fundamental elements (made up of spherical layers): ground, water, air and fire; and the Sun along with the fixed planets in their respective concentric spheres (also called nested spheres) revolving around the Earth.

Ptolemy's model:

Ptolemy's model of the universe was also geocentric, placing the Earth motionless in the center of the known universe and was the accepted model for a long time during the Middle Ages.

In this sense, ccording to Ptolemy's model of the universe our planet remains stationary while the other planets, the Moon and the Sun describe complicated orbits around it (called epicycles).

However, this model predicted with an acceptable accuracy for that time the motion of the known planets, until Copernicus appeared with the revolutionary Heliocentric model.

5 0

3 years ago

Which atom attains a stable valence electron configuration by bonding with another atom?

zloy xaker [14]

Answer:hydrogen .

Explanation:

6 0

3 years ago

A 3.0-L balloon containing helium gas, He, at 17 °C, was taken outdoors where the temperature was close to 0 °C. What happened t

kvv77 [185]

Answer:

The gas volume remains constant

Explanation:

The volume stayed constant however the balloon may seem to have shriveled or deflated upon walking outside due to the helium condensing in the cooler climate.

7 0

3 years ago

When determining the melting point of a substance, should one use a large or a small sample in the capillary? Should the sample

REY [17]

The glass capillary having one end closed and one open end is generally used for the determination of melting point of the sample. From the open end, the sample is put into the capillary, the sample must be firmly packed as the melting point is an intrinsic property that means it is independent of sample size. So, in order to determine the melting point of the sample small sample in the capillary is sufficient to measure the melting point of the sample. To obtain the more consistent value of melting point one must pack the sample firmly in the capillary.

Hence, when determining the melting point of a substance, one should use a small sample in the capillary and the sample should be firmly packed.

4 0

3 years ago