If nickel crystallized in a face-centered cubic structure, the six atoms on the faces of the unit cell would contribute three net nickel atoms, for a total of four atoms per unit cell.

Face-centered cubic structure:

(8 corners x 1/8) + (6 faces x 1/2) = 4 atoms

Because they have different numbers of atoms in a unit cell, each of these structures would have a different density. Let's therefore calculate the density for nickel based on each of these structures and the unit cell edge length for nickel given in the previous section: 0.3524 nm. In order to do this, we need to know the volume of the unit cell in cubic centimeters and the mass of a single nickel atom.
The volume (V) of the unit cell is equal to the cell-edge length (a) cubed.

V = a3 = (0.3524 nm)3 = 0.04376 nm3

Since there are 109 nm in a meter and 100 cm in a meter, there must be 107 nm in a cm.
We can therefore convert the volume of the unit cell to cm3 as follows.
The mass of a nickel atom can be calculated from the atomic weight of this metal and Avogadro's number.
The density of nickel, if it crystallized in a simple cubic structure, would therefore be 2.23 g/cm3, to three significant figures.

Simple cubic structure: Google cv templates free.

Because there would be twice as many atoms per unit cell if nickel crystallized in a body-centered cubic structure, the density of nickel in this structure would be twice as large.

Body-centered cubic structure:

There would be four atoms per unit cell in a face-centered cubic structure and the density of nickel in this structure would be four times as large.

Face-centered cubic structure:

The experimental value for the density of nickel is 8.90 g/cm3. The obvious conclusion is that nickel crystallizes in a face-centered cubic unit cell and therefore has a cubic closest-packed structure.
Estimates of the radii of most metal atoms can be found. Where do these data come from? How do we know, for example, that the radius of a nickel atom is 0.1246 nm?
Nickel crystallizes in a face-centered cubic unit cell with a cell-edge length of 0.3524 nm to calculate the radius of a nickel atom.
One of the faces of a face-centered cubic unit cell is shown in the figure below.
According to this figure, the diagonal across the face of this unit cell is equal to four times the radius of a nickel atom.
The Pythagorean theorem states that the diagonal across a right triangle is equal to the sum of the squares of the other sides. The diagonal across the face of the unit cell is therefore related to the unit-cell edge length by the following equation.
Taking the square root of both sides gives the following result.
We now substitute into this equation the relationship between the diagonal across the face of this unit cell and the radius of a nickel atom:
Solving for the radius of a nickel atom gives a value of 0.1246 nm:
A similar approach can be taken to estimating the size of an ion. Let's start by using the fact that the cell-edge length in cesium chloride is 0.4123 nm to calculate the distance between the centers of the Cs+ and Cl- ions in CsCl.
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CsCl crystallizes in a simple cubic unit cell of Cl- ions with a Cs+ ion in the center of the body of the cell, as shown in the figure below.
Before we can calculate the distance between the centers of the Cs+ and Cl- ions in this crystal, however, we have to recognize the validity of one of the simplest assumptions about ionic solids: The positive and negative ions that form these crystals touch.
We can therefore assume that the diagonal across the body of the CsCl unit cell is equivalent to the sum of the radii of two Cl- ions and two Cs+ ions.
Earth space colonies download free. The three-dimensional equivalent of the Pythagorean theorem suggests that the square of the diagonal across the body of a cube is the sum of the squares of the three sides.
Taking the square root of both sides of this equation gives the following result.
If the cell-edge length in CsCl is 0.4123 nm, the diagonal across the body in this unit cell is 0.7141 nm.
The sum of the ionic radii of Cs+ and Cl- ions is half this distance, or 0.3571 nm.
If we had an estimate of the size of either the Cs+ or Cl- ion, we could use the results to calculate the radius of the other ion. The ionic radius of the Cl- ion is 0.181 nm. Substituting this value into the last equation gives a value of 0.176 nm for the radius of the Cs+ ion.
The results of this calculation are in reasonable agreement with the value of 0.169 nm known for the radius of the Cs+ ion. The discrepancy between these values reflects the fact that ionic radii vary from one crystal to another. The tabulated values are averages of the results of a number of calculations of this type.

Example

Specify a font named 'myFirstFont', and specify the URL where it can be found:
@font-face {
font-family: myFirstFont;
src: url(sansation_light.woff);
}
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More 'Try it Yourself' examples below.

Definition and Usage

With the @font-face rule, web designers do not have to use one of the 'web-safe' fonts anymore.
In the @font-face rule you must first define a name for the font (e.g. myFirstFont), and then point to the font file.

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To use the font for an HTML element, refer to the name of the font (myFirstFont) through the font-family property:

Browser Support

The @font-face rule is supported in Edge, Chrome, Firefox, Safari, and Opera.
The numbers in the table specifies the first browser version that fully supports the font format.
Font format
TTF/OTF9.0*4.03.53.110.0
WOFF9.05.03.65.111.1
WOFF214.036.039.010.026.0
SVGNot supportedNot supportedNot supported3.2Not supported
EOT6.0Not supportedNot supportedNot supportedNot supported
*The font format only works when set to be 'installable'.

Syntax


Font descriptorValuesDescription
font-familynameRequired. Defines the name of the font.
srcURLRequired. Defines the URL(s) where the font should be downloaded from
font-stretchnormal
condensed
ultra-condensed
extra-condensed
semi-condensed
expanded
semi-expanded
extra-expanded
ultra-expanded
Optional. Defines how the font should be stretched. Default value is 'normal'
font-stylenormal
italic
oblique
Optional. Defines how the font should be styled. Default value is 'normal'
font-weightnormal
bold
100
200
300
400
500
600
700
800
900
Optional. Defines the boldness of the font. Default value is 'normal'
unicode-rangeunicode-rangeOptional. Defines the range of unicode characters the font supports. Default value is 'U+0-10FFFF'

More Examples

Example

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You must add another @font-face rule containing descriptors for bold text:
@font-face {
font-family: myFirstFont;
src: url(sansation_bold.woff);
font-weight: bold;
}
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The file 'sansation_bold.woff' is another font file, that contains the bold characters for the Sansation font.
Browsers will use this whenever a piece of text with the font-family 'myFirstFont' should render as bold.
This way you can have many @font-face rules for the same font.

Related Pages

CSS tutorial: CSS Web Fonts

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