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Gallium crystallizes in a primitive cubic unit cell The length of an edge of this cube is 362 rm pm What is the radius of a gallium atom?
The radius of a gallium atom in a primitive cubic unit cell can be calculated using the formula: r = (√3/4) * a, where r is the radius of the atom and a is the edge length of the unit cell. Plugging in the values: r = (√3/4) * 362 pm = 314 pm. So, the radius of a gallium atom is approximately 314 picometers.
Gallium crystallizes in a primitive cubic unit cell The length of an edge of this cube is 362 pm What is the radius of a gallium atom?
For a primitive cubic unit cell, the radius of an atom (r) can be calculated as r = √3 x a / 4, where a is the length of the edge of the cube. Plugging in the values, with a = 362 pm, the radius of a gallium atom is approximately 125.3 pm.
What is a reason why the structure of a unit cell of LiF differs from that of a unit cell LiCl?
The difference in the structure of the unit cells of LiF and LiCl arises from the different ionic sizes of fluorine and chlorine ions. LiF has a smaller F^- ion compared to the Cl^- ion in LiCl, leading to a shorter lattice parameter and a more compact arrangement of ions in the unit cell of LiF.
The expected value for 'i' for NaCl is?
The expected value for 'i' is 2 for NaCl because it dissociates into two ions (Na+ and Cl-) when dissolved in water. This means one formula unit of NaCl produces 2 ions in solution.
Silver has a face-centered cubic unit cell How many atoms of rm Ag are present in each unit cell?
There are a total of 4 silver (Ag) atoms present in each face-centered cubic unit cell.
What is the calculations of the density of gold based on the unit cell. The face-centered gold crystal has an edge length of 407 pm?
The face-centered cubic unit cell of gold contains 4 atoms. The volume of the unit cell can be calculated using the formula V = a^3, where a is the edge length. Then, the mass of the unit cell can be calculated using the density of gold. Finally, the density of gold can be calculated by dividing the mass of the unit cell by its volume.
How many composite Cl- ions are present in a single unit cell of a NaCl crystal lattice?
In a single unit cell of a NaCl crystal lattice, there are four composite Cl- ions.
How many composite Na ions are present in a single unit cell of a NaCl crystal lattice?
In a single unit cell of a NaCl crystal lattice, there are six composite Na ions present.
How I can transform 4 mm NaCl to ppm?
mm is unit of length, ppm is a non-SI unit of concentration.
Silver crystallizes in a face-centered cubic arrangement a silver atomis at the edge of each lattice point the length of the edge of the unit cell is 0.4086nm what is the atomic radius of silver?
0.1445 nm
What is the length of an edge of a cube with a volume of 343 cubic milliliters?
There is no such unit as a 'cubic milliliter'.A cube with an edge of 7 length units has a volume of 343 of the same unit cubed.
Gallium crystallizes in a primitive cubic unit cell The length of an edge of this cube is 362 rm pm What is the radius of a gallium atom?
The radius of a gallium atom in a primitive cubic unit cell can be calculated using the formula: r = (√3/4) * a, where r is the radius of the atom and a is the edge length of the unit cell. Plugging in the values: r = (√3/4) * 362 pm = 314 pm. So, the radius of a gallium atom is approximately 314 picometers.
How many molecules present in unit cell of NaCl crystal?
There are four molecules present in the unit cell of a NaCl crystal. These consist of one Na+ ion and one Cl- ion, forming the basic repeating unit structure of the crystal lattice.
IF the edge length of the unit cell is 468.9 pm what is the density of CdO in g per cm3?
To calculate the density of CdO, we first need to determine the volume of the unit cell. The volume of a cubic unit cell is calculated by V = a^3, where "a" is the edge length. Converting the edge length from picometers to centimeters (1 pm = 1 x 10^-12 cm), we find a = 4.689 x 10^-8 cm. Calculating V = (4.689 x 10^-8 cm)^3 will give the volume. From there, we can divide the molar mass of CdO by the volume to determine the density in g/cm^3.
Nickel has an fcc unit cell with an edge length of 350.7pm calculate the radius of a nickel atom?
The face diagonal of an fcc unit cell can be calculated using the relationship: face diagonal = √2 * edge length. Given the edge length of 350.7 pm, the face diagonal would be 495.2 pm. Since the atom sits on the body diagonal, the radius of a nickel atom can be calculated as half the body diagonal, which is (1/√3) * face diagonal. Calculating this gives a radius of approximately 202.1 pm.
Gallium crystallizes in a primitive cubic unit cell The length of an edge of this cube is 362 pm What is the radius of a gallium atom?
For a primitive cubic unit cell, the radius of an atom (r) can be calculated as r = √3 x a / 4, where a is the length of the edge of the cube. Plugging in the values, with a = 362 pm, the radius of a gallium atom is approximately 125.3 pm.
If the atomic radius of aluminum is 0.143 nm calculate the valume of its unit cell in cubic meters?
To calculate the volume of the unit cell of aluminum, we need to determine the type of unit cell it forms. Aluminum crystallizes in a face-centered cubic (FCC) structure. In an FCC unit cell, there are 4 atoms. The formula for the volume of an FCC unit cell is given by: V = (a^3) / (4), where 'a' is the edge length of the unit cell. Given that the atomic radius of aluminum is 0.143 nm, the edge length 'a' can be calculated as 2 * √2 * r, where r is the atomic radius. Substituting the values, we get 'a' = 2 * √2 * 0.143 nm = 0.404 nm. Converting this to meters (1 nm = 1 x 10^-9 m), we get 'a' = 0.404 x 10^-9 m. Finally, substituting this value into the formula for the volume of an FCC unit cell, we get V = (0.404 x 10^-9)^3 / 4 = 6.6 x 10^-29 m^3.
