How many atoms are there per unit cell in metallic tungsten if it forms a face centered cubic unit cell?
Q. How many atoms are there per unit cell in metallic tungsten if it forms a face centered cubic unit cell? 9 14 2 4 8 I'm not sure how to answer this, so please help! Thanks!
Asked by Sarah K - Fri Feb 22 20:32:56 2008 - - 2 Answers - 0 Comments
A. 4 Take eight dice and put them together with a square of four on the bottom and a square of four on the top. Consider the eight corners of the dice that all meet in the middle, buried in the stack. If there is an atom at that spot, it is shared by all eight cubes. But there is the same situation by the corners of any cubic cell. That means that any corner of a cubic cell has 1/8 atom. Because there are eight corners to a cube, 8 x 1/8 = 1 atom per cell owing to corners. Next consider faces. If there is an atom in each face, it is half in and half out of another cell. So each face in a face-centered cubic structure has 1/2 atom per face. Because there are six faces to a cube, 6 x 1/2 = 3 atoms per cell owing to faces. 1 + 3 = 4 atoms per… [cont.]
Answered by steve_geo1 - Fri Feb 22 21:01:25 2008
Q. How many atoms are there per unit cell in metallic tungsten if it forms a face centered cubic unit cell? 9 14 2 4 8 I'm not sure how to answer this, so please help! Thanks!
Asked by Sarah K - Fri Feb 22 20:32:56 2008 - - 2 Answers - 0 Comments
A. 4 Take eight dice and put them together with a square of four on the bottom and a square of four on the top. Consider the eight corners of the dice that all meet in the middle, buried in the stack. If there is an atom at that spot, it is shared by all eight cubes. But there is the same situation by the corners of any cubic cell. That means that any corner of a cubic cell has 1/8 atom. Because there are eight corners to a cube, 8 x 1/8 = 1 atom per cell owing to corners. Next consider faces. If there is an atom in each face, it is half in and half out of another cell. So each face in a face-centered cubic structure has 1/2 atom per face. Because there are six faces to a cube, 6 x 1/2 = 3 atoms per cell owing to faces. 1 + 3 = 4 atoms per… [cont.]
Answered by steve_geo1 - Fri Feb 22 21:01:25 2008
Does elemental gold have a face-centered cubic structure og a body centered cubic structure?
Q. The radius of gold is 144 pm, and the density is 19,32 g/cm3. Does elemental gold have a face-centered cubic structure or a body-centered cubic structure?
Asked by sillyhilly1 - Mon Nov 17 07:32:38 2008 - - 1 Answers - 2 Comments
A. *** first.. let's figure out what bcc and fcc cells look like *** if gold had a face centered cubic structure, then there would be 4 atoms per unit cell. right? 8 corner atoms. each corner atom exists in 8 different cells so 1/8 of each corner atom is in a particular unit cell. 8 x 1/8 = 1.. 6 face atoms. each one is in 2 cells so 6 x 1/2 = 3.. 1 from the corners + 3 from the faces = 4 atoms / cell a body centered cubic structure has 1 atom from the corners + 1 in the center = 2 atoms / cell. *** next let's find the length of a "face centered gold unit cell *** if gold were face centered. each side would look like a square with the length of a diagonal = 1/2 the diameter of an atom + 1 diameter of an atom + 1/2 diameter of an… [cont.]
Answered by m w - Mon Nov 17 11:42:17 2008
Q. The radius of gold is 144 pm, and the density is 19,32 g/cm3. Does elemental gold have a face-centered cubic structure or a body-centered cubic structure?
Asked by sillyhilly1 - Mon Nov 17 07:32:38 2008 - - 1 Answers - 2 Comments
A. *** first.. let's figure out what bcc and fcc cells look like *** if gold had a face centered cubic structure, then there would be 4 atoms per unit cell. right? 8 corner atoms. each corner atom exists in 8 different cells so 1/8 of each corner atom is in a particular unit cell. 8 x 1/8 = 1.. 6 face atoms. each one is in 2 cells so 6 x 1/2 = 3.. 1 from the corners + 3 from the faces = 4 atoms / cell a body centered cubic structure has 1 atom from the corners + 1 in the center = 2 atoms / cell. *** next let's find the length of a "face centered gold unit cell *** if gold were face centered. each side would look like a square with the length of a diagonal = 1/2 the diameter of an atom + 1 diameter of an atom + 1/2 diameter of an… [cont.]
Answered by m w - Mon Nov 17 11:42:17 2008
Why is the formula of a body-centered cubic different from that of a face-centered cubic?
Q. Why is the formula of a body-centered cubic a = 4r / 3 while a face-centered cubic can use the pythagorean theorem a^2 + b^2 = c^2 in the form of 2d^2=4r. Why are the radii (4r) the same while the distances are different? Why is it because of the shape of the unit cell?
Asked by Bob C - Tue Jan 5 12:57:59 2010 - - 1 Answers - 0 Comments
A. In a body centered cubic unit cell, the corners and the body center are the lattice points.So when you are trying to find the relation between r and a , you have to consider the two spheres which are diagonally opposite and the sphere at the center. Now when you form a right angled triangle, one side is diagonal of a face which is equal to 2 a and the other side is a and the hypotonuse is 4r.using pythagorous theorem you can solve for a which is a = 4r / 3 the distances are the same, the difference is in how you are forming the right angled triangle to apply the pythagorous theorem
Answered by halogen - Tue Jan 5 13:40:32 2010
Q. Why is the formula of a body-centered cubic a = 4r / 3 while a face-centered cubic can use the pythagorean theorem a^2 + b^2 = c^2 in the form of 2d^2=4r. Why are the radii (4r) the same while the distances are different? Why is it because of the shape of the unit cell?
Asked by Bob C - Tue Jan 5 12:57:59 2010 - - 1 Answers - 0 Comments
A. In a body centered cubic unit cell, the corners and the body center are the lattice points.So when you are trying to find the relation between r and a , you have to consider the two spheres which are diagonally opposite and the sphere at the center. Now when you form a right angled triangle, one side is diagonal of a face which is equal to 2 a and the other side is a and the hypotonuse is 4r.using pythagorous theorem you can solve for a which is a = 4r / 3 the distances are the same, the difference is in how you are forming the right angled triangle to apply the pythagorous theorem
Answered by halogen - Tue Jan 5 13:40:32 2010
Given the density of a face-centered cubic cell, what is the length of a unit cell edge? What is the radius?
Q. Silver metal crystallizes in a face-centered cubic unit cell. The density of the solid is 10.5 g/cm3. What is the length of a unit cell edge? (in pm) What is the radius of a silver atom? (in pm)
Asked by C - Mon Dec 8 20:25:25 2008 - - 1 Answers - 0 Comments
A. The edge length is 409 pm, and the radius is 144.6 pm.
Answered by N - Mon Dec 8 23:59:12 2008
Q. Silver metal crystallizes in a face-centered cubic unit cell. The density of the solid is 10.5 g/cm3. What is the length of a unit cell edge? (in pm) What is the radius of a silver atom? (in pm)
Asked by C - Mon Dec 8 20:25:25 2008 - - 1 Answers - 0 Comments
A. The edge length is 409 pm, and the radius is 144.6 pm.
Answered by N - Mon Dec 8 23:59:12 2008
How is the coordination number of a face centered cubic 12?
Q. Do we count the center atoms only? So is it because of 4 closest atoms on each plane. Therefore, 4 * 3 = 12?
Asked by xyzabc123 - Mon Apr 13 17:47:25 2009 - - 1 Answers - 1 Comments
A. the coordination # is the max # of "nearest neighbors" in the cube... ok...look at this. pic.. now look at the cell on the right. now look at the top.. front.. left.. atom... it's nearest neighbors are NOT the other corner atoms. The nearest neighbor atoms are the face atoms.. the top face.. the left face.. and the front face.. makes 3... next... imagine placing an identical cube on the top of this unit cell. of course we wouldn't need to add all 14 atoms because the top face of the original cell would actually be the bottom face of the new cell.. right? so we only add another 9 atoms... anyway the easiest way to see this is if your original top front left atom is now the bottom left front atom of that cell on the LEFT...ok? … [cont.]
Answered by m w - Tue Apr 14 00:28:14 2009
Q. Do we count the center atoms only? So is it because of 4 closest atoms on each plane. Therefore, 4 * 3 = 12?
Asked by xyzabc123 - Mon Apr 13 17:47:25 2009 - - 1 Answers - 1 Comments
A. the coordination # is the max # of "nearest neighbors" in the cube... ok...look at this. pic.. now look at the cell on the right. now look at the top.. front.. left.. atom... it's nearest neighbors are NOT the other corner atoms. The nearest neighbor atoms are the face atoms.. the top face.. the left face.. and the front face.. makes 3... next... imagine placing an identical cube on the top of this unit cell. of course we wouldn't need to add all 14 atoms because the top face of the original cell would actually be the bottom face of the new cell.. right? so we only add another 9 atoms... anyway the easiest way to see this is if your original top front left atom is now the bottom left front atom of that cell on the LEFT...ok? … [cont.]
Answered by m w - Tue Apr 14 00:28:14 2009
How many atoms are in one unit cell of a face-centered cubic lattice?
Q. I can't figure this chemistry question. Could someone help me? And if you could, briefly explain to me how you know the answer.
Asked by abc123321bca - Thu Dec 10 02:09:54 2009 - - 2 Answers - 0 Comments
A. 4 *** face centered cubic looks like this. 8 corner atoms. 6 face atoms. Each corner atom exists in 8 adjacent cells.. so 1/8 of each corner atom is in any particular unit cell. Each face atom exists in 2 adjacent cells. so 1/2 of each face atom is in any given unit cell. total atoms / cell = 8 x 1/8 + 6 x 1/2 = 4.. this model demonstrates that. so atoms / cell = 4 *** here's an example of a problem involving face centered cubic cells and calculating density.
Answered by m w - Thu Dec 10 02:18:07 2009
Q. I can't figure this chemistry question. Could someone help me? And if you could, briefly explain to me how you know the answer.
Asked by abc123321bca - Thu Dec 10 02:09:54 2009 - - 2 Answers - 0 Comments
A. 4 *** face centered cubic looks like this. 8 corner atoms. 6 face atoms. Each corner atom exists in 8 adjacent cells.. so 1/8 of each corner atom is in any particular unit cell. Each face atom exists in 2 adjacent cells. so 1/2 of each face atom is in any given unit cell. total atoms / cell = 8 x 1/8 + 6 x 1/2 = 4.. this model demonstrates that. so atoms / cell = 4 *** here's an example of a problem involving face centered cubic cells and calculating density.
Answered by m w - Thu Dec 10 02:18:07 2009
What is the coordination number of an aluminum atom in the face centered cubic structure of aluminum?
Q. What is the coordination number of an aluminum atom in the face centered cubic structure of aluminum?
Asked by Aaron - Mon Dec 7 19:56:05 2009 - - 1 Answers - 0 Comments
A. In close packing of spheres as in cubic close packing (fcc) and hexagonal close packing the coordination number of an individual sphere is 12. There are six nearest neighbors in a plane about the sphere and at the same distance three above and three below it. It is hard to see in the unit cell of fcc because to see the planes you have to tip the cell and make one corner just a single atom in the first plane, then the successive planes become apparent (My Kingdom for a blackboard!). It is easy to see in the unit cell of hcp, however.
Answered by sjbiis - Mon Dec 7 23:45:18 2009
Q. What is the coordination number of an aluminum atom in the face centered cubic structure of aluminum?
Asked by Aaron - Mon Dec 7 19:56:05 2009 - - 1 Answers - 0 Comments
A. In close packing of spheres as in cubic close packing (fcc) and hexagonal close packing the coordination number of an individual sphere is 12. There are six nearest neighbors in a plane about the sphere and at the same distance three above and three below it. It is hard to see in the unit cell of fcc because to see the planes you have to tip the cell and make one corner just a single atom in the first plane, then the successive planes become apparent (My Kingdom for a blackboard!). It is easy to see in the unit cell of hcp, however.
Answered by sjbiis - Mon Dec 7 23:45:18 2009
how many atoms does a face-centered cubic lattice cell have?
Q. i know an unit cell has 4, but how about a latitce?
Asked by Matic - Wed Dec 24 14:10:25 2008 - - 4 Answers - 0 Comments
A. a face centered cubic cell has 8 corner atoms. each corner atom is in 8 adjacent cells. So 1/8 of each corner atom is in a particular unit cell.. 8 x 1/8 = 1 atom from the corners. it also has 6 faces. each face has 1 atom centered on the face. Each face atom is in two adjacent cells. so 1/2 of each face atom is in the unit cell. 6 x 1/2 = 3 1 + 3 = 4 atoms / cell. *** as to a lattice.. That just what we call many adjacent unit cells that are electronically held together via "lattice energy"
Answered by m w - Wed Dec 24 18:54:29 2008
Q. i know an unit cell has 4, but how about a latitce?
Asked by Matic - Wed Dec 24 14:10:25 2008 - - 4 Answers - 0 Comments
A. a face centered cubic cell has 8 corner atoms. each corner atom is in 8 adjacent cells. So 1/8 of each corner atom is in a particular unit cell.. 8 x 1/8 = 1 atom from the corners. it also has 6 faces. each face has 1 atom centered on the face. Each face atom is in two adjacent cells. so 1/2 of each face atom is in the unit cell. 6 x 1/2 = 3 1 + 3 = 4 atoms / cell. *** as to a lattice.. That just what we call many adjacent unit cells that are electronically held together via "lattice energy"
Answered by m w - Wed Dec 24 18:54:29 2008
What are two ionic compounds for simple, body-centered and face-centered cubic lattices?
Q. Omg i cant find the answer to this question anywhere in the internet! its for regular chemistry and not even honors or ap and its so hard to find! big project grade and its actually due tomorrow!
Asked by jajajajaja - Wed Jan 16 00:24:20 2008 - - 3 Answers - 0 Comments
A. In chemistry, an ionic compound is a chemical compound in which ions are held together in a lattice structure by ionic bonds. Usually, the positively charged portion consists of metal cations and the negatively charged portion is a halogen or polyatomic ion. Ions in ionic compounds are held together by the electrostatic force between oppositely charged bodies. Ionic compounds have a high melting and boiling point, and they have a high hardness and are very brittle. Ions can be single atoms, as in common table salt sodium chloride, or more complex groups such as calcium carbonate. But to be considered an ion, they must carry a positive or negative charge. Thus, in an ionic bond, one 'bonder' must have a positive charge and the other a… [cont.]
Answered by emb_ s - Wed Jan 16 00:43:08 2008
Q. Omg i cant find the answer to this question anywhere in the internet! its for regular chemistry and not even honors or ap and its so hard to find! big project grade and its actually due tomorrow!
Asked by jajajajaja - Wed Jan 16 00:24:20 2008 - - 3 Answers - 0 Comments
A. In chemistry, an ionic compound is a chemical compound in which ions are held together in a lattice structure by ionic bonds. Usually, the positively charged portion consists of metal cations and the negatively charged portion is a halogen or polyatomic ion. Ions in ionic compounds are held together by the electrostatic force between oppositely charged bodies. Ionic compounds have a high melting and boiling point, and they have a high hardness and are very brittle. Ions can be single atoms, as in common table salt sodium chloride, or more complex groups such as calcium carbonate. But to be considered an ion, they must carry a positive or negative charge. Thus, in an ionic bond, one 'bonder' must have a positive charge and the other a… [cont.]
Answered by emb_ s - Wed Jan 16 00:43:08 2008
Gold crystallizes in a face-centered cubic structure.?
Q. Gold crystallizes in a face-centered cubic structure. What is the edge length of the unit cell if the atomic radius of silver is 144 pm?
Asked by Evan - Tue Mar 2 03:10:05 2010 - - 1 Answers - 0 Comments
A. so you have written ''Gold crystallizes in a face-centered cubic structure. What is the edge length of the unit cell if the atomic radius of silver is 144 pm?" ...so are you asking about gold or silver ?? anyways the relationship between edge length 'a' and atomic radius 'r' in a FCC structure is : r = a/[2X(square root of 2)] = a/2.828 = 0.354a given r = 144 pm so a = 144/0.354 = 406.779 pm see this link---
Answered by ashish sinha - Tue Mar 2 03:52:42 2010
Q. Gold crystallizes in a face-centered cubic structure. What is the edge length of the unit cell if the atomic radius of silver is 144 pm?
Asked by Evan - Tue Mar 2 03:10:05 2010 - - 1 Answers - 0 Comments
A. so you have written ''Gold crystallizes in a face-centered cubic structure. What is the edge length of the unit cell if the atomic radius of silver is 144 pm?" ...so are you asking about gold or silver ?? anyways the relationship between edge length 'a' and atomic radius 'r' in a FCC structure is : r = a/[2X(square root of 2)] = a/2.828 = 0.354a given r = 144 pm so a = 144/0.354 = 406.779 pm see this link---
Answered by ashish sinha - Tue Mar 2 03:52:42 2010
finding molar mass through crystallization (face centered cubic)?
Q. I know how to do this problem but the damn unit conversions are really tedious. Best answer to first correct answer You are given a small bar of an unknown metal X. You find the density of the metal to be 11.5 g/cm3. An X-ray diffraction experiment measures the edge of the face-centered cubic unit cell as 4.06 *10^-10 m. Find the molar mass of this metal, to help you identify it.
Asked by Kyle L - Thu Jul 22 21:22:26 2010 - - 1 Answers - 0 Comments
A. problem #7 "I know how to do this problem . . ." Then quit whining and do the damn problem. Working through the units is good for your brain. Forget the 10 points.
Answered by ChemTeam - Thu Jul 22 21:59:43 2010
Q. I know how to do this problem but the damn unit conversions are really tedious. Best answer to first correct answer You are given a small bar of an unknown metal X. You find the density of the metal to be 11.5 g/cm3. An X-ray diffraction experiment measures the edge of the face-centered cubic unit cell as 4.06 *10^-10 m. Find the molar mass of this metal, to help you identify it.
Asked by Kyle L - Thu Jul 22 21:22:26 2010 - - 1 Answers - 0 Comments
A. problem #7 "I know how to do this problem . . ." Then quit whining and do the damn problem. Working through the units is good for your brain. Forget the 10 points.
Answered by ChemTeam - Thu Jul 22 21:59:43 2010
Platinum has a face-centered cubic unit cell. An early calculation of the diameter of platinum was 2.75 angstr
Q. Platinum has a face-centered cubic unit cell. An early calculation of the diameter of platinum was 2.75 angstroms. Based on this what would be the density of platinum? (grams per centimeter cubed) The atomic weight of Platinum = 195.09 g/mol.
Asked by liquidfir33 - Wed Apr 9 13:28:02 2008 - - 2 Answers - 0 Comments
A. Let us consider 1.000 cm^3 platinum cube. It must contain: {(1x10^8)/(2.75)}^3 unit cells, and hence 4*{(1x10^8)/(2.75)}^3 platinum atoms. Divide this by 6.022x10^23 to get the number of moles, further time the atomic weight to get the weight of this platinum cube. Do you know how to get the density now?
Answered by Hahaha - Fri Apr 11 21:10:05 2008
Q. Platinum has a face-centered cubic unit cell. An early calculation of the diameter of platinum was 2.75 angstroms. Based on this what would be the density of platinum? (grams per centimeter cubed) The atomic weight of Platinum = 195.09 g/mol.
Asked by liquidfir33 - Wed Apr 9 13:28:02 2008 - - 2 Answers - 0 Comments
A. Let us consider 1.000 cm^3 platinum cube. It must contain: {(1x10^8)/(2.75)}^3 unit cells, and hence 4*{(1x10^8)/(2.75)}^3 platinum atoms. Divide this by 6.022x10^23 to get the number of moles, further time the atomic weight to get the weight of this platinum cube. Do you know how to get the density now?
Answered by Hahaha - Fri Apr 11 21:10:05 2008
A metal crystallizes in a face centered cubic structure and has a density of 11.9 g/cm3. If the radius of the?
Q. A metal crystallizes in a face centered cubic structure and has a density of 11.9 g/cm3. If the radius of the metal atom is 138 pm, what is the identity of the metal? A) At B) Pd C) Mn D) Fe E) Cr
Asked by Sarah - Fri Mar 26 15:51:03 2010 - - 1 Answers - 0 Comments
A. are you watching closely? (g / cm ) x (cm / cell) x (cell / atoms) x (atoms / mole) = (g/mole) do you see that? cm /cm cancels, cell / cell cancels.. atoms / atoms cancel leaving g/mole and if you knew g/mole, you could ID the element right? so do you go about finding those 4 values? *** g/cm *** this was given.. 11.9 g/cm *** cm / cell *** look at this picture... that is a face centered cubic cell. each sphere represents an atom. now look at this picture that is an image of a FCC cell. notice that the cell has PARTIAL atoms in it? that is because each corner atom exists in 8 different adjacent cells. so 1/8 of each corner atom is in the unit cell. Each face atom exists in 2 adjacent unit cells. so 1/2 of each face… [cont.]
Answered by m w - Fri Mar 26 16:39:16 2010
Q. A metal crystallizes in a face centered cubic structure and has a density of 11.9 g/cm3. If the radius of the metal atom is 138 pm, what is the identity of the metal? A) At B) Pd C) Mn D) Fe E) Cr
Asked by Sarah - Fri Mar 26 15:51:03 2010 - - 1 Answers - 0 Comments
A. are you watching closely? (g / cm ) x (cm / cell) x (cell / atoms) x (atoms / mole) = (g/mole) do you see that? cm /cm cancels, cell / cell cancels.. atoms / atoms cancel leaving g/mole and if you knew g/mole, you could ID the element right? so do you go about finding those 4 values? *** g/cm *** this was given.. 11.9 g/cm *** cm / cell *** look at this picture... that is a face centered cubic cell. each sphere represents an atom. now look at this picture that is an image of a FCC cell. notice that the cell has PARTIAL atoms in it? that is because each corner atom exists in 8 different adjacent cells. so 1/8 of each corner atom is in the unit cell. Each face atom exists in 2 adjacent unit cells. so 1/2 of each face… [cont.]
Answered by m w - Fri Mar 26 16:39:16 2010
Nickel has an face-centered cubic structure with an edge length of 352 pm . What is the density of this metal?
Q. Nickel has an face-centered cubic structure with an edge length of 352 pm . What is the density of this metal?
Asked by Sarah - Tue Jan 29 16:23:58 2008 - - 2 Answers - 0 Comments
A. if it is pure nickel it should always have a density of 8.91g/cm3 regardless of its shape or size... density is specific to the element.
Answered by deidrarw - Tue Jan 29 16:28:40 2008
Q. Nickel has an face-centered cubic structure with an edge length of 352 pm . What is the density of this metal?
Asked by Sarah - Tue Jan 29 16:23:58 2008 - - 2 Answers - 0 Comments
A. if it is pure nickel it should always have a density of 8.91g/cm3 regardless of its shape or size... density is specific to the element.
Answered by deidrarw - Tue Jan 29 16:28:40 2008
Krypton crystallizes with a face-centered cubic unit cell of edge 559 pm.?
Q. a) What is the density of solid krypton? b) What is the atomic radius of krypton? c) What is the volume of one krypton atom? d) What percentage of the unit cell is empty space if each atom is treated as a hard sphere? Thank You, I have no idea how to do this.
Asked by amar1214 - Sun Nov 16 21:40:48 2008 - - 1 Answers - 0 Comments
A. a) There are 14 atoms ( 8 at the corners and 6 at the face centers) assigned to a fcc unit cell. Since the atoms at the corners are shared between eight adjacent unit cells cells and the atoms at the face centers are shared you have net 4 atoms per unit cell. To calculate the density divide the mass of the atoms per cell by the cell volume: = 4 m / a = 4 83.798 amu / (559pm) = 4 83.798 1.66054 10 kg / (559 10 m) = 3376.58kg/m b) Consider the diagonal of a face of the cell. Since the atoms are in direct contact in the hemispherical plane, the length of the diagonal equals quadruple radius of the atom. From basic geometry you know that the length of diagonal in square of side length a is 2 a Hence: 2 a = 4 r => r = a / ( 2… [cont.]
Answered by schmiso - Mon Nov 17 16:59:51 2008
Q. a) What is the density of solid krypton? b) What is the atomic radius of krypton? c) What is the volume of one krypton atom? d) What percentage of the unit cell is empty space if each atom is treated as a hard sphere? Thank You, I have no idea how to do this.
Asked by amar1214 - Sun Nov 16 21:40:48 2008 - - 1 Answers - 0 Comments
A. a) There are 14 atoms ( 8 at the corners and 6 at the face centers) assigned to a fcc unit cell. Since the atoms at the corners are shared between eight adjacent unit cells cells and the atoms at the face centers are shared you have net 4 atoms per unit cell. To calculate the density divide the mass of the atoms per cell by the cell volume: = 4 m / a = 4 83.798 amu / (559pm) = 4 83.798 1.66054 10 kg / (559 10 m) = 3376.58kg/m b) Consider the diagonal of a face of the cell. Since the atoms are in direct contact in the hemispherical plane, the length of the diagonal equals quadruple radius of the atom. From basic geometry you know that the length of diagonal in square of side length a is 2 a Hence: 2 a = 4 r => r = a / ( 2… [cont.]
Answered by schmiso - Mon Nov 17 16:59:51 2008
how do I solve for the Atomic packing constant of Face-centered cubic structure?
Q. how do I solve for the Atomic packing constant of Face-centered cubic structure?
Asked by danilo g - Sat Jan 13 05:24:17 2007 - - 2 Answers - 0 Comments
A. for face centered cubic structure the number of atoms per unit is 4*(1/8) +4*(1/2) no. of atoms per unit cell=4*1/8+4*1/2 =2.5 atom /cell AND volume of one atom = (4/3)*pi*r^3 =5.584 r^3 volume of 2.5 atoms( no.of atoms per unit cell) =2.5*5.584* r^3 =13.96 * r^3 where r is the radius of the atom NEXT you should calculate the volume of the cell from geometry you will find that each face is a square with diagonal =4r that means thatyou can get the lenght of the square side (L) (4r)^2=L^2+L^2 2L^2=16r^2 L=(8^0.5 )*r L=2.828 r the volume of the cell = L^3 =(2.82)^3 * r^3 =22.627 r^3 atomic packing factor(or constant) = (total volume of… [cont.]
Answered by HuMaN being - Sat Jan 13 09:10:08 2007
Q. how do I solve for the Atomic packing constant of Face-centered cubic structure?
Asked by danilo g - Sat Jan 13 05:24:17 2007 - - 2 Answers - 0 Comments
A. for face centered cubic structure the number of atoms per unit is 4*(1/8)
Answered by HuMaN being - Sat Jan 13 09:10:08 2007
how do i calculate the lattice parameter in a face centered cubic?
Q. density = 11.35 crystal structure = FCC molar mass = 207.2
Asked by Kristen - Sat Sep 27 17:34:33 2008 - - 1 Answers - 0 Comments
A. calculate the number of atoms / cell multiply by mass / atom to get mass / cell multiply by volume / mass to get volume / cell assume it's a cube and convert volume cell to Length of a side.. ready? face centered cubic.. there is an atom at each corner that is in a total of 8 cells. so 1/8 of each corner atom is in a particular cell. there are 8 corners so 8 x 1/8 = 1 atom per cell from the corners. There are 6 sides. each side has an atom that exists in two cells.. so atom is 1/2 in each cell. 6 x 1/2 = 3 atoms for our unit cell... 1 + 3 = 4 atoms / cell...right? molar mass = 207.2 g/mole.. so mass of 1 atom can be found by ... 1 atom x (1 mole / 6.022x10^23 atoms) x (207.2 g / mole) = 3.441x10^-22 g still with me? 1 cell x (4… [cont.]
Answered by m w - Sat Sep 27 17:53:27 2008
Q. density = 11.35 crystal structure = FCC molar mass = 207.2
Asked by Kristen - Sat Sep 27 17:34:33 2008 - - 1 Answers - 0 Comments
A. calculate the number of atoms / cell multiply by mass / atom to get mass / cell multiply by volume / mass to get volume / cell assume it's a cube and convert volume cell to Length of a side.. ready? face centered cubic.. there is an atom at each corner that is in a total of 8 cells. so 1/8 of each corner atom is in a particular cell. there are 8 corners so 8 x 1/8 = 1 atom per cell from the corners. There are 6 sides. each side has an atom that exists in two cells.. so atom is 1/2 in each cell. 6 x 1/2 = 3 atoms for our unit cell... 1 + 3 = 4 atoms / cell...right? molar mass = 207.2 g/mole.. so mass of 1 atom can be found by ... 1 atom x (1 mole / 6.022x10^23 atoms) x (207.2 g / mole) = 3.441x10^-22 g still with me? 1 cell x (4… [cont.]
Answered by m w - Sat Sep 27 17:53:27 2008
Nickel has a face-centered cubic unit cell. The density of nickel is 6.84 grams per centimeter cubed. What is?
Q. the atomic radius, in angstroms, of nickel? Atomic weight of Nickel = 58.71 g/mol
Asked by henrick - Tue Nov 6 00:09:17 2007 - - 2 Answers - 0 Comments
A. 1.36
Answered by stars - Wed Nov 7 13:21:52 2007
Q. the atomic radius, in angstroms, of nickel? Atomic weight of Nickel = 58.71 g/mol
Asked by henrick - Tue Nov 6 00:09:17 2007 - - 2 Answers - 0 Comments
A. 1.36
Answered by stars - Wed Nov 7 13:21:52 2007
Nickel, Ni, has a face-centered cubic structure with an edge length of 352 pm. What is the density of Ni?
Q. Nickel, Ni, has a face-centered cubic structure with an edge length of 352 pm. What is the density of Ni?
Asked by Drew - Wed Apr 22 17:41:51 2009 - - 1 Answers - 0 Comments
A. density nickel 8902 kg/m .
Answered by billrussell42 - Wed Apr 22 18:26:18 2009
Q. Nickel, Ni, has a face-centered cubic structure with an edge length of 352 pm. What is the density of Ni?
Asked by Drew - Wed Apr 22 17:41:51 2009 - - 1 Answers - 0 Comments
A. density nickel 8902 kg/m .
Answered by billrussell42 - Wed Apr 22 18:26:18 2009
Is it possible for a salt of formula AB3 to have a face-centered cubic unit cell of anions & cations in all...?
Q. the 8 available holes??? Please explain.
Asked by zeezee - Wed Nov 26 15:12:52 2008 - - 1 Answers - 0 Comments
Q. the 8 available holes??? Please explain.
Asked by zeezee - Wed Nov 26 15:12:52 2008 - - 1 Answers - 0 Comments
From Yahoo Answer Search: 'Face-centered cubic'
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... transformations [ face-centered cubic (fcc)]/ [hexagonal close-packed (hcp)] and /' [body-centered cubic (bcc) or body-centered tetragonal (bct)], ...
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hu, 26 Mar 2009 06:34:33 GM
A diamond is a transparent China Ring wholesale crystal of tetrahedrally bonded carbon atoms (sp3) that crystallizes into the . face centered cubic. diamond lattice structure. Diamonds have been adapted for many uses because of the ...
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