Density of Platinum in face centered cubic structure?
Q. What is the density (in g/cm^3) of platinum if it crystallizes in the face centered cubic structure with an atomic radius of 139 pm?? 1pm= 1x10^(-12)m thanks =)
Asked by tangt.0902 - Fri Jan 23 18:11:16 2009 - - 1 Answers - 0 Comments
A. Use the formula: D = M/v where D = Density, M is mass, and V= Volume Since you are given the radius, you have to find the mass of a cube using the formula l*w*h Peace ^_^
Answered by Bawan - Tue Jan 27 18:01:06 2009
Q. What is the density (in g/cm^3) of platinum if it crystallizes in the face centered cubic structure with an atomic radius of 139 pm?? 1pm= 1x10^(-12)m thanks =)
Asked by tangt.0902 - Fri Jan 23 18:11:16 2009 - - 1 Answers - 0 Comments
A. Use the formula: D = M/v where D = Density, M is mass, and V= Volume Since you are given the radius, you have to find the mass of a cube using the formula l*w*h Peace ^_^
Answered by Bawan - Tue Jan 27 18:01:06 2009
The coordination number of an aluminum atom in the face-centered cubic structure of aluminum is?
Q. The coordination number of an aluminum atom in the face-centered cubic structure of aluminum is?
Asked by Jordan M - Fri Nov 30 16:13:07 2007 - - 1 Answers - 0 Comments
A. The answer is 12. Any closest packed metal structure, of which face-centered cubic is a type, involves a coordination number of 12.
Answered by mnrlboy - Fri Nov 30 16:48:02 2007
Q. The coordination number of an aluminum atom in the face-centered cubic structure of aluminum is?
Asked by Jordan M - Fri Nov 30 16:13:07 2007 - - 1 Answers - 0 Comments
A. The answer is 12. Any closest packed metal structure, of which face-centered cubic is a type, involves a coordination number of 12.
Answered by mnrlboy - Fri Nov 30 16:48:02 2007
in a body centered cubic structure (cesium chloride) do the atoms touch?
Q. i must recreate a model of this bcc structure and wanted to know if the atoms touch one another or should i simply represent the bonds with toothpicks
Asked by Shiva B - Sun Mar 2 23:53:31 2008 - - 2 Answers - 0 Comments
A. if its a solid they touch. hence being a solid. you can probably still use toothpicks though
Answered by Nick L - Mon Mar 3 00:00:25 2008
Q. i must recreate a model of this bcc structure and wanted to know if the atoms touch one another or should i simply represent the bonds with toothpicks
Asked by Shiva B - Sun Mar 2 23:53:31 2008 - - 2 Answers - 0 Comments
A. if its a solid they touch. hence being a solid. you can probably still use toothpicks though
Answered by Nick L - Mon Mar 3 00:00:25 2008
what is a face centered cubic structure?
Q. why is it called that?
Asked by bebe - Sun Mar 21 11:45:38 2010 - - 1 Answers - 0 Comments
A. think of a single dice. with all sides being "1"... not 1,2,3,4,5 and 6. A dot right in the middle. now think of an atom at each corner and one where the dot is. it will look something like this. and actually, those atoms exist in adjacent cells..so each unit cell looks like this. and from that pic, you can see that there are 8 corner atoms. and 1/8 of each corner atom is in the cell. And there are 6 face atoms and 1/2 of each face atom is in the cell.. so the total atoms per cell is 8x1/8 + 6x1/2 = 4
Answered by m w - Sun Mar 21 12:00:55 2010
Q. why is it called that?
Asked by bebe - Sun Mar 21 11:45:38 2010 - - 1 Answers - 0 Comments
A. think of a single dice. with all sides being "1"... not 1,2,3,4,5 and 6. A dot right in the middle. now think of an atom at each corner and one where the dot is. it will look something like this. and actually, those atoms exist in adjacent cells..so each unit cell looks like this. and from that pic, you can see that there are 8 corner atoms. and 1/8 of each corner atom is in the cell. And there are 6 face atoms and 1/2 of each face atom is in the cell.. so the total atoms per cell is 8x1/8 + 6x1/2 = 4
Answered by m w - Sun Mar 21 12:00:55 2010
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
What is the density of tungsten, and how do you draw the diagram it's body-centred cubic structure?
Q. Tungsten forms a body-centred cubic crystal. a. Draw a diagram of the crystal structure and indicate the (110) plane. b. The atomic radius for tungsten is 0.137nm and its atomic weight is 183.5g/mol. What is the density of tungsten?
Asked by John - Mon Mar 29 06:57:45 2010 - - 1 Answers - 0 Comments
A. b. 19.25 g cm 3
Answered by Manu - Mon Mar 29 16:02:42 2010
Q. Tungsten forms a body-centred cubic crystal. a. Draw a diagram of the crystal structure and indicate the (110) plane. b. The atomic radius for tungsten is 0.137nm and its atomic weight is 183.5g/mol. What is the density of tungsten?
Asked by John - Mon Mar 29 06:57:45 2010 - - 1 Answers - 0 Comments
A. b. 19.25 g cm 3
Answered by Manu - Mon Mar 29 16:02:42 2010
How many atoms per unit cell are there in: (a) a body centred cubic structure?
Q. (b) a face centred cubic structure? Help with crystal structure!
Asked by Tony_0 - Tue Mar 24 09:17:19 2009 - - 3 Answers - 0 Comments
A. a) A BCC (Body Centered Cubic) structure has 1(in the center)+1(1/8 is the contribution of each part on the corners)=2 atoms/unit cell b)A FCC(face centered cubic) structure has 3(11/2*6 for faces)+1(1/8 is the contribution of each part on the corner)=4 atoms/unit cell
Answered by Viral J - Tue Mar 24 09:27:00 2009
Q. (b) a face centred cubic structure? Help with crystal structure!
Asked by Tony_0 - Tue Mar 24 09:17:19 2009 - - 3 Answers - 0 Comments
A. a) A BCC (Body Centered Cubic) structure has 1(in the center)+1(1/8 is the contribution of each part on the corners)=2 atoms/unit cell b)A FCC(face centered cubic) structure has 3(11/2*6 for faces)+1(1/8 is the contribution of each part on the corner)=4 atoms/unit cell
Answered by Viral J - Tue Mar 24 09:27:00 2009
I have a Chemistry Problem about Body-Centered Cubic Unit Cells.?
Q. Barium has an atomic radius of 217pm and a body centered cubic crystal structure. a) Calculate the edge length of the unit cell in pm. b) Calculate the volume of the unit cell in cm^3. c) Calculate the density of barium using the above result. I'm completely confused on where to even start with this, I know what the body centered cubic looks like and everything I just don't know where to start with each problem.
Asked by Alyshia T - Wed Feb 3 19:46:47 2010 - - 1 Answers - 0 Comments
A. read my answer here. go to the section where I talk about volume / cell. read that section and follow the links. from there... 4r = 3 x (edge length) make sure you convert to pm... *** V = (edge length) *** density is one of those terms in ( ) I started the problem with. find the other 4, rearrange the equation.. and plug and chug. *** questions?
Answered by m w - Thu Feb 4 00:04:45 2010
Q. Barium has an atomic radius of 217pm and a body centered cubic crystal structure. a) Calculate the edge length of the unit cell in pm. b) Calculate the volume of the unit cell in cm^3. c) Calculate the density of barium using the above result. I'm completely confused on where to even start with this, I know what the body centered cubic looks like and everything I just don't know where to start with each problem.
Asked by Alyshia T - Wed Feb 3 19:46:47 2010 - - 1 Answers - 0 Comments
A. read my answer here. go to the section where I talk about volume / cell. read that section and follow the links. from there... 4r = 3 x (edge length) make sure you convert to pm... *** V = (edge length) *** density is one of those terms in ( ) I started the problem with. find the other 4, rearrange the equation.. and plug and chug. *** questions?
Answered by m w - Thu Feb 4 00:04:45 2010
Structure and Properties of Solids?
Q. Barium has a body-centered cubic structure. If the atomic radius of barium is 222pm, calculate the density of solid barium
Asked by Sara - Wed Nov 19 17:38:20 2008 - - 1 Answers - 0 Comments
A. a body centered cubic cell has 8 corners. one atom at each corner. and 1 atom in the center. each corner atom exists in 8 different cells. so 1/8 of each corner atom is in a particular cell. So total atoms per cell = 8 corner atoms x 1/8 each + 1 center atom = 2 atoms / cell you ok to here?.. the next part is a bit complicated...ready? the diagonal of the cube (the line drawn between opposite corners of the cube through the center) has length = 1/2 diameter of 1 corner atom + the diameter of the center atom + 1/2 diameter of the other corner atom. that is because each corner atom is partially in each cell. so the length of the cube's diagonal = 4 x the radius of a barium atom.. look at this pic. it may help anyway, via… [cont.]
Answered by m w - Wed Nov 19 18:12:29 2008
Q. Barium has a body-centered cubic structure. If the atomic radius of barium is 222pm, calculate the density of solid barium
Asked by Sara - Wed Nov 19 17:38:20 2008 - - 1 Answers - 0 Comments
A. a body centered cubic cell has 8 corners. one atom at each corner. and 1 atom in the center. each corner atom exists in 8 different cells. so 1/8 of each corner atom is in a particular cell. So total atoms per cell = 8 corner atoms x 1/8 each + 1 center atom = 2 atoms / cell you ok to here?.. the next part is a bit complicated...ready? the diagonal of the cube (the line drawn between opposite corners of the cube through the center) has length = 1/2 diameter of 1 corner atom + the diameter of the center atom + 1/2 diameter of the other corner atom. that is because each corner atom is partially in each cell. so the length of the cube's diagonal = 4 x the radius of a barium atom.. look at this pic. it may help anyway, via… [cont.]
Answered by m w - Wed Nov 19 18:12:29 2008
Calculating the density of a closest packed solid?
Q. Sodium has a body-centered cubic structure. If the atomic radius of sodium is 186 pm, calculate the density of solid sodium.
Asked by Chris Y - Thu Nov 20 03:55:52 2008 - - 1 Answers - 0 Comments
A. The bcc unit cell has eight atoms at the corners of the cubic cell and one atom right in the centre. However, for the 8 atoms at the corners, they are shared by more than one unit cell. Only 1/8 of each atom actually is in this unit cell. Total number of atoms in one unit cell = 1/8*8 + 1 = 2 Mass of one Na atom = 23/(6.022*10^23) = 3.82 x 10^-23 g Mass of one unit cell = Mass of 2 Na atoms = 7.64 x 10^-23 g So let's find out the volume of one unit cell. Looking along the diagonal of the cell, the length of the diagonal is r+2r+r = 4r. What's the length of the unit cell? By Pythagoras theorem, l^2 + l^2 + l^2 = diagonal^2. l = sqrt ((4r)^2/3) = 2.30 r = 4.28 10^-10 m Volume of unit cell = l^3 = 7.83 10^-29 m^3 Density = Mass/Volume = [cont.]
Answered by Fiona F - Thu Nov 20 04:17:16 2008
Q. Sodium has a body-centered cubic structure. If the atomic radius of sodium is 186 pm, calculate the density of solid sodium.
Asked by Chris Y - Thu Nov 20 03:55:52 2008 - - 1 Answers - 0 Comments
A. The bcc unit cell has eight atoms at the corners of the cubic cell and one atom right in the centre. However, for the 8 atoms at the corners, they are shared by more than one unit cell. Only 1/8 of each atom actually is in this unit cell. Total number of atoms in one unit cell = 1/8*8 + 1 = 2 Mass of one Na atom = 23/(6.022*10^23) = 3.82 x 10^-23 g Mass of one unit cell = Mass of 2 Na atoms = 7.64 x 10^-23 g So let's find out the volume of one unit cell. Looking along the diagonal of the cell, the length of the diagonal is r+2r+r = 4r. What's the length of the unit cell? By Pythagoras theorem, l^2 + l^2 + l^2 = diagonal^2. l = sqrt ((4r)^2/3) = 2.30 r = 4.28 10^-10 m Volume of unit cell = l^3 = 7.83 10^-29 m^3 Density = Mass/Volume = [cont.]
Answered by Fiona F - Thu Nov 20 04:17:16 2008
Ap Chemistry cubic cell question.?
Q. A salt, MY, crystallizes in a body-centered cubic structure with a Y- anion at each ocube corner an an M+ cation atthe cubecenter. Assuminf that Y- anions touch each otehr and the M+ cation at the center, and the radius of Y- is 1.50X102 pm the radius of m+ is a-62pm b-110pm c-124 pm d- 220 pm Please explain how to get to the answer.
Asked by ny_spinner_dan - Fri Jan 25 13:25:13 2008 - - 1 Answers - 0 Comments
A. If you draw this thing (all ions touching), you'll see that the length of each side of the cube is 2 x radius of anion (300pm total). Now you have to see how this can be compared to the radius of the cation. Notice that the only "simple" way is to find the length of the body-diagonal of the cube (corner to corner). The length of this diagonal is: D = sqrt(300^2 + 300^2 + 300^2) = 519.6pm This distance includes one full cation diameter and two anion radii (1 anion diameter), so subtract the anion radii from 519.6 to get the cation diameter: 519.6-300=219.6 pm. But they want the radius, so divide by 2: 109.8 or 110 pm (B).
Answered by anotherhumanmale - Fri Jan 25 13:41:59 2008
Q. A salt, MY, crystallizes in a body-centered cubic structure with a Y- anion at each ocube corner an an M+ cation atthe cubecenter. Assuminf that Y- anions touch each otehr and the M+ cation at the center, and the radius of Y- is 1.50X102 pm the radius of m+ is a-62pm b-110pm c-124 pm d- 220 pm Please explain how to get to the answer.
Asked by ny_spinner_dan - Fri Jan 25 13:25:13 2008 - - 1 Answers - 0 Comments
A. If you draw this thing (all ions touching), you'll see that the length of each side of the cube is 2 x radius of anion (300pm total). Now you have to see how this can be compared to the radius of the cation. Notice that the only "simple" way is to find the length of the body-diagonal of the cube (corner to corner). The length of this diagonal is: D = sqrt(300^2 + 300^2 + 300^2) = 519.6pm This distance includes one full cation diameter and two anion radii (1 anion diameter), so subtract the anion radii from 519.6 to get the cation diameter: 519.6-300=219.6 pm. But they want the radius, so divide by 2: 109.8 or 110 pm (B).
Answered by anotherhumanmale - Fri Jan 25 13:41:59 2008
what is the metalic crystal structure of lead?
Q. what is the metalic crystal structure of just pure lead? is it cubic body centered, cubic face centered, or hexagonal?
Asked by Claudine - Wed Jun 17 06:09:34 2009 - - 1 Answers - 0 Comments
A. Lead metal has a face centered cubic structure.
Answered by Colin - Wed Jun 17 07:16:57 2009
Q. what is the metalic crystal structure of just pure lead? is it cubic body centered, cubic face centered, or hexagonal?
Asked by Claudine - Wed Jun 17 06:09:34 2009 - - 1 Answers - 0 Comments
A. Lead metal has a face centered cubic structure.
Answered by Colin - Wed Jun 17 07:16:57 2009
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
how to calculate the unit cell length of these?
Q. Calculate the unit cell length in terms of atomic radius for: (a) a face centred cubic structure (b) a body centred cubic structure could you please show working or tell me which formula to use cheers
Asked by magic_ian - Thu Aug 20 04:16:24 2009 - - 1 Answers - 0 Comments
A. See the reference. It shows the idea. For the fcc, the diagonal of a face of the cube is r+2r+r=4r so the edge of the cube is 4r/ 2 = (2 2)r For the bcc the body diagonal of the cube is r+2r+r=4r so the edge of the cube is 4r/ 3 = (4 3 / 3)r
Answered by goober - Thu Aug 20 05:52:34 2009
Q. Calculate the unit cell length in terms of atomic radius for: (a) a face centred cubic structure (b) a body centred cubic structure could you please show working or tell me which formula to use cheers
Asked by magic_ian - Thu Aug 20 04:16:24 2009 - - 1 Answers - 0 Comments
A. See the reference. It shows the idea. For the fcc, the diagonal of a face of the cube is r+2r+r=4r so the edge of the cube is 4r/ 2 = (2 2)r For the bcc the body diagonal of the cube is r+2r+r=4r so the edge of the cube is 4r/ 3 = (4 3 / 3)r
Answered by goober - Thu Aug 20 05:52:34 2009
Cubic Structures Help! Will Rate for sure for the best answer?
Q. Part A Polonium is a radioactive metal that is used as a heat source in space satellites. Polonium has a simple cubic unit cell. How many atoms of Po are present in each unit cell? Part B Silver has a face-centered cubic unit cell. How many atoms of Ag are present in each unit cell? Part C Chromium has a body-centered cubic unit cell. How many atoms of Cr are present in each unit cell?
Asked by Jeeva - Wed May 21 20:51:34 2008 - - 1 Answers - 0 Comments
A. This is so hard without diagrams... But I'll try! A Draw a cube. Put a Polonium atom on each of the corners. The unit cell contains eight corners of the eight atoms. A "corner" of a sphere is one eighth... like the bottom front right bit. The content of the cell is (8 x 1/8) = 1. B Again draw a cube. Put a Silver atom on each corner and one on each of the six faces. The content is (8 x 1/8) + (6 x 1/2) = 4. C This time put one on each corner and one in the centre. The content is (8 x 1/8) + 1 = 2. Try it with models. A corner atom will be in eight different cubes. One on the face will be split in half with one in each cube. One in the centre is completely within that cell. I hope this is clear... Toodles!
Answered by boringalice - Wed May 21 21:16:40 2008
Q. Part A Polonium is a radioactive metal that is used as a heat source in space satellites. Polonium has a simple cubic unit cell. How many atoms of Po are present in each unit cell? Part B Silver has a face-centered cubic unit cell. How many atoms of Ag are present in each unit cell? Part C Chromium has a body-centered cubic unit cell. How many atoms of Cr are present in each unit cell?
Asked by Jeeva - Wed May 21 20:51:34 2008 - - 1 Answers - 0 Comments
A. This is so hard without diagrams... But I'll try! A Draw a cube. Put a Polonium atom on each of the corners. The unit cell contains eight corners of the eight atoms. A "corner" of a sphere is one eighth... like the bottom front right bit. The content of the cell is (8 x 1/8) = 1. B Again draw a cube. Put a Silver atom on each corner and one on each of the six faces. The content is (8 x 1/8) + (6 x 1/2) = 4. C This time put one on each corner and one in the centre. The content is (8 x 1/8) + 1 = 2. Try it with models. A corner atom will be in eight different cubes. One on the face will be split in half with one in each cube. One in the centre is completely within that cell. I hope this is clear... Toodles!
Answered by boringalice - Wed May 21 21:16:40 2008
Why does the atomic structure of copper allows it to be drawn into a wire.?
Q. With the aid of clear and annotated diagrams explain why the structure of copper allows it to be drawn into a wire, anyone can help please i found what the atomic structure of copper is but I do not understand how it works to make the material ductile the type is "face centered cubic" but how does it work to make it ductile? Hmm i do not understand what you mean, can you please be more specific
Asked by Aivaras S - Fri Nov 30 16:07:31 2007 - - 1 Answers - 1 Comments
A. fcc is most compact structure and has high energy . this structure tends to reduce its energy (thermodynamic principles) and this is because to be drawn even to a wire.
Answered by eshaghi_2006 - Fri Nov 30 16:48:13 2007
Q. With the aid of clear and annotated diagrams explain why the structure of copper allows it to be drawn into a wire, anyone can help please i found what the atomic structure of copper is but I do not understand how it works to make the material ductile the type is "face centered cubic" but how does it work to make it ductile? Hmm i do not understand what you mean, can you please be more specific
Asked by Aivaras S - Fri Nov 30 16:07:31 2007 - - 1 Answers - 1 Comments
A. fcc is most compact structure and has high energy . this structure tends to reduce its energy (thermodynamic principles) and this is because to be drawn even to a wire.
Answered by eshaghi_2006 - Fri Nov 30 16:48:13 2007
Why does the atomic structure of plain carbon steel used for structural work?
Q. I am not sure how to explain it and i think its got Body Centered Cubic formation? but how to explain why its good for structural work?
Asked by Aivaras S - Sun Dec 2 05:52:09 2007 - - 1 Answers - 0 Comments
A. A crystalline structure is good for a material that is tough because under stress the atoms in a metallic crystalline structure will easily change places allowing the material to deform or bend under a load without breaking. When carbon atoms are added they get in the way not allowing some of the atoms in the crystalline structure to move. This gives the metal more strength because it will not bend as much under stress due to the carbon atoms getting in the way. This also makes the metal more brittle, so choosing the carbon content to make the metal stronger, but not too brittle is important. Which is why low carbon steel is generally used in structural members (they will have decent strength but will be able to bend some so that when wind… [cont.]
Answered by Dan J - Sun Dec 2 21:30:10 2007
Q. I am not sure how to explain it and i think its got Body Centered Cubic formation? but how to explain why its good for structural work?
Asked by Aivaras S - Sun Dec 2 05:52:09 2007 - - 1 Answers - 0 Comments
A. A crystalline structure is good for a material that is tough because under stress the atoms in a metallic crystalline structure will easily change places allowing the material to deform or bend under a load without breaking. When carbon atoms are added they get in the way not allowing some of the atoms in the crystalline structure to move. This gives the metal more strength because it will not bend as much under stress due to the carbon atoms getting in the way. This also makes the metal more brittle, so choosing the carbon content to make the metal stronger, but not too brittle is important. Which is why low carbon steel is generally used in structural members (they will have decent strength but will be able to bend some so that when wind… [cont.]
Answered by Dan J - Sun Dec 2 21:30:10 2007
which of the following are true statements? a) A simple cubic unit cell contains a total of one atom b) A bo
Q. which of the following are true statements ? a) A simple cubic unit cell contains a total of one atom b) A body-centered cubic unit cell contains a total of one and a half atoms c) A face-centered cubic unit cell contains a total of three atoms d) In a hexagonal close-packed structure each sphere has twelve nearest neighbors there is more than one answer.
Asked by patti_xd - Mon Nov 5 22:57:53 2007 - - 3 Answers - 0 Comments
A. a d
Answered by stars - Tue Nov 6 21:37:23 2007
Q. which of the following are true statements ? a) A simple cubic unit cell contains a total of one atom b) A body-centered cubic unit cell contains a total of one and a half atoms c) A face-centered cubic unit cell contains a total of three atoms d) In a hexagonal close-packed structure each sphere has twelve nearest neighbors there is more than one answer.
Asked by patti_xd - Mon Nov 5 22:57:53 2007 - - 3 Answers - 0 Comments
A. a d
Answered by stars - Tue Nov 6 21:37:23 2007
What is the range number of atoms of 8 side by side connected unit cells in a metal with : (see details)?
Q. What is the range number of atoms of 8 side by side connected unit cells in a metal with : a) simple cubic structure b) face-centered-cubic structure give your explanations for the least/most shared and unshared atoms c)White also the number of octants for each structure
Asked by Warm Ice - Thu Mar 29 23:57:02 2007 - - 1 Answers - 0 Comments
A. (a) The range of atoms for the simple cubic crystal structure is a 2x2x2 formation with 27 atoms. (See Here all the atome are equally shared. (b) The range of atoms for the FCC crystal structure is a 2x2x2 formation with 63 atoms. (See Here the atoms at the corners are most shared, whereas the atoms at the center of each face is least shared.
Answered by PhysicsDude - Sat Mar 31 21:49:05 2007
Q. What is the range number of atoms of 8 side by side connected unit cells in a metal with : a) simple cubic structure b) face-centered-cubic structure give your explanations for the least/most shared and unshared atoms c)White also the number of octants for each structure
Asked by Warm Ice - Thu Mar 29 23:57:02 2007 - - 1 Answers - 0 Comments
A. (a) The range of atoms for the simple cubic crystal structure is a 2x2x2 formation with 27 atoms. (See Here all the atome are equally shared. (b) The range of atoms for the FCC crystal structure is a 2x2x2 formation with 63 atoms. (See Here the atoms at the corners are most shared, whereas the atoms at the center of each face is least shared.
Answered by PhysicsDude - Sat Mar 31 21:49:05 2007
Calculate the length of the edge of the conventional face-centred cubic unit cell of NaBr?
Q. -Calculate the length of the edge of the conventional face-centred cubic unit cell of NaBr -Calculate the density of solid NaBr. -Calculate the shortest distance (centre to centre) between two sodium cations in solid NaBr. Link is posted for the picture of the structure
Asked by Marie Denise S - Thu Sep 11 23:05:08 2008 - - 1 Answers - 0 Comments
A. I ( and many other persons ) can not access the linked web site ( web pages ) . But to make life easier, just stop considering this question as a chemistry question. Actually it is a mathematics question ( or to be more specific - a geometric question ) dealing with Touching Spheres - Diameter - Radius - Cubic - Triangle - Distance - etc. Once you change your point of view, these problems will be easier.
Answered by Real Chemist - Fri Sep 12 07:29:21 2008
Q. -Calculate the length of the edge of the conventional face-centred cubic unit cell of NaBr -Calculate the density of solid NaBr. -Calculate the shortest distance (centre to centre) between two sodium cations in solid NaBr. Link is posted for the picture of the structure
Asked by Marie Denise S - Thu Sep 11 23:05:08 2008 - - 1 Answers - 0 Comments
A. I ( and many other persons ) can not access the linked web site ( web pages ) . But to make life easier, just stop considering this question as a chemistry question. Actually it is a mathematics question ( or to be more specific - a geometric question ) dealing with Touching Spheres - Diameter - Radius - Cubic - Triangle - Distance - etc. Once you change your point of view, these problems will be easier.
Answered by Real Chemist - Fri Sep 12 07:29:21 2008
From Yahoo Answer Search: 'centered cubic structure'
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