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7 Jun 2010: NBB = z1. 2Nx2. NAB NBA = zN(1 x)x. (2). where NAB represents both A–B and B–A bonds which cannot be. ... B–B bonds 12zNax2 and A–B B–A bonds zNa(1 x)x where z.

7 Jun 2010: 2). where NAB. represents both A–B and B–A bonds which cannot be dis-. ... a(1 x)2,. B–B bonds 12zN. ax2 and A–B B–A bonds zN.

7 Jun 2010: where NAB represents both A–B and B–A bonds which cannot be distinguished. ... 2, B–B bonds 12 zNax2 and A–B bonds zNa(1 x)x. where z is the co–ordination number.

7 Jun 2010: energy as an AA and a BB bond is broken to create 2(AB)bonds. ... energy in the process of forming AB bonds. The activity is equalto the mole fraction in an ideal solution (Fig.

7 Jun 2010: bonds. An ideal solution is formed when there is no change in energy in. ... the process of forming A B bonds. The activity is equal to the mole.

7 Jun 2010: B–B bonds respectively, and the third and fourth terms the numbers of A–B and B–A. ... bonds may then be obtained by setting ln{}/NAB = 0 (Christian, 1975; Lupis,.

7 Jun 2010: AB BA 1N N zNx x+ = where NAB represents both A–B and B–A bonds which cannot be distinguished. ... N is the total number of atoms. Notice that the unlike bonds A–B and B–A cannot be distinguished and hence have been consolidated.

7 Jun 2010: The number of B-B bonds is similarly 212 AzN x , and number of A-B B-A bonds is A (1 )zN x x. ... the factor of two in the last term is to count A-B and B-A bonds.

7 Jun 2010: bonds.4. In ordered crystals, A atoms prefer to be next to B atoms: the enthalpy of ordering HO. ... finding a B atom andx(1 x) is that of finding an A atom next to a B atom.

7 Jun 2010: b) Explain how the number of allowed electrons in a band can be. ... solution. [20%]. Calculate the number of A-A, B-B, and A-B B-A bonds in a random mixture of A and B atoms, with the total number of atoms