Question

The compound cscl has zns structure and its length of the unit cell 500 cm calculate its density

Answer 1

Explanation:

Since ZnS has a face-centered cubic structure,  it contains 4 sphere

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4.

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volume

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxm

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cm

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cmM = 63+35.5 = 98.5

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cmM = 63+35.5 = 98.5Na= 6.023×10^23

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cmM = 63+35.5 = 98.5Na= 6.023×10^23the density of CuCl is as follows

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cmM = 63+35.5 = 98.5Na= 6.023×10^23the density of CuCl is as follows =[{(63+35.5)×4}÷{(6.023×10^23)×(5×10^-8)³}] g/cc

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cmM = 63+35.5 = 98.5Na= 6.023×10^23the density of CuCl is as follows =[{(63+35.5)×4}÷{(6.023×10^23)×(5×10^-8)³}] g/cc=[394÷(752.875x10^-1)]

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cmM = 63+35.5 = 98.5Na= 6.023×10^23the density of CuCl is as follows =[{(63+35.5)×4}÷{(6.023×10^23)×(5×10^-8)³}] g/cc=[394÷(752.875x10^-1)]=0.523327x10 g/cc

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cmM = 63+35.5 = 98.5Na= 6.023×10^23the density of CuCl is as follows =[{(63+35.5)×4}÷{(6.023×10^23)×(5×10^-8)³}] g/cc=[394÷(752.875x10^-1)]=0.523327x10 g/cc=5.23 g/cc

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cmM = 63+35.5 = 98.5Na= 6.023×10^23the density of CuCl is as follows =[{(63+35.5)×4}÷{(6.023×10^23)×(5×10^-8)³}] g/cc=[394÷(752.875x10^-1)]=0.523327x10 g/cc=5.23 g/ccIf u still have any doubt please ask in comment section or just DM

Since ZnS has a face-centered cubic structure,  it contains 4 sphere so Z=(6x(1/2))+(8x(1/8)) = 4. Density id given by = mass/volumemass=zxmvolume=Na x a³So, density=(z x m)÷(Na x a³)where a=500 pm = 5x10^-8 cmM = 63+35.5 = 98.5Na= 6.023×10^23the density of CuCl is as follows =[{(63+35.5)×4}÷{(6.023×10^23)×(5×10^-8)³}] g/cc=[394÷(752.875x10^-1)]=0.523327x10 g/cc=5.23 g/ccIf u still have any doubt please ask in comment section or just DMRead more on Brainly.in - https://brainly.in/question/3646637#readmore