Shweta thakur Asked a Question
July 7, 2020 7:18 pmpts 30 pts
=6.04 x 103 Ex. 10. The density f a particular crystal of LiF is 2.65 g/cc. X-ray analysis show length of gold atoms that Li' and F 1onS are arranged in a cubic array at a spacing of 2.01 A. From these data calculate the apparent Avogadro costant. Solution: See Example 29, Chapter 1.
  • 3 Answer(s)
  • Shares
  • Priyanshu kumar Best Answer correlate this with this vieeo explanation
    Likes(0) Reply(0)
  • Priyanshu kumar thankyou
    see this
    • cropped5909672459007468679.jpg
    Likes(0) Reply(2)
    Priyanshu kumar
    got shweta here lif is placed on the edge length at a gap of 2.01 A ...
  • Priyanshu kumar thankyou
    see as there is spacing between the edge length of cube is not same as the LiF
    Likes(0) Reply(0)
  • Ashutosh singh
    No, Shweta.
    Likes(0) Reply(9)
    Ashutosh singh
    Some questions just want to understand your depth of knowledge about it..rather than being formula based..we use formula for atoms, not molecules... please note it down! That densi...
    Show more
  • Ashutosh singh
    Hey Shweta, refer this!
    • cropped2121167341.jpg
    • cropped1099690637.jpg
    Likes(0) Reply(2)
    Shweta thakur
    ok sir but sir here can't we use density direct formula