<span>Answer:
H-C-N H-N-C C-H-N
Notice that C-H-N is the same as N-H-C just written backwards. ( i.e. they have the same connectivtiy.) You can exclude the last one with H in the middle since H has two bonds and 4 electrons around it. At this point you couldn't differentiate between the first two, so I would give you the connectivity in such a problem, which in this case is H-C-N.</span>
Answer:
0. 414
Explanation:
Octahedral interstitial lattice sites.
Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.
The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.
The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.
A chemical structure of a molecule includes the arrangement of atoms and the chemical bonds that hold the atoms together. The 2-HEPTANONE molecule contains a total of 21 bond(s) There are 7 non-H bond(s), 1 multiple bond(s), 4 rotatable bond(s), 1 double bond(s) and 1 ketone(s) (aliphatic).
There are 2 sig figs in 0.092
