You would have to look for the mass of the sample and the volume of the sample.
balance by putting 2 next to HF
Double replacement i believe
Explanation:
Formula to calculate hybridization is as follows.
Hybridization = ![\frac{1}{2}[V+N-C+A]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5BV%2BN-C%2BA%5D)
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
So, hybridization of 
is as follows.
Hybridization = ![\frac{1}{2}[V + N - C + A]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5BV%20%2B%20N%20-%20C%20%2B%20A%5D)
= ![\frac{1}{2}[2 + 2]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B2%20%2B%202%5D)
= 2
Hybridization of is sp. Therefore, is a linear molecule. There will be only two electron groups through which Be is attached.
Similarly, hybridization of 
is calculated as follows.
Hybridization =
= ![\frac{1}{2}[8 + 2]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B8%20%2B%202%5D)
= 5
Therefore, hybridization of is ![sp^{3}d. Therefore, [tex]XeF_{2}](https://tex.z-dn.net/?f=sp%5E%7B3%7Dd.%20Therefore%2C%20%5Btex%5DXeF_%7B2%7D)
is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.
Thus, we can conclude that out of the given options is the correct examples of linear molecules for five electron groups.
<span>Scientists ignore the forces of attraction between particles in a gas under ordinary conditions</span><span> because the particles in a gas are apart and moving fast, rather than clustered and moving slow, therefore the forces of attraction are too weak to have a visible effect.</span>
