Answer is 0.289nm.
Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.
wt % of Fe in Fe-V alloy = 85%
wt % of V in Fe-V alloy = 15%
We need to calculate edge length of the unit cell having bcc structure.
Using density formula,
For calculating edge length,
For calculating 
, we use the formula
Similarly for calculating 
, we use the formula
From the periodic table, masses of the two elements can be written
Specific density of both the elements are
Putting and 
formula's in edge length formula, we get
![a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}](https://tex.z-dn.net/?f=a%3D%5Cleft%20%5B%5Cfrac%7BZ%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B%28wt%5C%25%29_%7BFe%7D%7D%7BM_%7BFe%7D%7D%2B%5Cfrac%7B%28wt%5C%25%29_%7BFe%7D%7D%7BM_%7BFe%7D%7D%7D%20%20%5Cright%20%29%7D%7BN_A%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B%28wt%5C%25%29_V%7D%7B%5Crho_V%7D%2B%5Cfrac%7B%28wt%5C%25%29_V%7D%7B%5Crho_V%7D%7D%20%20%5Cright%20%29%7D%20%20%5Cright%20%5D%5E%7B1%2F3%7D)
![a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}](https://tex.z-dn.net/?f=a%3D%5Cleft%20%5B%5Cfrac%7B2atoms%2F%5Ctext%7Bunit%20cell%7D%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B85%5C%25%7D%7B55.85g%2Fmol%7D%2B%5Cfrac%7B15%5C%25%7D%7B50.941g%2Fmol%7D%7D%20%20%5Cright%20%29%7D%7B%286.023%5Ctimes10%5E%7B23%7Datoms%2Fmol%29%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B85%5C%25%7D%7B7.874g%2Fcm%5E3%7D%2B%5Cfrac%7B15%5C%25%7D%7B6.10g%2Fcm%5E3%7D%7D%20%20%5Cright%20%29%7D%20%20%5Cright%20%5D%5E%7B1%2F3%7D)
By calculating, we get
When the reaction equation is:
HF ↔ H+ + F-
and when the Ka expression
= concentration of products/concentration of reactions
so, Ka = [H+][F-]/[HF]
when we assume:
[H+] = [F-] = X
and [HF] = 0.35 - X
So, by substitution:
6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X
∴ X = 0.015 M
∴[H+] = X = 0.015
when PH = -㏒[H+]
∴PH = -㏒0.015
= 1.8
Space -filling models are also known as Calotte models. These are 3 dimensional models that depict the spatial relationships between atoms.
The space-filling model of water molecule shows the 3 D structure of the water molecule. This model CLEARLY REVEALS THE OXYGEN ATOM WHICH IS LOCATED CENTRALLY WITH TWO HYDROGEN ATOMS IN THE ADJACENT SIDES. THE MODEL ALSO SUGGESTS HOW THE WATER MOLECULES ATTRACT EACH OTHER.
