Copper would be the last to melt because it has the highest melting point!
Chemical potential energy: chemical potential of a species is energy that can be absorbed or released due to a change of the particle number of the given species, in a chemical reaction or phase transition
Gasoline used as kinetic energy: the various chemicals that make up gasoline contain a large amount of chemical potential energy that is released when the gasoline is burned in a controlled way in the engine of the car. The release of that energy does two things. Some of the potential energy is transformed into work, which is used to move the car
Dynamite used as kinetic energy: the dynamite being used was most likely made of nitroglycerin. Once the dynamite explodes from a percussion force (then breaking of weak bonds to releasing the raw atom) the energy is then converted to thermal, kinetic, and sound energy.
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
