An alloy maybe could be used as a solid but i really don't think that would be it so i would sayether from an alloy or dirt
Its about 11.5hg because if you divide it with the atom it would result to 11.5hg
Answer:

Explanation:
Hello.
In this case, given the formula:

Whereas E is the energy, h the Planck's constant and u the frequency of the photon. Thus, solving for it, we obtain:

Or also:

Best regards.
Answer: The equilibrium constant for the overall reaction is 
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
a) 
![K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
b) 
![K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
For overall reaction on adding a and b we get c
c) 
![K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5E%5Cfrac%7B5%7D%7B2%7D%7D)
![K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_c%3DK_a%5Ctimes%20K_b%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%5Ctimes%20%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
The equilibrium constant for the overall reaction is 