<u>Answer:</u>
<u>For a:</u> The edge length of the unit cell is 314 pm
<u>For b:</u> The radius of the molybdenum atom is 135.9 pm
<u>Explanation:</u>
To calculate the edge length for given density of metal, we use the equation:

where,
= density = 
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal (molybdenum) = 95.94 g/mol
= Avogadro's number = 
a = edge length of unit cell =?
Putting values in above equation, we get:
![10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm](https://tex.z-dn.net/?f=10.28%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%20%28a%29%5E3%7D%5C%5C%5C%5Ca%5E3%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%2010.28%7D%3D3.099%5Ctimes%2010%5E%7B-23%7D%5C%5C%5C%5Ca%3D%5Csqrt%5B3%5D%7B3.099%5Ctimes%2010%5E%7B-23%7D%7D%3D3.14%5Ctimes%2010%5E%7B-8%7Dcm%3D314pm)
Conversion factor used:
Hence, the edge length of the unit cell is 314 pm
To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

where,
R = radius of the lattice = ?
a = edge length = 314 pm
Putting values in above equation, we get:

Hence, the radius of the molybdenum atom is 135.9 pm
Answer:
Coefficients are the numbers in front of the formulas.
Answer:
a) The equilibrium will shift in the right direction.
b) The new equilibrium concentrations after reestablishment of the equilibrium :
![[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M](https://tex.z-dn.net/?f=%5BSbCl_5%5D%3D%280.370-x%29%20M%3D%280.370-0.0233%29%20M%3D0.3467%20M)
![[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BSbCl_3%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)
![[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)
Explanation:

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
On increase in amount of reactant

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of
is increasing .So, the equilibrium will shift in the right direction.
b)

Concentration of
= 0.195 M
Concentration of
= 
Concentration of
= 
On adding more
to 0.370 M at equilibrium :

Initially
0.370 M
At equilibrium:
(0.370-x)M
The equilibrium constant of the reaction = 

The equilibrium expression is given as:
![K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSbCl_3%5D%5BCl_2%5D%7D%7B%5BSbCl_5%5D%7D)

On solving for x:
x = 0.0233 M
The new equilibrium concentrations after reestablishment of the equilibrium :
![[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M](https://tex.z-dn.net/?f=%5BSbCl_5%5D%3D%280.370-x%29%20M%3D%280.370-0.0233%29%20M%3D0.3467%20M)
![[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BSbCl_3%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)
![[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)
Answer:
Role is defined below
Explanation:
A small GTP-binding protein, is an important module of the signal transduction pathway used by growth factors to initiate cell growth and differentiation. Cellular activation with growth factors such as epidermal growth factor (EGF) induces Ras to move from an inactive state linked to GDP to an active state linked to GTP. In recent times, a mixture of genetic and biochemical studies has resulted in the elucidation of a signaling pathway that leads from growth factor receptors to Ras. After joining EGF, the EGF receptor tyrosine kinase is activated, which leads to receptor auto phosphorylation in multiple tyrosine residues. Signaling proteins with homology domains Src 2 (SH2) then bind to these phosphorylated residues in tyrosine, initiating multiple signaling cascades. Distinct of these SH2 area proteins, Grb2, exists in the cytoplasm in a preformed complex with a second protein, Son of Sevenless (Sos), which can catalyze the Ras GTP / GDP exchange. After stimulation of the growth factor, the phosphorylated EGF receptor with tyrosine binds to the Grb2 / Sos complex and translocates it to the plasma membrane. It is believed that this translocation brings Sos closer to Ras, which leads to the activation of Ras. In dissimilarity, the insulin receptor does not bind Grb2 directly, but rather induces the tyrosine phosphorylation of two proteins, the substrate-1 insulin receptor and Shc, which bind to the Grb2 / Sos complex. Once Ras is activated, a cascade of protein kinases that are important in a myriad of growth factor responses is stimulated.