The group on the periodic table that would have 0 electronegativity due to the fact that their valence shell is full, i.e, have a full octet would be the inert or noble gases. They have a total of 8 electrons in their valence shell and are thus inert and cannot strongly attract electrons toward itself, from neighbouring atom electrons as it does not need to.
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit.
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust. Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress.
I hope this helped! These are COMPLEX questions though! =D
Energy increases with increasing frequency.
8.38e -21Q^2 -1.07e -23Q^2 +3.15e +19
= 10.46e -44Q^2 + 19
44Q^2 = 10.46e + 19
Q^2 = 523/2200e + 19/44
Q1= ≈ -1.03828
Q2= ≈ 1.03828
Answer : The correct option is, (C) 1.7
Explanation :
First we have to calculate the moles of 
and 
.
The balanced chemical reaction will be:
0.01 mole of dissociate to give 0.01 mole of 
ion and 0.02 mole of 
ion
and
0.03 mole of dissociate to give 0.03 mole of 
ion and 0.03 mole of 
ion
That means,
0.02 moles of ion neutralize by 0.02 moles of ion.
The excess moles of ion = 0.03 - 0.02 = 0.01 mole
Total volume of solution = 100 + 300 = 400 ml = 0.4 L
Now we have to calculate the concentration of ion.
Now we have to calculate the pH of the solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
Therefore, the pH of the solution is, 1.7
