11.7 g hope this helps and have a great day
Electronegativity is the property of the element to attract electrons to its nucleus. The trend of electronegativity in the periodic table is decreasing from right to left and decreasing from top to bottom. hence from the given elements, the element with the highest electronegativity should be A.beryllium
Answer : The correct option is, +91 kJ/mole
Solution :
The balanced cell reaction will be,

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)

![E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5E0_%7Bcell%7D%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)

Now we have to calculate the standard Gibbs free energy.
Formula used :

where,
= standard Gibbs free energy = ?
n = number of electrons = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = -0.47 V
Now put all the given values in this formula, we get the Gibbs free energy.

Therefore, the standard Gibbs free energy is +91 kJ/mole
Answer : The lewis dot structure includes the lone pair of electrons in any element and is helpful for defining the bond formation using the electrons.
In the molecule of HOI hydrogen is to the left of oxygen; oxygen is in middle and Iodine is at right of oxygen.
The picture is attached for better understanding.
Answer:
The independent variable is graphed on the x-axis. The dependent variable, which changes in response to the independent variable, is graphed on the y-axis. Controlled variables are usually not graphed because they should not change.