Explanation:
It is known that for a body centered cubic unit cell there are 2 atoms per unit cell.
This means that volume occupied by 2 atoms is equal to volume of the unit cell.
So, according to the volume density
![5 \times 10^{26} atoms = 1 [tex]m^{3}](https://tex.z-dn.net/?f=5%20%5Ctimes%2010%5E%7B26%7D%20atoms%20%3D%201%20%5Btex%5Dm%5E%7B3%7D)
2 atoms = 
= 
Formula for volume of a cube is 
. Therefore,
Volume of the cube =
As lattice constant (a) = 
= 
Therefore, the value of lattice constant is .
And, for bcc unit cell the value of radius is as follows.
r = 
Hence, effective radius of the atom is calculated as follows.
r =
= 
= 
Hence, the value of effective radius of the atom is .
Answer:
Carbon dioxide is a product (cellular respiration)
Carbon dioxide is a reactant(photosynthesis)
Carried out in animals(cellular respiration)
Carried out in plants(both)
Chemical reaction(cellular respiration)
Oxygen is a product(photosynthesis)
Oxygen is a reactant(cellular respiration)
Produces usable energy source(photosynthesis)
Your welcome!!! Plz mark brainliest.
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
