<span>Boyle's Law, because it describes what will happen in the relationship between the pressure and volume of the gas.</span>
We’ve all noticed that science has been accelerating and a very fast rate resulting in what has been called information overload and more recently filter failure. There are now more researchers add more papers than ever which has led to the importance of bibliometric measures Bibliometric as a Field is a fairly new discipline but it has seen in impressive growth in years due to computation and data storage. Journal metrics can play an important role for editors. There are many different metrics
Answer:
The magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.
Explanation:
Given;
Radius of circular loop, R = 3.00 cm = 0.03 m
Current in the loop, I = 12.0 A
Magnetic field at the center of circular loop is given as;
B = μ₀I / 2R
Where;
μ₀ is constant = 4π x 10⁻⁷ T.m/A
R is the radius of the circular loop
I is the current in the loop
Substitute the given values in the above equation and calculate the magnitude of the magnetic field;
B = (4π x 10⁻⁷ x 12)/ 0.03
B = 5.0272 x 10⁻⁴ T
Therefore, the magnitude of the magnetic field B at the center of the loop is 5.0272 x 10⁻⁴ T.
Answer : The cell potential for this cell 0.434 V
Solution :
The balanced cell reaction will be,
Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
![E^o_{[Cu^{2+}/Cu]}=0.34V](https://tex.z-dn.net/?f=E%5Eo_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D0.34V)
![E^o_{[Ag^{+}/Ag]}=0.80V](https://tex.z-dn.net/?f=E%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D0.80V)
![E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5Eo%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)
Now we have to calculate the concentration of cell potential for this cell.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCu%5E%7B2%2B%7D%5D%5BAg%5D%5E2%7D%7B%5BCu%5D%5BAg%5E%2B%5D%5E2%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this cell 0.434 V
