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
No the only thing that affects it is how it is built
False.
Theories are hypotheses that have been shown to accurately and predictably explain results obtained through repeated experimentation, to the point where the hypothesis can be assumed to be true. Then, such a hypothesis would be considered a theory.
These objects would be classified as extreme trans Neptunian object (ETNO).
Explanation:
ETNO’s are the objects lying beyond the planet Neptune and orbiting the Sun. They follow a highly eccentric path which is tilted. ETNO has been grouped into three major according to their respective perihelia.
Within this region (beyond Neptune’s orbit), a hypothetical planet has been discovered. It was discovered following its gravitational effect on the other objects of Kuiper Belt (region beyond the orbit of the Neptune- the last planet of our Solar system)
The Planet is assumed to be around 2 times the Earth’s size and around 10 times heavier than Earth.
<span>To find the acceleration we are given two facts to begin. The impact at 16 km/h and the dent of 6.4 cm, or 0.064 meters. In solving the problem uniform acceleration is assumed, which would mean the avg speed during the impact was 8 km/hr by taking 16/2. We know distance = rate*time (d=r*t) . So t = d / r, so 0.64/8 = 0.008hr for t. Now we can solve for acceleration by taking a = 16 / 0.008 = 2000 km/hr.</span>