Answer : The standard cell potential of the reaction is, -1.46 V
Explanation :
The given balanced cell reaction is,
Here, chromium (Cr) undergoes oxidation by loss of electrons and act as an anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
The standard values of cell potentials are:
Standard reduction potential of lead ![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)
Standard reduction potential of chromium ![E^0_{[Cr^{3+}/Cr]}=1.33V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D%3D1.33V)
Now we have to calculate the standard cell potential for the following reaction.
![E^0=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)
Therefore, the standard cell potential of the reaction is, -1.46 V
Answer:
The answer to your question is V2 = 434.7 l
Explanation:
Data
Volume 1 = V1 = 240 l Volume 2 = ?
Temperature 1 = T1 = 479°K Temperature 2 = T2 = 293°K
Pressure 1 = P1 = 300 KPa Pressure 2 = P2 = 101.325 Kpa
Process
1.- Use the combined gas law to solve this problem
P1V1/T1 = P2V2/t2
-Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (300)(240)(293) / (479)(101.325)
3.- Simplification
V2 = 21096000 / 48534.675
4.- Result
V2 = 434.7 l
Answer:
Rutherford overturned Thomson's model in 1911 with his well-known gold foil experiment, in which he demonstrated that the atom has a tiny, high- mass nucleus. In his experiment, Rutherford observed that many alpha particles were deflected at small angles while others were reflected back to the alpha source.
D) 59 (protons+neutrons=mass)
The most important thing to determine whether a change is physical or chemical is whether the substance was changed. When the substance being changed, it is chemical change. Otherwise, it is physical.
