Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Answer:
The final temperature will be "12.37°".
Explanation:
The given values are:
mass,
m = 0.125 kg
Initial temperature,
c = 22.0°C
Time,
Δt = 4.5 min
As we know,
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
Answer:
The scaling factor is 5.
Explanation:
Hello there!
In this case, since the scaling factor is defined as the ratio of the molar mass of the molecular formula (complete) to the empirical formula (simplified), it is possible to compute it for the empirical formula of CH2O whose molar mass is 30 g/mol (12+2+16) as shown below:
Therefore, we can also infer that the molecular formula would be:
Best regards!
Thermodynamics, Nuclear Physics, Quantum Physics, Astronomy and Astrophysics
