Evidence could be gathered in the future that contradicts original theories. <span>Phenomena cannot be proven by conclusive evidence in science because, as of now, the evidence isn't conclusive. It is speculation. Just as a phenomena cannot be proven, it also cannot be disproven. </span>
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:
H₂O is the limiting reactant
Theoretical yield of 240 g Al₂O₃ and 14 g H₂
Explanation:
Find how many moles of one reactant is needed to completely react with the other.
6.5 mol Al × (3 mol H₂O / 2 mol Al) = 9.75 mol H₂O
We need 9.75 mol of H₂O to completely react with 6.5 mol of Al. But we only have 7.2 mol of H₂O. Therefore, H₂O is the limiting reactant.
Now find the theoretical yield:
7.2 mol H₂O × (1 mol Al₂O₃ / 3 mol H₂O) × (102 g Al₂O₃ / mol Al₂O₃) ≈ 240 g Al₂O₃
7.2 mol H₂O × (3 mol H₂ / 3 mol H₂O) × (2 g H₂ / mol H₂) ≈ 14 g H₂
Since the data was given to two significant figures, we must round our answer to two significant figures as well.
Answer: Neutrons vary from each other in their element
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