5 moles because molarity signifies number of moles dissolved in One litre of water
The rate law depicts the effect of concentration on reaction rate. Second mechanism 2NO(g) ⇄ N₂O₂(g) [fast], N₂O₂(g) + O₂(g) → 2NO₂(g) [slow] is most reasonable. Thus, option b is correct.
<h3>What is rate law?</h3>
Rate law and equation give the rate at which the reaction takes place under the influence of the concentration of the reactants. The balanced chemical reaction is given as,
2NO(g) + O₂(g) → 2NO₂(g)
The rate of the equation is given as,
rate = k [NO]² [O₂]
In a multi-step chemical reaction, the slowest step is the rate-determining step. The second mechanism is given as,
2NO (g) → N₂O₂ (g) [fast]
N₂O₂(g) +O₂(g) → 2NO₂ (g) [slow]
Rate is given as,
rate = k [N₂O₂] [O₂]
Therefore, option b. the second mechanism is the most reasonable.
Learn more about rate law, here:
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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
1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)
Ksp{Zn(OH)₂}=1.2*10⁻¹⁷
2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)
Ksp{Zn(CN)₂}=2.6*10⁻¹³
Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}
Zn(OH)₂ precipitates first
