The reaction is second order in AB, so:
![v=k[AB]^2](https://tex.z-dn.net/?f=v%3Dk%5BAB%5D%5E2)
. In the statement, we obtain that
![[AB]=0.104~M](https://tex.z-dn.net/?f=%5BAB%5D%3D0.104~M)
and, at 25 ºC,
. Then:
![v=k[AB]^2\\\\ v=0.0164\cdot0.104^2\\\\ v=0.0164\cdot0.010816\\\\ v\approx0.000177=1.77\times10^{-4}~mol/s](https://tex.z-dn.net/?f=v%3Dk%5BAB%5D%5E2%5C%5C%5C%5C%0Av%3D0.0164%5Ccdot0.104%5E2%5C%5C%5C%5C%0Av%3D0.0164%5Ccdot0.010816%5C%5C%5C%5C%0Av%5Capprox0.000177%3D1.77%5Ctimes10%5E%7B-4%7D~mol%2Fs)
Now, we'll calculate the number of mols of the products in the gas. Using the Ideal Gas Law:
Since each AB molecule forms one of A and one of B,
. Hence:
.
We'll consider that in the beginning there was not A or B. So,
. Furthermore, since the ratio of AB to A and to B is 1:1,
.
Calculating the time by the expression of velocity:
![v=\dfrac{|\Delta[AB]|}{\Delta t}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_{AB}|}{V}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_A||}{250~mL}\\\\ 1.77\cdot10^{-4}=\dfrac{1}{\Delta t}\cdot\dfrac{0.0026~mol}{0.25~L}\\\\ \Delta t=\dfrac{0.0026}{0.25\cdot1.77\cdot10^{-4}}\\\\ \boxed{\Delta t\approx58.76~s}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B%7C%5CDelta%5BAB%5D%7C%7D%7B%5CDelta%20t%7D%3D%5Cdfrac%7B1%7D%7B%5CDelta%20t%7D%5Ccdot%5Cdfrac%7B%7C%5CDelta%20n_%7BAB%7D%7C%7D%7BV%7D%3D%5Cdfrac%7B1%7D%7B%5CDelta%20t%7D%5Ccdot%5Cdfrac%7B%7C%5CDelta%20n_A%7C%7C%7D%7B250~mL%7D%5C%5C%5C%5C%0A1.77%5Ccdot10%5E%7B-4%7D%3D%5Cdfrac%7B1%7D%7B%5CDelta%20t%7D%5Ccdot%5Cdfrac%7B0.0026~mol%7D%7B0.25~L%7D%5C%5C%5C%5C%0A%5CDelta%20t%3D%5Cdfrac%7B0.0026%7D%7B0.25%5Ccdot1.77%5Ccdot10%5E%7B-4%7D%7D%5C%5C%5C%5C%0A%5Cboxed%7B%5CDelta%20t%5Capprox58.76~s%7D)
Answer:
Decreasing the diameter of the channel by removing aa's. Since Na+ is chemically similar to K+, one can assume the difference must be due to the size of the atom. The K+ ion is larger than the Na+ ion, so reducing the diameter of the channel can allow Na+ to enter while preventing K+ entry. This explains clearly and perfectly how reducing the diameter reduces the
transport of K+ in favor of Na+ transport.
Answer is: 20 ions, left side.
Unbalanced half reaction: C₅O₅²⁻(g) → CO₃²⁻(aq).
1) There are 5 carbon atoms on the left side of half reaction and 1 carbon atom on right, so first add coefficient 5 in fron of carbon dioxide to balance carbon atoms: C₅O₅²⁻(g) → 5CO₃²⁻(aq).
2) Because there are more oxygen atoms on the right, add OH⁻ ions on the left side of half reaction and water on the right: OH⁻(aq) + C₅O₅²⁻(g) → 5CO₃²⁻(aq) + H₂O(l).
3) Balance oxygen (25 atoms on boths side) and hydrogen (20 atoms on both side of half reaction) atoms:
Balanced half reaction: 20OH⁻(aq) + C₅O₅²⁻(g) → 5CO₃²⁻(aq) + 10H₂O(l).
Answer:
according to me option a is the correct one
