Answer:
ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq)
Explanation:
The reactions are:
(1) ClO⁻(aq) + H₂O(l) → HClO(aq) + OH⁻(aq) [fast]
(2) I⁻(aq) + HClO(aq) → HIO(aq) + Cl⁻(aq) [slow]
(3) OH⁻(aq) + HIO(aq) → H₂O(l) + IO⁻(aq) [fast]
By adding up the 3 equations we get:
ClO⁻(aq) + <em>H₂O(l)</em> + I⁻(aq) + <em>HClO(aq)</em> + <em>OH⁻(aq)</em> + <em>HIO(aq)</em> → <em>HClO(aq)</em> + <em>OH⁻(aq)</em> + <em>HIO(aq)</em> + Cl⁻(aq) + <em>H₂O(l)</em> + IO⁻(aq)
And by canceling common terms on both sides, we can get the overall equation:
ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq)
Therefore, the overall equation is ClO⁻(aq) + I⁻(aq) → Cl⁻(aq) + IO⁻(aq).
I hope it helps you!
Answer:
0.57 M
Explanation:
rate = change in concentration /time
Initial concentration of Cl2O5 = 1.16 M
Let the concentration of Cl2O5 after 5.70 seconds be y
rate = (1.16 - y)/5.7
The reaction follows a first order
Therefore, rate = ky = 0.184y
0.184y = (1.16 - y)/5.7
0.184y × 5.7 = 1.16 - y
1.0488y + y = 1.16
2.0488y = 1.16
y = 1.16/2.0488 = 0.57 M
Concentration of Cl2O5 after 5.70 seconds is 0.57 M
Answer:
0.052636002587839 moles
Explanation:
im not sure so sorry if u get it wrong...
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PH + pOH = 14
pH = 14 - pOH = 14 - 2 = 12
A pOH of 2 is equivalent to a pH of 12. A pH greater than 7 is basic, and 12 is greater than 7. Therefore, a solution with a pOH of 2 is basic.
Thus, the correct answer choice is B.