A chemist adds 370.0mL of a 1.41/molL potassium iodide KI solution to a reaction flask. Calculate the millimoles of potassium io
1 answer:
Answer: The millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.
Explanation:
Given: Volume of KI = 370.0 mL (1 mL = 0.001 L) = 0.37 L
Molarity of KI solution = 1.41 mol/L
Now, moles of KI (potassium iodide) is calculated as follows.
Convert moles into millimoles as follows.
1 mol = 1000 millimoles
0.5217 mol =
This can be rounded off to the value 522 millimoles.
Thus, we can conclude that the millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.
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Answer:
a) [H₃O⁺] = 1.8x10⁻⁵ M
b) pH = 4.75
c) % rxn = 3.5x10⁻³ %
Explanation:
a) The dissociation reaction of HCN is:
HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)
0.5 M - x x x
The dissociation constant from the above reactions is given by:
By solving the above quadratic equation we have:
x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]
Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.
b) The pH is equal to:
Then, the pH of the HCN solution is 4.75.
c) The % reaction is the % ionization:
Therefore, the % reaction or % ionization is 3.5x10⁻³ %.
I hope it helps you!