Answers are:
1) The balanced oxidation half reaction: 2I⁻(aq) → I₂(s) + 2e⁻.
Iodine is oxidized (lost electrons) from -1 to neutral charge (0).
2) The balanced reduction half-reaction: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻.
Hydrogen is reduced (gain electrons) from +1 to neutral charge.
3) The oxidation <span>reaction takes place at the anode.</span>
Answer:
0.32M
Explanation:
<u>Step 1:</u> Balance the reaction
K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3
We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution
SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3
<u>Step 2: </u>Calculate concentration
To find the concentration of the barium cation we use the following equation:
Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>
<u />
<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3
[Ba2+] = 0.32 M
The concentration of Barium ion in solution is 0.32 M
Try putting this in biology category not chemistry so more ppl can help :)
Answer:
Yes, Pb3(PO4)2.
Explanation:
Hello there!
In this case, according to the given balanced chemical reaction, it is possible to use the attached solubility series, it is possible to see that NaNO3 is soluble for the Na^+ and NO3^- ions intercept but insoluble for the Pb^3+ and PO4^2- when intercepting these two. In such a way, we infer that such reaction forms a precipitate of Pb3(PO4)2, lead (II) phosphate.
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