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weeeeeb [17]
3 years ago
6

Why does the solution decolorize on standing after the equivalence point has been reached?

Chemistry
2 answers:
Arisa [49]3 years ago
8 0

The equivalence point of a titration is equal to its stoichiometric equivalents of analyte and titrant.

Depending on the concentration of titrant we could be adding little excess of it and this may result in persistence of color of solution. After continuous stirring for a while the excess titrant may react with dissolved CO₂ in air and thus decolorizing the solution.

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Colt1911 [192]3 years ago
3 0

Answer:

The potassium permanganate slowly fades in color as it is dissolved in the water.

Explanation:

Hello,

Perhaps, your talking about a titration that was carried out by using potassium permanganate as the titrant, thus, bringing to bear the fact that the equivalence point of a titration is equal to its stoichiometric equivalents of analyte and titrant, one could state that as the titration is carried out, the potassium permanganate slowly fades in color as it is dissolved in the water after the equivalence point is reached since the formed species become colorless as well.

Best regards.

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How many liters of oxygen gas, at standard
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Explanation:

  • For the balanced reaction:

<em>4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)​.</em>

It is clear that 4 mol of Fe react with 3 mol of O₂ to produce 2 mol of Fe₂O₃.

  • Firstly, we need to calculate the no. of moles of 35.8 grams of Fe metal:

no. of moles of Fe = mass/molar mass = (35.8 g)/(55.845 g/mol) = 0.64 mol.

  • Now, we can find the no. of moles of O₂ is needed to react with the proposed amount of Fe:

<em><u>Using cross multiplication:</u></em>

4 mol of Fe is needed to react with → 3 mol of O₂, from stichiometry.

0.64 mol of Fe is needed to react with → ??? mol of O₂.

∴ The no. of moles of O₂ needed = (3 mol)(0.64 mol)/(4 mol) = 0.48 mol.

  • Finally, we can get the volume of oxygen using the information:

<em>It is known that 1 mole of any gas occupies 22.4 L at standard P and T (STP).</em>

<em></em>

<em><u>Using cross multiplication:</u></em>

1 mol of O₂ occupies → 22.4 L, at STP conditions.

0.48 mol of O₂ occupies → ??? L.

∴ The no. of liters of O₂ = (0.48 mol)(22.4 L)/(1 mol) = 10.752 L.

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