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2.68×10^-3 moles of a solution containing an ion A^n acquire 1.61×10^-3 moles of MnO4^- for the oxidation of A^ n to AO3 ^- in acidic medium.what is the value of n ? A)3 B)2 C)5 D)4?
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2.68×10^-3 moles of a solution containing an ion A^n acquire 1.61×10^...
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2.68×10^-3 moles of a solution containing an ion A^n acquire 1.61×10^...
To determine the value of n in the equation A^n + MnO4^- -> AO3^- in acidic medium, we need to use the stoichiometry of the reaction.

Let's start by determining the number of moles of MnO4^- required to oxidize A^n to AO3^- based on the given information:

Moles of MnO4^- = 1.61×10^-3 moles

Next, we need to find the molar ratio between MnO4^- and A^n. This can be determined by balancing the oxidation half-reaction, which involves the transfer of electrons:

MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O

From this balanced equation, we can see that 1 mole of MnO4^- requires 5 moles of electrons. Therefore, the molar ratio between MnO4^- and A^n is 5:1.

Now, we can calculate the number of moles of A^n by multiplying the moles of MnO4^- by the molar ratio:

Moles of A^n = (1.61×10^-3 moles MnO4^-) * (1 mole A^n / 5 moles MnO4^-)
= 3.22×10^-4 moles A^n

Finally, we can calculate the value of n by comparing the moles of A^n with the given amount of A^n in the solution:

n = (moles of A^n) / (moles of solution)
= (3.22×10^-4 moles) / (2.68×10^-3 moles)
≈ 0.12

Since n represents a whole number, we can round 0.12 to the nearest whole number, which is 0. Therefore, the value of n is 0.

Answer: The value of n is 0.
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2.68×10^-3 moles of a solution containing an ion A^n acquire 1.61×10^-3 moles of MnO4^- for the oxidation of A^ n to AO3 ^- in acidic medium.what is the value of n ? A)3 B)2 C)5 D)4?
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