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Add coin into nitric acid and sulfuric acid. Nitric should be 60% weight of the coin and sulfuric should be 1.6 times. It will dissolve in minutes. Heat it up to boil unused nitric. Add some urea to neutralize acids. Add iron fillings to remove copper. Filter and melt the copper powder. You will have pure copper. Take rest of the solution, dilute it a bit, ...

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Boron mostly favours the +3 oxidation state. However, boron does form some compounds in +1 O.S. but they are unstable and rearranges to stable molecule (such as boron monofluoride in @Oscar's answer which polymerizes by itself to compounds containing 10-12 boron atoms). Apart from the monofluoride, I found one example of boron(I) complex which is a radical ...

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We may not have all boron atoms equivalently bonded to carbon in $\ce{B4C}$, so we might not be able to assign all boron atoms an equal oxidation state of +1. For that matter, assuming a carbon oxidation state of -4 might also be a reach. A better candidate for boron(I) is boron monofluoride, $\ce{BF}$, which is unstable but isolable: Boron monofluoride can ...

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Oxyanions have a central atom surrounded by a number of more electronegative O atoms and the central atom is significantly positively charged. They are subject to reduction. There are three main factors first the electronegativity of the central atom the more electronegative the stronger the oxidant, second the more O's the stronger the oxidant and third the ...

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Essentially, it's because nitric acid is a strong acid. At concentrations where we generally perform aqueous chemistry, it is essentially completely dissociated to give $\ce{H^+(aq)}$ and $\ce{NO3^-}$. So the nitrogen-bearing species that does the oxidizing is primarily the dissociated ion $\ce{NO3^-}$. As you balance the reaction you will also see the $\ce{... 1 Balancing a reaction equation is a sequential process. One approach is that you identify the starting materials by their formulae, and the reaction products by theirs. you balance the equation such the same number of atoms on the left hand side equates the number of atoms on the left hand side. Depending on the reaction, there are some groups of atoms, or ... 1 In contrary to acid-base reactions(1), a redox reaction is such a reaction, which can be formally divided into two half-reactions. At least one reactant in such a half-reaction explicitly formally acts as a donor or acceptor of one or more electrons. $$\ce{Cl2(aq) + 2 Fe^2+(aq) -> 2 Cl-(aq) + 2 Fe^3+(aq)}$$ can be formally divided to:$\$\ce{Cl2(aq) + 2 e- ...

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You first focus on the groups that undergo oxidation or reduction. The sulfate does neither, so you can set it aside, just like the potassium cations. The half reactions you wrote, however, need some work. The oxidation state of manganese in permanganate is +7, so there is a transfer of 5 electrons. For bromine to bromide, the oxidation state changes by one, ...

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