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ch4aika [34]
3 years ago
5

Consider the following balanced redox reaction (do not include state of matter in your answers):2CrO2−(aq) + 2H2O(l) + 6ClO−(aq)

→2CrO42−(aq) + 3Cl2(g) + 4OH−(aq)(a) Which species is being oxidized? (b) Which species is being reduced? (c) Which species is the oxidizing agent? (d) Which species is the reducing agent? (e) From which species to which does electron transfer occur?
Chemistry
2 answers:
Valentin [98]3 years ago
6 0

Answer:

E i think

Explanation:

zhenek [66]3 years ago
5 0

Answer:

Explanation:

Hello,

At first, it is necessary to establish the oxidation states for all the species involved in the reaction:

2(Cr^{+3}O_2^{-2})^{-}(aq) + 2H_2^{+1}O^{-2}(l) + 6(Cl^{+1}O^{-2})^{-1}(aq) -->2(Cr^{+6}O_4^{-2})^{-2}(aq) + 3Cl_2^0(g) + 4(O^{-2}H^{+1})^-(aq)

In such a way, we answer to the questions as follows:

(a) Oxidized species is the one that has an increase in its oxidation state, this is chromium which passes from +3 to +6.

(b) Reduced species is the one that has a decrease in its oxidation state, this is chlorine which passes from +1 to 0.

(c) Oxidizing agent is the same reduced species, this is chlorine as it removes electrons.

(d) Reducing agent is the same oxidized species, this is chromium as it gains electrons.

(e) Electron transfer occur from chlorine to chromium as chlorine removes electrons and chromium gains them.

Best regards.

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stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,
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Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

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Answer:

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Explanation:

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Answer:

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If you need more help on this subject, don't be afraid to ask me. I'm willing to help.

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