Oxalic acid, H2C2O4, reacts

Chemistry

1 answer:

Dominik [7]3 years ago

6 0

Answer:0.005M

Explanation:

First deduce the oxidation and reduction half equations and from that obtain the balanced redox reaction equation. From that, the number of moles of reacting species are seen from the stoichiometry of the reaction from which the number of moles of oxalate is obtained and substituted to obtain the molar concentration of oxalate.

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The reaction of 2-chloropropane with sodium hydroxide can occur via both SN1 and SN2 mechanisms.

strojnjashka [21]

Answer:

Explanation:

(a) Part 1: $S_N1$ reaction. This is a nucleophilic substitution reaction in which we have two steps. Firstly, chlorine, a good leaving group, leaves the carbon skeleton to form a relatively stable secondary carbocation. This carbocation is then attacked by the hydroxide anion, our nucleophile, to form the final product.

To summarize, this mechanism takes places in two separate steps. The mechanism is attached below.

Part 2: $S_N2$ reaction. This is a nucleophilic substitution reaction in which we have one step. Our nucleophile, hydroxide, attacks the carbon and then chlorine leaves simultaneously without an intermediate carbocation being formed.

The mechanism is attached as well.

(b) The rate determining step is the slow step. Formation of the carbocation has the greatest activation energy, so this is our rate determining step for $S_N1$ . For $S_N2$ , we only have one step, so the rate determining step is the attack of the nucleophile and the loss of the leaving group.