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defon
3 years ago
8

Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction i

dentification Sn2 (aq) 2e-Sn(s) reduction Al(s)Al3 (aq) 3e- oxidation (2) Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients.
Chemistry
1 answer:
Natasha2012 [34]3 years ago
5 0

Answer: Oxidation reaction:  Al(s)\rightarrow Al^{3+}(aq)+3e^-

Reduction reaction:  Sn^{2+}(aq)+2e^-\rightarrow Sn(s)

Overall redox reaction : 2Al(s)+3Sn^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Sn(s)

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

Al(s)\rightarrow Al^{3+}(aq)+3e^-

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

Sn^{2+}(aq)+2e^-\rightarrow Sn(s)

Overall redox reaction : 2Al(s)+3Sn^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Sn(s)

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Answer : The molar mass of an acid is 266.985 g/mole

Explanation : Given,

Mass of an acid (HX) = 4.7 g

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First we have to calculate the moles of NaOH.

\text{Moles of }NaOH=\text{Molarity of }NaOH\times \text{Volume of solution}=0.54mole/L\times 0.0326L=0.017604mole

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In the titration, the moles of an acid will be equal to the moles of NaOH.

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Now we have to calculate the molar mass of and acid.

\text{Moles of an acid}=\frac{\text{Mass of an acid}}{\text{Molar mass of an acid}}

Now put all the given values in this formula, we get:

0.017604mole=\frac{4.7g}{\text{Molar mass of an acid}}

\text{Molar mass of an acid}=266.985g/mole

Therefore, the molar mass of an acid is 266.985 g/mole

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How many ml of a 2.0 m nabr solution are needed to make 200.0 ml of 0.50 m nabr?
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<span>V1 = 50 mL needed</span>

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