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rewona [7]
3 years ago
15

1. Which of the following statements describe cations? Select all that apply.

Chemistry
2 answers:
Wittaler [7]3 years ago
5 0

Answer:

<h3><em>N</em><em>E</em><em>G</em><em>A</em><em>T</em><em>I</em><em>V</em><em>E</em><em>L</em><em>Y</em><em> </em><em>C</em><em>H</em><em>A</em><em>R</em><em>G</em><em>E</em><em>D</em><em> </em><em>I</em><em>O</em><em>N</em><em>S</em></h3>

Explanation:

<h2><em>I</em><em>H</em><em>O</em><em>P</em><em> </em><em>N</em><em>A</em><em>K</em><em>A</em><em>T</em><em>U</em><em>L</em><em>O</em><em>N</em><em>G</em><em> </em><em>#</em><em>L</em><em>I</em><em>K</em><em>E</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>F</em><em>O</em><em>L</em><em>L</em><em>O</em><em>W</em></h2>
Crazy boy [7]3 years ago
5 0
Hey :)

Cations are positively charged ions and they have lost electrons

Hope this helps!
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When ethyl-2-methylpropanoate is treated with sodium ethoxide in ethanol, there is no observable reaction. Explain this observat
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Answer:

No net change in reaction occurs in this nucleophilic acyl subtitution reaction

Explanation:

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Here ethoxide group replaces an ethoxide group from ester through addition-ellimination pathway.

So, ultimately, the product of this reaction is identical with reactant i.e. ethyl-2-methylpropanoate is reproduced.

Hence one might observe no change during reaction as product and reactant of this reaction are same.

Mechanistic pathway has been shown below.

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Choose the correct coefficients to balance the following equation:
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3. Balance each of the following redox reactions in the listed aqueous environment. (4 pts each, 8 pts total) Crs)NO3 (a) Cr (a)
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Explanation:

(a)   The given reaction equation is as follows.

        Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g) (acidic)

So, here the reduction and oxidation-half reactions will be as follows.

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

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As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.

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(b)   The given reaction equation is as follows.

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So, here the reduction and oxidation-half reactions will be as follows.

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Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.

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Therefore, balancing the whole reaction equation in the basic medium as follows.

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