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Sergio [31]
3 years ago
12

In a chemical reaction, the element 13al will most preferably ________. in a chemical reaction, the element 13al will most prefe

rably ________. gain three electrons and become positively charged lose five electrons and become positively charged lose three electrons and become positively charged gain five electrons and become negatively charged
Chemistry
1 answer:
Rasek [7]3 years ago
6 0
Let us calculate the structure of the electric shells of the Al atom. It has an atomic number of 13, so it has 13 electrons. The first 2 go to the first hell. The next 8 need to go to the second shell and the last 3 ones would go to the outermost shell. The outer shell, that is the most important one for chemical reactions, has thus 3 electrons. An atom always tries to have a completed outer shell (with either 2 or 8 atoms). It is easier for a cell to have a charge of +3 than a charge of -5 (smaller absolute value) and thus the Aluminum atom will try to get rid of the 3 electrons. In this process, it loses negative charge thus it will become positively charged. Hence, the correct answer is that it will prefer to lose 3 electrons and become positively charged.
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For the following balanced reaction:

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We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:

Kp=\frac{(P PCl_{3})(P Cl_{2})}{(P PCl_{5})}

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What would happen if the sand dunes in an area were destroyed?
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Answer:

<u>Our beaches would be unprotected</u>

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What would be the major product if 1,4-dibromo-4-methylpentane was allowed to react with:
Levart [38]

Answer : The correct answer for a) 4-bromo-2-iodo-4-methyl pentane and b)5-bromo-2-ethoxy-2-methyl pentane.

A) Reaction with NaI :

Reaction of alkyl halide with NaI is known as Finkelstein Reaction . The acetone is used as solvent . It involves bimolecular nucleophillic substitution rmechanism (SN²) . There is replecement of one halogen with other occurs .

The incoming Nucleophile(Nu⁻) (halide) attacks on carbon from back side , while the leaving group (halide) leaves the compound from front side , simultaneously. The product so formed have is inverted .(Image)

NaI releases I⁻ ion which act as nucelophile and attacks on C1 carbon and Br⁻ from C1 carbon is released . Out of two bromines at C1 and C4 carbons , C1 is primary carbon which is less sterically hindered while C-4 is tertiary carbon and sterically hindered . So it is easy for incoming Nu⁻ to attack on C1 carbon .So Br⁻ is repleaced by I⁻.

1,4-dibromo-4-methylpentane + NaI → 4-bromo-1-iodo-4-methylpentane

The product formed from reaction between 1,4-dibromo-4-methylpentane and NaI is 4-bromo-1-iodo-4-methylpentane . (Image)

B) Reaction with AgNO3 :

Reaction of alkyl halide with AgNO3 in ethanol takes place via SN¹ ( unimolecular nucleophilic substitution ) mechanism . In this leaving group(halide) leaves from alkyl halide forming an intermediate carbocation species . The incoming Nu⁻ attack on this carbocation.

AgNO3 reacts releases Ag⁺ion which abstract Br⁻ of C-4 carbon from 1,4-dibromo-4-methylpentane. THis forms tertiary carbocation which is more stable than carbocation formed by removal of Br from C-1 . The ethanol being more Nucleophilic than NO₃⁻ (from AgNO₃), attacks on this carbocation .(Image )

The product formed as a result is 5-bromo-2-ethoxy-2-methyl pentane.

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2 years ago
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