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Aleks04 [339]
3 years ago
8

Is a 93.3% solution dilute or concentrated? Dilute Concentrated

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
3 0

Answer:

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  • <u><em>Concentrated</em></u>

Explanation:

Concentration measures the amount of solute in a solution. There are many expressions of concentration. Some of then are percentage (mass/mass, volume/mass, volume/volume), molarity, molality, mole fraction, among others.

When a solution has a high concentration it is said that it is <em>concentrated; </em>when a solution has a low concentration is is said that is is diluted.

Concentrated solutions expressed in percentage typically have about 80 - 90% (or more) of solute.

Diluted solutions expressed in percentage, tipylcally have about 10% - 20% or less.

But they are not fixed limits. You might say that a 85% solution is concentrated. Acids at 75 % sure are concentrated.

Hence, a 93.3% solution is concentrated, definitely.

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What is the pH of an acidic solution?
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The answer is A, between 0 and 7.

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

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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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