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NeX [460]
3 years ago
14

Zinc metal is produced by heating solid zinc sulfide with solid wine witte resulting in liquid zinc and sulfur dioxide gas Write

a balanced equation for the reaction using complete formulas for the compounds with phase labels (Use the lowest possible coefficients. Use the pull down boxes to specify states such as (aq) or (b). If a box is not needed, leave it blank.) Submit Answer Try Another Version item attempts remaining
Chemistry
1 answer:
elena-s [515]3 years ago
4 0

Answer:

The balance chemical equation can be given as:

ZnS(s)+ZnSO_4(aq)\overset{heat}\rightarrow 2Zn(s)+2SO_2(g)

Explanation:

When zinc sulfide is allowed to react with zinc sulfate it gives zinc metal and sulfur dioxide gas as a product.The balance chemical equation can be given as:

ZnS(s)+ZnSO_4(aq)\overset{heat}\rightarrow 2Zn(s)+2SO_2(g)

According to reaction we can see when 1 mol of zinc sulfide reacts with 1 mol of zinc sulfate it forms 2 moles of zinc metal and 2 moles of sulfur dioxide gas.

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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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3 years ago
How was Bohr’s atomic model different from Rutherford’s atomic model?
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C is the answer.

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3 0
3 years ago
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