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Illusion [34]
3 years ago
6

Monochlorination of propane yields two constitutional isomers, and dichlorination yields four. Trichlorination yields five const

itutional isomers. Draw all five trichloropropane isomers.

Chemistry
2 answers:
RSB [31]3 years ago
8 0

Answer:

The trichloro isomers are [figure attached for structures]

a) 1,1,1-trichloropropane

b) 1,1,2-trichloropropane

c) 1,1,3-trichloropropane

d) 1,2,2-trichloropropane

e) 1,2,3-trichloropropane

Explanation:

Constitutional isomers are the compounds with same molecule formula but different arrangements of atoms on the chain.

The monochloro propane isomers are :

a) 1-chloropropane

b) 2-chloropropane

The dichloro isomers are :

a) 1,2-dichloropropane

b) 1,3-dichloropropane

c) 1,1-dichloropropane

d) 2,2-dichloropropane

The trichloro isomers are [figure attached for structures]

a) 1,1,1-trichloropropane

b) 1,1,2-trichloropropane

c) 1,1,3-trichloropropane

d) 1,2,2-trichloropropane

e) 1,2,3-trichloropropane

GaryK [48]3 years ago
6 0
By Tri chlorination, three chlorine atoms can replace the 3 hydrogen atoms in propane molecule to give tri-chlorinated products according to the attached picture.

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Diasaccharides e.g maltose are reducing sugars.their standard test is BENEDICT test .

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Answer:

[Zn²⁺] = 4.78x10⁻¹⁰M

Explanation:

Based on the reaction:

ZnBr₂(aq) + K₂CO₃(aq) → ZnCO₃(s) + 2KBr(aq)

The zinc added produce the insoluble ZnCO₃ with Ksp = 1.46x10⁻¹⁰:

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<em>Moles ZnB₂ (Molar mass: 225.2g/mol) = Moles Zn²⁺:</em>

6.63g ZnBr₂ * (1mol / 225.2g) = 0.02944moles Zn²⁺

<em>Moles K₂CO₃ = Moles CO₃²⁻:</em>

0.100L * (0.60mol/L) = 0.060 moles CO₃²⁻

Moles CO₃²⁻ in excess: 0.0600moles CO₃²⁻ - 0.02944moles =

0.03056moles CO₃²⁻ / 0.100L = 0.3056M = [CO₃²⁻]

Replacing in Ksp expression:

1.46x10⁻¹⁰ = [Zn²⁺] [0.3056M]

<h3>[Zn²⁺] = 4.78x10⁻¹⁰M</h3>

4 0
3 years ago
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