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IrinaK [193]
2 years ago
15

An alkene X undergoes ozonolysis and gives two compounds Y and Z of molecular formula CHO. Y and Z. are functional isomers of ea

ch other Write the two-steps chemical equation for the conversion of X into Y and Z. 121 Write the structural formula of Y and 7. Why are they called functional isomers? [3] ili. What happens when hydrogen gas in the presence of nickel catalyst is passed over X? (1] TV. What is the application of ozonolysis in the organic reaction mechanism? 11 How can you prove chemically the compound X is unsaturated? [1]
Chemistry
1 answer:
AleksAgata [21]2 years ago
3 0

The answers to the given questions are given below.

<h3>What is Ozonolysis ?</h3>

A reaction done with Ozone to determine the position of C-C double bond in unsaturated Hydrocarbons.

It is given in the question

An alkene X undergoes ozonolysis and gives two compounds Y and Z of molecular formula CHO

Y and Z. are functional isomers of each other

  • The two-steps chemical equation for the conversion of X into Y and Z.

\rm \\(CH_3)_2 -C= CH-CH_2-CH_3  + Zn - > CH_3 -CO-CH_3 +  CH0-CH_2-CH_3\\\\Y \;is \;CH_3 -CO-CH_3 \\\\and \;Z \;is \;CHO-CH_2-CH_3

  • Here X is 2 methyl pent-2 ene
  • On reaction with ozone , ozonide is formed

                         \rm Y \;is \;CH_3 -CO-CH_3 \\\\and \;Z \;is \;CHO-CH_2-CH_3

  • These are functional isomers as they have same molecular weight , but different functional groups.

  • When hydrogen gas in the presence of nickel catalyst is passed over  2 methyl pent-2 ene

  • A process called catalytic Hydrogenation will take place , and the alkene will get converted into alkane , 2-methyl pentane

  • The ozonolysis process is useful in the pharmaceutical industry

  • The unsaturation can be tested  by bromine water test

To know more about Ozonolysis

brainly.com/question/14356308

#SPJ1

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8 0
4 years ago
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mass of pentane : = 30.303 g

moles of Al₂(CO₃)₃ : = 0.147

<h3>Further explanation</h3>

Given

1. Reaction

C₅H₁₂+8O₂→6H₂O+5CO₂.

45.3 g water

2. 2AlCl₃ + 3MgCO₃ → Al₂(CO₃)₃ + 3MgCl₂

37.2 MgCO₃

Required

mass of pentane

moles of Al₂(CO₃)₃

Solution

1. mol water = 45.3 : 18 g/mol = 2.52

From equation, mol ratio of C₅H₁₂ : H₂O = 1 : 6, so mol pentane :

= 1/6 x mol H₂O

= 1/6 x 2.52

= 0.42

Mass pentane :

= mol x MW

= 0.42 x 72.15 g/mol

= 30.303 g

2. mol MgCO₃ : 37.2 : 84,3139 g/mol = 0.44

mol Al₂(CO₃)₃ :

= 1/3 x mol MgCO₃

= 1/3 x 0.44

= 0.147

4 0
3 years ago
3. Gold-198 decays with a half-life of 2.7 days according to the equation:198Au → 0−1β + 198Cd
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a. k=0.256/day

b.sample of Au-198 remains after seven days : 1.67 g

<h3>Further explanation</h3>

1. A half-life of 2.7 days⇒t1/2=2.7 days

The half-life can be expressed in a decay constant( λ)

\tt \displaystyle t_ {1/2} = {\dfrac {\ln (2)} {\lambda}}\\\\\lambda(or~k)=\dfrac{0.693}{2.7}=0.256/day

2. We can use formula : (integrated rate law) :

\tt ln[A]=-kt+ln[Ao]\\\\ln(\dfrac{Ao}{A})=kt

Ao=10 g

t=7 days

k=0.256/day

\tt ln(\dfrac{10}{A})=0.256\times 7\\\\ln(\dfrac{10}{A})=1.792\\\\e^{1.792}=\dfrac{10}{A}\\\\6=\dfrac{10}{A}\rightarrow A=1.67~g

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