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Sauron [17]
2 years ago
6

Select all the statements that correctly describe the chair conformation of cyclohexane. Multiple select question. The bond angl

es are near 109.50. Each carbon atom has one axial and one equatorial hydrogen atom. All hydrogen atoms on adjacent carbon atoms are staggered. The chair conformation is less stable than the boat conformation.
Chemistry
1 answer:
Yuki888 [10]2 years ago
8 0

All the statements except Option D is true for the chair confirmation of cyclo hexane.

<h3>Describe the Chair confirmation of Cyclohexane  ?</h3>

The flexibility of Cyclo hexane allows for formation of a confirmation ,

This confirmation has lowest energy among the confirmations that can be made for Cyclohexane.

The bond angles are near 109.50 degree,

Each carbon atom has one axial and one equatorial hydrogen atom and All hydrogen atoms on adjacent carbon atoms are staggered.

Therefore Option A , B , C are all true statements about the Chair Confirmation of Cyclohexane.

To know more about Chair Confirmation

brainly.com/question/14852324

#SPJ1

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Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

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Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

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Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
3 years ago
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