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masha68 [24]
3 years ago
13

The two alternative chair conformations of cis-1-bromo-2-methylcyclohexane differ in their Gibbs free energy. Using the data for

ΔG (Axial-Equatorial) for monosubstituted cyclohexanes at room temperature (25ºC): Axial → Equatorial Group ΔG° (kJ/mol) Group ΔG° (kJ/mol) -0.8 -5.9 -2.4 -7.3 -3.9 1,2-gauche 3.8 Calculate the absolute value of the difference in the Gibbs free energy between the alternative chair conformations. kJ/mol Which group in this compound is in axial position in the energetically preferred chair conformation? _______

Chemistry
1 answer:
horsena [70]3 years ago
5 0

Answer:

a.) 4.9kj/mol

b.) -Br group

Explanation:

first all of your question did not make mention of the groups in the table.

here it is:-

axial group     ΔG°(kj/mol)    group     ΔG°kj/mol

CN                   -O.8                NH₂          -5.9

Br                     -2.4                 CH₃           -7.3

OH                    -3.9            1,2-gauche      3.8

The 6 axial groups are bonded one on each carbon and also they are parallel and also alternate from up to bottom.

a.)

to get the absolute ΔG;

ΔG = -2.4-(-7.3)

= -2.4+7.3

= 4.9KJ/MOL

b.)

the axial group in this compound which is in the axial position in the energetically preferred chain conformation is the -Br group

Please check attachment for diagram

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sergeinik [125]

Answer:

82.04g

Explanation:

To solve this problem, you need to find the molar mass of HCl, which is the sum of the molar mass of H plus the molar mass of Cl. The molar mass of H is 1.008g/mol while the molar mass of Cl is 35.45g/mol. Thusly the molar mass of HCl is 36.458g/mol. You have 2.25 moles of HCl, so multiply the number of moles by the molar mass to get the mass in g. In this case it is 2.25mol * 36.458g/mol = 82.04g.

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3 0
3 years ago
1. Metallic strontium crystallizes in a face-centered cubic lattice, with one Sr atom per lattice point. If the edge length of t
Zarrin [17]

Answer:

r=215pm

N_{Mn}=20

Explanation:

From the question we are told that:

Edge length of the unit cell l=608pm

a)

Generally the equation for The relationship between edge length and radius is mathematically given by

4r=\sqrt{2a}

Therefore

4r=\sqrt{2*608}

r=\frac{\sqrt{2*608}}{4}

r=215pm

b)

From the question we are told that:

Density \rho=7.297

Edge length of l=630.0 pm=>630*10^-{10}

Therefore Volume  is given as

V=l^3

V=630*10^-{10}^3

V=2.50047*10^{−22}

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m=Volume*density

m=V*\rho

m=2.50047*10^{−22}*7.297

m=1.83*10^{-21}g

Therefore Molarity is given as

n=\frac{M}{Molar M}

n=\frac{1.83*10^{-21}g}{55}

n=3.32*10^{-23}

Finally The atoms in a unit cell is

N_{Mn}=Moles*Avogadro\ constant

N_{Mn}=3.32*10^{-23}*6.023*10^{23}

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7 0
3 years ago
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Answer:

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4 0
4 years ago
What happens in a chemical reaction with the individual elements?
Gemiola [76]
Chemical reactions involve electrons and where they move (if they move).
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Angelina_Jolie [31]

Answer:

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Explanation:

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