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9966 [12]
4 years ago
10

Which of the statements below correctly describes the chair conformations of trans-1,4-dimethylcyclohexane? A. The higher energy

chair conformation contains one axial methyl group and one equatorial methyl group. B. The lower energy chair conformation contains one axial methyl group and one equatorial methyl group. C. The two chair conformations are of equal energy. D. The higher energy chair conformation contains two axial methyl groups. E. The lower energy chair conformation contains two axial methyl groups.
Chemistry
1 answer:
Taya2010 [7]4 years ago
5 0

Answer:

D. The higher energy chair conformation contains two axial methyl groups.

Explanation:

The diaxial trans-1,4-dimethylcyclohexane has the highest energy (the least stable) and the diequatorial chair conformation has the lowest energy (the most stable).

The diaxial conformation has a higher energy due to the interaction between the methyl groups.

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At what the pressure, in atmospheres, that O2, gas at 75°C has a density of 2.00 g/L​
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Which of the following statements is true about the food chain below? Grass ⇒ grasshopper ⇒ bird ⇒ hawk The amount of energy tra
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In a food chain we arrange the energy in the form of a pyramid.

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As you move down through the family of alkali metals, the first ionization energy ________ because
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3 years ago
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1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
anastassius [24]

Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol (P^o_1) = 20.9 torr

Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

x_1 = mole fraction of 1‑Propanol

p_1=(0.540)\times (20.9torr)=11.3torr

and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

3 0
3 years ago
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