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3 years ago

Why is a cis-1,3-disubstituted cyclohexane more stable thanits trans isomer?

Chemistry

1 answer:

anzhelika [568]3 years ago

5 0

Answer:

Explanation has been given below

Explanation:

In diaxial conformation of cis-1,3-disubstituted cyclohexane, 4 gauche-butane interactions along with syn-diaxial interaction are present. Hence it readily gets converted to diequitorial conformation where no such gauche-butane interaction is present
In two possible conformations of trans-1,3-disubstituted cyclohexane, 2 gauche-butane interactions are present in each of them.
Hence cis-1,3-disubstituted cyclohexane exists almost exclusively in diequitorial form. But trans-1,3-disubstituted cyclohexane has no such option.

Trans-1,3-disubstituted cyclohexane experiences gauche butane interaction in each of the two conformations.
Therefore cis-1,3-disubstituted cyclohexane is more stable than trans conformation

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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m

Nataly_w [17]

The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

Trial [A] [B] Rate 
 (M) (M) (M/s) 
1 0.50 0.010 3.0×10−3 
2 0.50 0.020 6.0×10−3 
3 1.00 0 .010 1.2×10−2

Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0 =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0 = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s

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Maru [420]

Answer:

Explanation:

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Explain why uranium must be enriched to be used in a nuclear power plant.

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The nuclear fuel used in a nuclear reactor needs to have a higher concentration of the U 235 isotope than that which exists in natural uranium ore. U235 when concentrated (or "enriched") is fissionable in light-water reactors (the most common reactor design in the USA).

Explanation:

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In IR spectroscopy, we normally talk about "frequencies" when in reality we are referring to wavenumbers. What is the mathematic

Svetach [21]

Answer:

Here's what I get.

Explanation:

(b) Wavenumber and wavelength

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$\bar \nu = \dfrac{1}{\lambda}$

Thus, if λ = 3 µm,

$\bar \nu = \dfrac{1}{3 \times 10^{-6} \text{ m}}= 3.3 \times 10^{5}\text{ m}^{-1} = \textbf{3300 cm}^{-1}$

(a) Wavenumber and frequency

Since

λ = c/f and 1/λ = f/c

the relation between wavenumber and frequency is

$\bar \nu = \mathbf{\dfrac{f}{c}}$

Thus, if f = 90 THz

$\bar \nu = \dfrac{90 \times 10^{12} \text{ s}^{-1}}{3 \times 10^{8} \text{ m$\cdot$ s}^{-1}}= 3 \times 10^{5} \text{ m}^{-1} = \textbf{3000 cm}^{-1}$

(c) Units

(i) Frequency

The units are s⁻¹ or Hz.

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The SI base unit is metres, but infrared wavelengths are usually measured in micrometres (roughly 2.5 µm to 20 µm).

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