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kompoz [17]
3 years ago
15

The structure of PF3(C6H5)2 is trigonal bipyramidal, with one equatorial and two axial F atoms which interchange positions when

heated. Describe the low- and high- temperature 31P and 19F NMR spectra.
Engineering
1 answer:
Iteru [2.4K]3 years ago
6 0

Answer:

For 31 P NMR spectra

<u>low temperature </u>

there is two types of 19f seen in low temperature and they are

  • one at equitorial
  • one at axial

therefore in low temperature the 31p couples with the two types of 19F seen ( b_{f} and c_{f}to form a triplet and this couples more with a_{f} to form a doublet. i.e. one (1) peak

<u>High temperature </u>

At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak

For 19 P NMR spectra

<u>low temperature </u>

In low temperature a_{f}, b_{f}  , c_{f} is fixed  and the environment where b_{f} and c _{f} is the same hence a peak is formed and another peak is formed by a_{f} that makes the number of peaks = 2 peaks

<u>High temperature </u>

In high temperature a_{f}, b_{f}  , c_{f}  exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all

Explanation:

For 31 P NMR spectra

<u>low temperature </u>

there is two types of 19f seen in low temperature and they are

  • one at equitorial
  • one at axial

therefore in low temperature the 31p couples with the two types of 19F seen ( b_{f} and c_{f}to form a triplet and this couples more with a_{f} to form a doublet. i.e. one (1) peak

<u>High temperature </u>

At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak

For 19 P NMR spectra

<u>low temperature </u>

In low temperature a_{f}, b_{f}  , c_{f} is fixed  and the environment where b_{f} and c _{f} is the same hence a peak is formed and another peak is formed by a_{f} that makes the number of peaks = 2 peaks

<u>High temperature </u>

In high temperature a_{f}, b_{f}  , c_{f}  exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all

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Kryger [21]

Answer:

I know it is C)Virtual reality

Explanation:

Look at the clues

story about putting on a headset ( virtual reality head set!)

seeing a digital world (A virtual reality world)

they could walk around in (Fake walking you are basically jogging in place)

explore in order to see what ancient Benin looked like (Looking at a real place only digitally)

as if they were really there ( they think they are actually there)

The only reason I know all of this is because I have done virtual reality multiple times and I LOVED it SUPER fun ( I was doing archery) :) Hope this helps!

6 0
2 years ago
Read 2 more answers
Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo
rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

KE = 300 KJ

Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

7 0
2 years ago
When measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial re
kozerog [31]

Answer:

10.5

Explanation:

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2 years ago
How much cornfield area would be required if you were to replace all the oil consumed in the United States with ethanol from cor
zaharov [31]

Answer:

2377.35 km

Explanation:

Given the following;

1. A cornfield is 1.5% efficient at converting radiant energy into stored chemical potential energy;

2. The conversion from corn to ethanol is 17% efficient;

3. A 1.2:1 ratio for farm equipment to energy production

4. A 50% growing season and,

5. 200 W/m2 solar insolation.

As per our assumptions,1.2/1 is the ratio for farm equipment to energy production,

So USA need around 45.45% (1/(1+1.2) replacement of fuel energy production which is nearly about = 0.4545*10^{20} J/year = \frac{0.4545*10^{20}}{365*24*3600}=1.44121*10^{12} J/sec

Growing season is only part of year ( Given = 50%),

Net efficiency = 1.5%*17%*50%=0.015*0.17*0.5=0.001275 = 0.1275%

Hence , Actual Energy replacement (Efficiency),

=\frac{1.44121*10^{12}}{0.001275} = 1.13*10^{15} J/sec=1.13*10^{15} W

As per assumption (5),

\because 200 W/m2 solar insolation arequired,

So USA required corn field area = 1.13*10^{15}/200 = 5.65*10^{12} m^{2}

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4 0
3 years ago
1.0•10^-10 standard form
Drupady [299]

Answer:

1.0 * 10^{-10} = 0.0000000001

Explanation:

Given

1.0 * 10^{-10}

Required

Convert to standard form

1.0 * 10^{-10}

From laws of indices

a^{-x} = \frac{1}{a^x}

So, 1.0 * 10^{-10} is equivalent to

1.0 * 10^{-10} = 1.0 * \frac{1}{10^{10}}

1.0 * 10^{-10} = 1.0 * \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}* \frac{1}{10}

1.0 * 10^{-10} = 1.0 * \frac{1}{10000000000}

1.0 * 10^{-10} = 1.0 * 0.0000000001

1.0 * 10^{-10} = 0.0000000001

Hence, the standard form of 1.0 * 10^{-10} is 0.0000000001

3 0
3 years ago
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