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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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Drupady [299]

Answer:

450,000m = 450km = 4.5E5

32,600,000W = 32.6MW = 3.26E7

59,700,000,000cal = 59.7Gcal = 5.97E10

0.000000083s = 83ns = 8.3E-8

35,000Ω = 35kΩ = 3.5E4

Explanation:

Giga   = 1,000,000,000

Mega = 1,000,000

kilo     = 1,000

unit    = 1

deci   = .1

centi  = .01

milli    = .001

micro = .000001

nano = .0000000001

pico  = .000000000001

You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.

7 0
3 years ago
List one advantage and one disadvantage of the use of the commutator?
Mariulka [41]

Answer and  Explanation:

Commutator are used in DC machine commutator is mainly used for the reversing the direction of the current .It is connected to the armature of the DC generator or motor

ADVANTAGE OF COMMUTATOR The main advantage of the commutator in DC motor is to keep keep the direction of the toque always in the same direction by changing the current direction

DISADVANTAGE OF COMMUTATOR : The main disadvantage is due to the friction between the commutator and brushes there is a friction loss.

3 0
3 years ago
Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value
masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0
3 years ago
For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration
Amiraneli [1.4K]

Answer:

135 hour

Explanation:

It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.

We have to find the time necessary to achieve the same concentration at a 6 mm position.

we know that \frac{x_1^2}{Dt}=constant where x is distance and t is time .As the temperature is constant so D will be also constant

So \frac{x_1^2}{t}=constant

then \frac{x_1^2}{t_1}=\frac{x_2^2}{t_2} we have given x_1=2 mm\ ,t_1=15 hour\ ,x_2=6\ mm and we have to find t_2 putting all these value in equation

\frac{2^2}{15}=\frac{6^2}{t_2}

so t_2=135\ hour

5 0
4 years ago
Explain three (3) modes of heat transfer in air conditioning system.
LenKa [72]

Answer:

1. Conduction

2. Convection

3. Radiation

Explanation:

The 3 modes of heat transfer i an air conditioning system:

1. Conduction:

The transfer of heat by conduction  takes place in solid and is when the conduction takes place as a result of direct contact in between the interacting material which transfer the heat energy from particle to particle thus conducting the heat through out the system.

2. Convection:

The other mode for the transfer of heat which takes place especially in fluids - gases and liquids is through the technique of convection in which the transfer of heat takes place by the circular motion of the atoms and molecules of the fluid which carries the heat energy and results in the distribution of the heated fluid in the entire system thus transferring all the heat energy in the entire system.

3. Radiation:

The third mode of heat transfer in the air conditioning system is through radiation. This method transfers the heat by making use of the electro-magnetic radiation in the infra red spectrum where the waves of the spectrum transfers the heat energy with the help of a medium or without any medium at all.

Thus making the radiation method of heat transfer as the only method out of the three methods which does not require the material medium for the transfer of heat energy.

4 0
3 years ago
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