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alina1380 [7]
2 years ago
6

What is responsible for the differences in chemical shift observed in carbon 4 in 4-fluoroheptane and carbon 4 in heptane?

Chemistry
1 answer:
Papessa [141]2 years ago
4 0

The chemical shift of a given proton is decided specifically with the aid of interactions with the nearby electrons. The most necessary thing to recognize is that when electrons are subjected to an external magnetic field, they structure their own small prompted magnetic fields in opposition to the exterior field.

Consider the methane molecule ( CH4 ) in which the four equal protons have a chemical shift of 0.23 ppm (this is a price we can look up in any chemistry reference source). The valence electrons around the methyl carbon, when subjected to B0, generate their own very small triggered magnetic area that opposes  B0 . This induced field, to a small however full-size degree, shields the close by protons from experiencing the full force of  B0 , an impact regarded as local diamagnetic shielding. In different words, the methane protons do now not quite journey the full force of  B0  - what they journey is known as  Beff , or the high-quality field, which is slightly weaker than  B0  due to the have an effect on of the close by electrons.

In nuclear magnetic resonance (NMR) spectroscopy, the chemical shift is the resonant frequency of an atomic nucleus relative to a preferred in a magnetic field. Often the position and range of chemical shifts are diagnostic of the shape of a molecule.

Learn more about chemical shifts here:

brainly.com/question/4289021

#SPJ4

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Answer: The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.947 atm

P_2 = final pressure of gas = 0.987 atm

V_1 = initial volume of gas = 150 ml

V_2 = final volume of gas = 144 ml    

T_1 = initial temperature of gas = 25^0C=(25+273.15)K=298.15K

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Now put all the given values in the above equation, we get:

\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}

T_2=298.31K=(298.31-273.15)^0C=25.16^0C

The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

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