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Natali [406]
2 years ago
7

In an NMR experiment, shielding refers to the reduced impact of the ______on a nucleus due to the presence of ______around the n

ucleus. Shielding can be_____by the presence of more electronegative groups around the nucleus. A shielded nucleus appears further ______ whereas a deshielded nucleus appears further _____in an NMR spectrum.
Chemistry
1 answer:
suter [353]2 years ago
5 0

Answer:

In an NMR experiment, shielding refers to the reduced impact of the APPLIED MAGNETIC FIELD on a nucleus due to the presence of ELECTRON DENSITYaround the nucleus.

Shielding can be DECREASED by the presence of more electronegative groups around the nucleus.

A shielded nucleus appears further UPFIELD whereas a deshielded nucleus appears further downfield in an NMR spectrum.

Explanation:

1 There is reducece because alot of the electronegative atom binds more and thus reduce the electron density around the nucleus.

While a reduction in electron density bring about more nuclear magnetic field thus it resonates at higher frequency resulting downfield peak.

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2 years ago
Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of
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<u>Answer:</u> The electronic configuration of the elements are written below.

<u>Explanation:</u>

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

For the given options:

  • <u>Option a:</u>  Carbon (C)

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is 1s^22s^22p^2

  • <u>Option b:</u>  Phosphorus (P)

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is 1s^22s^22p^63s^23p^3

  • <u>Option c:</u>  Vanadium (V)

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is 1s^22s^22p^63s^23p^64s^23d^3

  • <u>Option d:</u>  Antimony (Sb)

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3

  • <u>Option e:</u>  Samarium (Sm)

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6

Hence, the electronic configuration of the elements are written above.

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2 years ago
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What mass of CaSO3 must have been present initially to produce 14.5 L of SO2 gas at a temperature of 12.5°C and a pressure of 1.
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When the reaction equation is:

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we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

to get the moles of SO2 we are going to use the ideal gas equation:

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n is the moles' number (which we need to calculate)

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and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

so, by substitution:

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so we can easily get the mass of CaSO3:

when mass = moles * molar mass

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