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LenKa [72]
2 years ago
11

Can anyone explain to me the octet rule in detail with examples my teacher didn't do the greatest job of making us understand.

Chemistry
1 answer:
iogann1982 [59]2 years ago
4 0
In chemistry, the octet rule explains how atoms of different elements combine to form molecules. In a chemical formula, the octet rule strongly governs the number of atoms for each element in a molecule; for example, calcium fluoride is CaF2 because two fluorine atoms and one calcium satisfy the rule. Hope this helps
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Write the chemical equation for this reaction. Use the picture I provide!
Lesechka [4]

Answer:

H2O

H2O

Explanation:

because the are only two hydrogen that can react to Oxygen

3 0
2 years ago
Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of
dybincka [34]

<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.

4 0
3 years ago
What is the value of the equilibrium constant for this redox reaction? 2Cr3+(aq) + 3Sn2+ (aq) 2Cr(s) + 3Sn4+(aq) E = -0.89 v
charle [14.2K]
<span>b. square root of 3 over 3 is the answer to your question!!! I hope I helped!!!!!!!!!!                                                                 XoXo -Marcey<3!  :D
</span>
8 0
3 years ago
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Yanka [14]
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3 0
3 years ago
Which one of the following would have the largest dispersion forces?
Sergio039 [100]

Answer:

A) CH3CH2SH

Explanation:

Dispersion forces are weak attractions found between non-polar and polar molecules. The attractions here can be attributed to the fact that a  non-polar molecule sometimes become polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant. If this happens, the molecule has a temporary dipole. This dipole can induce the neighbouring molecules to be distorted and form dipoles as well. The attractions between these dipoles constitute the Dispersion Forces.

Therefore; the greater the molar mass of a compound or molecule, the higher the Dispersion Force. This implies that the compound or molecule with the highest molar mass have the largest dispersion forces.

Now; for option (A)

CH3CH2SH

The molar mass is :

= (12 + (1 × 3 ) +12 + (1 ×2) + 32+1)

= (12 + 3+ 12 + 2 + 32 + 1)

= 62 g/mol

For option (B)

CH3NH2

The molar mass is:

= (12 + (1 ×  3 ) +14 + (1 ×  2)

= (12 + 3 + 14 + 2)

= 31 g/mol

For option (C)

CH4

The molar mass is :

= 12 + (1 × 4)

= 12 + 4

= 16 g/mol

For option (D)

CH3CH3

The molar mass is :

= 12 + ( 1 × 3 ) + 12 + ( 1 × 3)

= 12 + 3 + 12 + 3

= 30 g/mol

Thus ; option (A) has the highest molar mass, as such the largest dispersion force is A) CH3CH2SH

7 0
3 years ago
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