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allochka39001 [22]
3 years ago
15

Which of the following molecules would you expect to have a dipole moment of zero? a,CH2 Ch3

Chemistry
1 answer:
Vadim26 [7]3 years ago
4 0

Answer: The molecule CH_{3}-CH_{3} is expected to have a dipole moment of zero.

Explanation:

The product of magnitude of the charge calculated in electrostatic units is called dipole moment.

Formula for dipole moment is as follows.

Dipole moment = Charge (in esu) \times distance (in cm)

Non-polar molecules have zero dipole moment.

For example, CH_{3}-CH_{3} is a non-polar molecule so its dipole moment is zero.

H_{2}C=O is a polar molecule so it will have dipole moment.

CH_{2}Cl_{2} is a polar molecule so it will have dipole moment.

NH_{3} has nitrogen atom as more electronegative than hydrogen atom. So, net dipole moment will be in the direction of nitrogen atom.

Thus, we can conclude that the molecule CH_{3}-CH_{3} is expected to have a dipole moment of zero.

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How many double bonds does CCL2H2 have?
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None. Both chlorines and both hydrogens are single-bonded to the central carbon atom; the molecule is comprised of four single bonds and no double bonds.

Hope this helps!
4 0
3 years ago
According to Hund's rule of maximum spin multiplicity, how many singly-occupied orbitals are there in the valence shells of the
leva [86]

Answer:

A) carbon  - 2

B) cobalt  - 3

C) sulfur   - 2

D) fluorine   - 1

E) titanium   - 2

F) germanium  - 2

Explanation:

Hund's rule of maximum multiplicity:-

Firstly, every orbital which is present in the sublevel is singly occupied and then the orbital is doubly occupied.  

(A) Carbon.

The electronic configuration is -  

1s^22s^22p^2

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, Carbon has 2 singly occupied orbitals.

(B) Cobalt.

The electronic configuration is -  

1s^22s^22p^63s^23p^63d^{7}4s^2

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 4 electrons will be paired in 2 orbitals and 3 orbitals will be singly filled in cobalt.

(C) Sulfur.

The electronic configuration is -  

1s^22s^22p^63s^23p^4

Thus, 3s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 2 electrons will be paired in 1 orbital and 2 orbitals will be singly filled in sulfur.

D) fluorine

The electronic configuration is -  

1s^22s^22p^5

Thus, 2s orbital is fully filled and p orbital can singly filled 3 electrons. Thus, 4 electrons will be paired in 2 orbitals and 1 orbital will be singly filled in fluorine.

E) Titanium

The electronic configuration is -  

1s^22s^22p^63s^23p^63d^{2}4s^2

Thus, 4s orbital is fully filled and d orbital can singly filled 5 electrons. Thus, 2 orbitals will be singly filled in titanium.

F) Germanium

The electronic configuration is -  

1s^22s^22p^63s^23p^63d^{10}4s^24p^2

Thus, 4s, 3d orbitals are fully filled and p orbital can singly filled 3 electrons. Thus, Germanium has 2 singly occupied orbitals.

4 0
3 years ago
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4 years ago
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Andrej [43]
<h3><u>moles of H2SO4</u></h3>

Avogadro's number (6.022 × 1023) is defined as the number of atoms, molecules, or "units of anything" that are in a mole of that thing. So to find the number of moles in 3.4 x 1023 molecules of H2SO4, divide by 6.022 × 1023 molecules/mole and you get 0.5646 moles but there are only 2 sig figs in the given so we need to round to 2 sig figs. There are 0.56 moles in 3.4 x 1023 molecules of H2SO4

Note the way this works is to make sure the units are going to give us moles. To check, we do division of the units just like we were dividing two fractions:

(molecules of H2SO4) = (molecules of H2SO4)/1 and so we have 3.4 x 1023/6.022 × 1023 [(molecules of H2SO4)/1]/[(molecules of H2SO4)/(moles of H2SO4)]. Now, invert the denominator and multiply:

<h3 />
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A..........is the correct answer

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