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3 years ago

5

What intermolecular forces are present in hydrogen chloride gas?

1 answer:

vivado [14]3 years ago

8 0

The dipole-dipole forces are present in hydrogen chloride gas. Their interaction in HCI is fairly weak. Hope that answerd your question ☺️☺️

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Which of the following statements is subjective? Van Gogh painted beautiful works of art. The planets revolve around the sun. Bl

mezya [45]

The "Van Gogh" one is subjective, because it expresses opinion and is not a true, hard fact. :)

4 0

3 years ago

Read 2 more answers

If the acceleration is 600 m/s/s and the mass is 300 kg, what force is acting on the object?

vampirchik [111]

Explanation:

Acceleration (a) = 600 m/s²

Mass (m) = 300 kg

Force (F) = ?

We know

F = m * a

= 300 * 600

=180000 Newton

The force acting on the object is 18000 Newton.

5 0

2 years ago

Read 2 more answers

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

An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.

Tju [1.3M]

Answer:

a) $\Delta S$

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Explanation:

a) The change of entropy for a reversible process:

$\delta S=\frac{\delta Q}{T}$

$\Delta S=\frac{Q}{T}$

The energy balance:

$\delta U=[tex]\delta Q- \delta W$

If the process is isothermical the U doesn't change:

$0=[tex]\delta Q- \delta W$

$\delta Q= \delta W$

$Q= W$

The work:

$W=\int_{V1}^{V2}P*dV$

If it is an ideal gas:

$P=\frac{n*R*T}{V}$

$W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV$

Solving:

$W=n*R*T*ln(V2/V1)$

Replacing:

$\Delta S=\frac{n*R*T*ln(V2/V1)}{T}$

$\Delta S=n*R*ln(V2/V1)}$

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

$\Delta S$

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.

7 0

3 years ago

babunello [35]

D.mno4- is reuced it loses h atom

7 0

3 years ago