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sammy [17]
3 years ago
13

The strongest intermolecular interactions between hydrogen sulfide (H2S) molecules arise from: Group of answer choices a) disulf

ide linkages b) dipole-dipole forces c) hydrogen bonding d) London dispersion forces.
Chemistry
2 answers:
Bezzdna [24]3 years ago
7 0

Answer:

c) hydrogen bonding.

Explanation:

Like water molecules which are H2O, hydrogen disulfide also has a bent shape. However, unlike water, it cannot have hydrogen bonding. Why? Because hydrogen bonding only occurs with nitrogen, oxygen, and fluorine. I haven't heard of disulfide linkages. And yes, H2S has London dipersion forces. All polar molecules have LDFs. But greater than that, H2S has dipole-dipole forces. Event though it is a dipole-dipole bond, it is still highly electronegative.

jok3333 [9.3K]3 years ago
4 0

Answer:

hydrogen bonding which is Option C

Explanation:

un

Unlike water, it cannot have hydrogen bonding.

Resson? Because hydrogen bonding only occurs with nitrogen, oxygen, and fluorine.

There is know disulfide linkages. And yes, H2S has London dipersion forces. All polar molecules have LDFs. But greater than that, H2S has dipole-dipole forces. Event though it is a dipole-dipole bond, it is still highly electronegative.

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4 years ago
Find the ΔH for the reaction below, given the following reactions and subsequent ΔH values: PCl5(g) → PCl3(g) + Cl2(g)
nalin [4]

Answer:

The value of \Delta H for the desired reaction will be 249.75 KJ.

Explanation:

The desired reaction is shown below

\textrm{PCl}_{5}\left ( g \right )\rightarrow \textrm{PCl}_{3}\left ( g \right )+\textrm{Cl}_{2}\left ( g \right )

The desired reaction can be obtained by adding the given reactions and then dividing both sides by 4.

P_{4}\left ( s \right )+6Cl_{2}\left ( g \right )\rightarrow 4PCl_{3}\left ( g \right ) \\4PCl_{5}\left ( g \right )\rightarrow P_{4}\left ( s \right )+10Cl_{2}\left ( g \right )

Net Enthalpy change for the desired reaction

\displaystyle \frac{3438-2439}{4} \textrm{ KJ} = 249.75 \textrm{ KJ}

\Delta H = 249.75 \textrm{ KJ}

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