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kykrilka [37]
3 years ago
13

Which member of the following pairs has the larger London dispersion forces?

Chemistry
1 answer:
Virty [35]3 years ago
4 0

Answer:

H₂S; CO₂; SiH₄

Explanation:

London dispersion forces are larger in molecules that are large and have more atoms or electrons.

A. H₂O or H₂S

H₂S. S is below O in the Periodic Table, so it is the larger atom. Its electrons are more polarizable.

B. CO₂ or CO

CO₂. CO₂ has more atoms. It is also linear, so the molecules can get close to each other and maximize the attractive forces.

C. CH₄ or SiH₄

CH₄. Si is below C in the Periodic Table, so it is the larger atom. Its electrons are more polarizable.

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Sodium(na) is a highly reactive metal and chlorine is a poisonous gas. Which compound do they form when they chemically combine?
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Calculate ΔHrxn for the following reaction: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) Use the following reactions and given ΔH′s. 2Fe(s)+3/
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Answer:

The answer to your question is:  ΔHrxm = -23.9 kJ

Explanation:

Data:

2Fe(s)+3/2O2(g)→Fe2O3(s),  ΔH = -824.2 kJ     (1)

    CO(g)+1/2O2(g)→CO2(g)         ΔH = -282.7 kJ   (2)

Reaction:

                          Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)  

We invert (1) and change the sign of  ΔH

        Fe2O3(s)   →  2Fe(s)+3/2O2(g)  ΔH = 824.2 kJ

We multiply (2) by 3

      3(  CO(g)+1/2O2(g)→CO2(g)         ΔH = -282.7 kJ)   (2)

      3CO(g)+3/2O2(g)→3CO2(g)         ΔH = -848.1 kJ

We add (1) and (2)

Fe2O3(s)   →  2Fe(s)+3/2O2(g)  ΔH = 824.2 kJ

3CO(g)+3/2O2(g)→3CO2(g)         ΔH = -848.1 kJ

Fe2O3(s) +  3CO(g)+3/2O2(g)  →  2Fe(s)+3/2O2 + 3CO2(g)    

Simplify

    Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)    and ΔHrxm = -23.9 kJ

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3 years ago
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