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4 years ago

Which lists the elements in order from least conductive to most conductive?

Chemistry

2 answers:

TEA [102]4 years ago

6 0

Answer:

It's A) nitrogen (N), antimony (Sb), bismuth (Bi)

Explanation: Just got it from the test

AfilCa [17]4 years ago

5 0

nitrogen (N), antimony (Sb), bismuth (BI)

Explanation:

Electrical conductance increases from Nitrogen to antimony and then Bismuth. Electrical conductance increases with available free electrons.

The elements listed belongs to Group V on the periodic table.
Group V elements are Nitrogen, Phosphorus, Arsenic, Antimony, Bismuth.
As we go down the group, electrical conductance increases.
This is because the radius of the atom becomes larger and the pull on the outer electrons become very weak.
Outer electrons becomes free to facilitate electrical conductance.
We can also see that metallicity i.e electropositivity increases down the group.

Learn more:

Metallic bonds brainly.com/question/6071838

#learnwithBrainly

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3 years ago

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The answer for this question is:

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4 years ago

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H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles

abruzzese [7]

Answer: The value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]$

We are given:

$\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = $\frac{-14.1}{1}\times 2.20=-31.02J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

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3 years ago

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telo118 [61]

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