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kicyunya [14]
3 years ago
8

At STP, fluorine is a gas and iodine is a solid. This observation can be explained by the fact that fluorine has

Chemistry
2 answers:
motikmotik3 years ago
6 0
<span>The answer is 4. The molecules of each material entice each other over dispersion (London) intermolecular forces. Whether a substance is a solid, liquid, or gas hinge on the stability between the kinetic energies of the molecules and their intermolecular magnetisms. In fluorine, the electrons are firmly apprehended to the nuclei. The electrons have slight accidental to stroll to one side of the molecule, so the London dispersion powers are comparatively weak. As we go from fluorine to iodine, the electrons are far from the nuclei so the electron exhausts can more effortlessly misrepresent. The London dispersion forces developed to be increasingly stronger.</span>
Ierofanga [76]3 years ago
3 0

Answer:

F THE OTHER WRONG ANSWER ITS <u><em>2</em></u>

Explanation:

ITS 2 BLESSING!!!!

MAN RESPECT

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What type of reactive intermediate is formed in the reaction of an alkene with aqueous acid to give an alcohol?
Mariana [72]

Answer:

The intermediate is carbocation.

Explanation:

The reaction of alkene with an acid to give alcohol is an electrophilic addition reaction.

In electrophilic reaction, the alkene is being attacked by an alkene.

in case of aqueous acid reaction with alkene , the electrophile is [H⁺].

When the proton attacks the alkene it generates carbocation.

The carbocation generated is formed based on stability of carbocation.

The mechanism is shown in the figure.

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I think the awnser would be C. becuase Pure Substances are made of mostly the same material and properties Such as: Nitrogen Gas, Oxygen Gas, Carbon, Iorn, Water, Sugar and Salt.
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Darya [45]

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Consider the reaction FeO (S) + CO(g) &lt;-----&gt; Fe(s) + CO2(g) for which KP is found to have the following values:
Svetach [21]

Explanation:

\Delta G^o=-RT\ln K_1

where,

R = Gas constant = 8.314J/K mol

T = temperature = 600^oC=[273.15+600]K=873.15 K

K_p = equilibrium constant at 600°C =  0.900

Putting values in above equation, we get:

\Delta G^o=-(8.314J/Kmol)\times 873.15 K\times \ln (0.900 )

\Delta G^o=764.85 J/mol

The ΔG° of the reaction at 764.85 J/mol is 764.85 J/mol.

Equilibrium constant at 600°C = K_1=0.900

Equilibrium constant at 1000°C = K_2=0.396

T_1=[273.15+600]K=873.15 K

T_2=[273.15+1000]K=1273.15 K

\ln \frac{K_2}{K_1}=\frac{\Delta H^o}{R}\times [\frac{1}{T_1}-\frac{1}{T_2}]

\ln \frac{0.396}{0.900}=\frac{\Delta H^o}{8.314 J/mol K}\times [\frac{1}{873.15 K}-\frac{1}{1273.15 K}]

\Delta H^o=-18,969.30 J/mol

The ΔH° of the reaction at 600 C is -18,969.30 J/mol.

ΔG° = ΔH° - TΔS°

764.85 J/mol = -18,969.30 J/mol - 873.15 K × ΔS°

ΔS° = -22.60 J/K mol

The ΔS° of the reaction at 600 C is -22.60 J/K mol.

FeO (s) + CO(g)\rightleftharpoons Fe(s) + CO_2(g)

Partial pressure of carbon dioxide = p_1=P\times \chi_1

Partial pressure of carbon monoxide = p_2=P\times \chi_2

Where \chi_1\& \chi_2 mole fraction of carbon dioxide and carbon monoxide gas.

The expression of K_p is given by:

K_p=\frac{p_1}{p_2}=\frac{P\times \chi_1}{P\times \chi_2}

0.900=\frac{\chi_1}{\chi_2}

\chi_1=0.900\times \chi_2

\chi_1+\chi_2=1

0.9\chi_2+\chi_2=1

1.9\chi_2=1

\chi_2=\frac{1}{1.9}=0.526

\chi_1=1-\chi_2=1-0.526=0.474

Mole fraction of carbon dioxide at 600°C is 0.474.

6 0
3 years ago
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