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Illusion [34]
2 years ago
15

How does mass affect potential energy?

Chemistry
1 answer:
Andreyy892 years ago
8 0

What i would say: The amount of gravitational potential energy an object has depends on its height and mass. The heavier the object and the higher it is above the ground, the more gravitational potential energy it holds. Gravitational potential energy increases as weight and height increases.

Hope this helps! :)

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What is the relationship between the period and the orbital radius? Select one: a. It is a direct relationship- As the period in
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You take out your best silver spoons and notice that they are very dull and have some black spots, chemical or physical change?
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Consider the reaction FeO (S) + CO(g) <-----> 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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