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3 years ago

11

Two friends are arguing about the structure of the universe. Caleb says that the universe is made of many galaxies that are made

of millions of stars and many galaxies that are made of millions of planets. Maria says that he is wrong because galaxies like the Milky Way are made up of millions of universes with millions of stars.

1 answer:

kiruha [24]3 years ago

6 0

I think that Maria is wrong

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Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

HELP PLEASE!!!!!!!!! What if you started with 100 grams of wood and only have 95 grams of ashes produced, is mass still conserve

Lerok [7]

My guess is smoke. Smoke is matter.

7 0

3 years ago

DENIUS [597]

I believe that it is B

3 0

3 years ago

How can a student use the diagram to calculate the atomic mass for the helium atom

djyliett [7]

Answer:

Add G and H to determine the atomic mass for helium

Explanation:

5 0

3 years ago

A dull metal object has a density of 8.8 G/ML and a volume of 20 ML calculate the mass

Alex

Answer:

Mass = 0.000176 gram

Steps:

m = V × ρ

= 20 milliliter × 8.8 gram/cubic meter

= 2.0E-5 cubic meter × 8.8 gram/cubic meter

= 0.000176 gram

Explanation:

8 0

3 years ago