What happens when molten material is heated from Earth’s core?

These models show the electron structures of two different nonmetal elements. Element 1 at left has a purple circle at center wi

Daniel [21]

Answer:

Element 2

Explanation:

If we look at the model stated for element 1, it is clear that element 1 must be a noble gas. It has eight electrons in its outermost shell this implies that it has already attained a complete octet of electrons and is reluctant towards chemical reaction.

The second element belongs to group 16 since it has six electrons on its outermost shell. It is certainly more reactive than element 1 which is a noble gas.

7 0

3 years ago

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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

Apart from boiling point and flammability, what other quality does the size of the molecules in hydrocarbons affect?

Sonbull [250]

Answer:

Viscosity

Explanation:

Sijui kuexpalain lakini

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3 years ago

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Explain why the pbcl2 dissolved when water was added

lara [203]

PbCl2 would not dissolve because it is insoluble based on the solubility rules for substances that will dissolve in water. This compound would instead form a solid precipitate at the bottom of the container.

4 0

3 years ago

Which of the following gases would have the greatest kinetic energy at 300 K?

tekilochka [14]

Answer:

D. All of them would have the same kinetic energy

Explanation:

The expression for the kinetic energy of the gas is:-

$K.E.=\frac{3}{2}\times K\times T$

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

T is the temperature

<u>Since, kinetic energy depends only on the temperature. Thus, at same temperature, at 300 K, all the gases which are $N_2,\ NH_3\ and\ Ar$ will posses same value of kinetic energy.</u>

4 0

3 years ago