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Keith_Richards [23]
3 years ago
12

The __lithosphere__ combines the two crusts, oceanic and continental, with the coolest part of the __________________. The asthe

nosphere allows the tectonic plates to slide on the putty substance of the mantle and the mesosphere is the ___________________ part of the mantle. The outer core is hot enough to turn the metal into a ________________. The inner core is the hottest but the extreme pressure makes it a ________________.
Please help me
Chemistry
2 answers:
zmey [24]3 years ago
8 0

Answer:

The lithosphere And the upper mantle combines the two crusts, oceanic and continental, with the coolest part of the Crust?. The asthenosphere allows the tectonic plates to slide on the putty substance of the mantle and the mesosphere is the Smallest? part of the mantle. The outer core is hot enough to turn the metal into a Liquid The inner core is the hottest but the extreme pressure makes it a Solid.

klio [65]3 years ago
4 0

Answer:

....,............

........

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Sodium metal and water react to create sodium hydrogen and hydrogen gas through the unbalanced equation.
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Answer:

Theoretical yield = 2.5 g

Explanation:

Given data:

Mass of sodium = 79.7 g

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Theoretical yield of hydrogen gas = ?

Solution:

Chemical equation:

2Na + 2H₂O → 2NaOH + H₂

Number of moles of sodium:

Number of moles = mass/ molar mass

Number of moles = 79.7 g / 23 g/mol

Number of moles = 3.5 mol

Number of moles of water:

Number of moles = mass/ molar mass

Number of moles = 45.3 g / 18g/mol

Number of moles = 2.5 mol

Now we will compare the moles of hydrogen gas with water and sodium.

                        H₂O           :             H₂

                           2             :              1

                          2.5           :            1/2×2.5 =1.25 mol

                     

                           Na           :              H₂

                             2            :               1

                           3.5           :             1/2×3.5 =1.75 mol

water will be limiting reactant.

Theoretical yield:

Mass = number of moles × molar mass

Mass =  1.25 mol  × 2 g/mol

Mass = 2.5 g

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

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Answer:

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Explanation:

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