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Temka [501]
3 years ago
7

Based on what you know about the relative temperatures of Earth’s geological layers, what can be said about the crust, mantle, a

nd core of other terrestrial planets?
For any terrestrial planet, the
would be the coolest layer, and the
would be the hottest layer.
Chemistry
1 answer:
Brrunno [24]3 years ago
8 0

Answer:

Any terrestrial planet has highest temperature at its core and least temperature at its crust.  

HOPE THIS HELPS! <3

Explanation:

Considering the composition of earth, there are three main layers whose composition and temperature. Here as follows-

<em>a) Inner Core – The inner core is made up of iron and nickel and has a temperature of 7200 degree Celsius. </em>

<em> </em>

<em>b) Molten Outer core – It is also made up of iron and nickel and has a temperature of 4300 degree Celsius. </em>

<em> </em>

<em>c) Inner mantle – It is semi rigid and made up of silicon, oxygen, magnesium, iron, aluminum, and calcium and has a temperature of 3700 degree Celsius </em>

<em> </em>

<em>d) Crust – It is made up of water, basalt and granite and has a temperature of 0 degree Celsius </em>

<em />

Looking at the composition and the temperature of the different layers, it shows that any terrestrial planet has highest temp in the core and the least at its crust!

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a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

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<u>Step 2:</u> The balanced equation

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molarity = moles / volume

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Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

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Explanation:

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