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torisob [31]
3 years ago
12

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

c data are given below. 2 Ag2O(s) → 4 Ag(s) + O2(g) ΔH°f(kJ/mol) –31.1 -- -- S°(J/K·mol) 121.3 42.55 205.07 What are the values of ΔH°, ΔS° and ΔG°?
Physics
1 answer:
yuradex [85]3 years ago
6 0

Answer : The values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

Explanation :

The given balanced chemical reaction is,

2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)

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

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

\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]

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=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]

\Delta H^o=62.2kJ=62200J

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_{product}

\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]

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=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]

\Delta S^o=132.67J/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 298 K.

\Delta G^o=(62200J)-(298K\times 132.67J/K)

\Delta G^o=22664.34J=22.66kJ

Therefore, the values of \Delta H^o,\Delta S^o\text{ and }\Delta G^o are 62.2kJ,132.67J/K\text{ and }22.66kJ respectively.

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Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

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The average velocity is given by;

V_{avg} = \frac{Final \ position\  - \ Initial \ position}{final \ time\  - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\  - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr

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3 years ago
A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a
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Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec            

Explanation:

We have given acceleration of the car a_1=5.9m/sec^2

Acceleration of the stock car a_2=3.6m/sec^2

When 1st car overtakes the second car then distance traveled by both the car will be same

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As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So \frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6

After solving t = 3.56 sec

(b) From second equation of motion s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m

(c) From first equation pf motion v = u+at

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

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Genrish500 [490]

Answer:

36.408cm3

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Since we acknowledge that density is d= m/v, once we switch it up to maintain v as the number to be found it will change to v=m/d. Therefore, 275.32/7.562 is 36.408 and the unit is cm cube!

Hope that helped!!

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21. Calculate the acceleration of the bus from point D to E. Show your work.
Marat540 [252]

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Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

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In this problem, we want to measure the acceleration of the bus from point D to point E. We have:

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Therefore, the acceleration between D and E is

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22) This question is the same as 21), so the result is the same.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

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#LearnwithBrainly

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Answer:

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