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WINSTONCH [101]
3 years ago
12

Be able to describe the relationship between Freedom of Movement and phase change.

Chemistry
1 answer:
lisabon 2012 [21]3 years ago
7 0

Ion knoe Wdym by “be able to describ’ so ima put it in my own words idr lol:)

if you talm bout some kentic energy or sum ok but other Dan dat ion knoe tbh

I can explain how transferring kinetic energy in and out of a substance can cause a change

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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
In a 51.0-g aqueous solution of methanol ch4o the mole fraction of methanol is 0.270. what is the mass of each component?
gladu [14]
Mole fraction<span> is another way of expressing the concentration of a solution or mixture. It is equal to the </span>moles<span> of one component divided by the total </span><span>moles.

mass CH4O = 0.270 (51) = 13.77 g methanol

water = 51 - 13.77 = 37.23 g water

Hope this answers the question. Have a nice day.</span>
4 0
3 years ago
What have chemists done to help pople conserve energy
Ainat [17]
Chemists have developed insulation.
4 0
3 years ago
Arrange in terms of increasing atomic radius. From smallest to largest.<br> S, F, Sr, Na, Ga
BaLLatris [955]

Answer:

F<S<Na<Ga<Sr

Explanation:

Atomic radius increased from the right of the periodic table to left, and from the top to the down

5 0
3 years ago
The US mail delivery considered a network true or false
Svetllana [295]
True
Hope this helps!!❤️
6 0
3 years ago
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