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natka813 [3]
3 years ago
12

1) ΔS depends not merely on q but on ______ . Although there are many possible paths that could take a system from its initial t

o final state, there is always only one _____ isothermal path between two states. Thus, ΔS _____ _______ the path taken between states. Options: q rev, reversible,irreversible, has only one particular value, regardless of, can have different values, which depends on, q sys2)Which system has the greatest entropy?1 mol of H2(g) at STP1 mol of H2(g) at 100∘C 0.5 atm1 mol of H2O(s) at 0∘C1 mol of H2O(l) at 25∘C3)During _____ , the temperature _____ but the entropy change can be large as molecules _____ their degrees of freedom and motion. Options: a phase change, remains constant, increases, heating, raises, reaction, decrease, falls
Chemistry
1 answer:
Rus_ich [418]3 years ago
4 0

Answer:

1) qrev/ reversible/ has only one particular value/ regardless of;

2) 1 mol of H₂(g) at 0ºC

3) a phase change/ remains constant/ increase

Explanation:

1) The change can occur in a reversible or an irreversible process. When the change is reversible, the system and the surroundings can be restored to their original state by exactly reversing the change. In an irreversible process, that is not possible.

The entropy variation (ΔS) is calculated only in the reversible process because, in the irreversible one, it's difficult to determine it. So, ΔS depends on qrev, which is the heat for the reversible process.

There is only one reversible isothermal path between two states, that's why it's easy to calculate ΔS for it.

The value of ΔS doesn't depend on the path, but only on the initial and the final states, so ΔS has only one particular value regardless of the path taken between states.

2) The entropy is the measure of the disorganization of a system, so when the molecules are more distant and vibrating, the entropy is higher (Sgas > Sliquid > Ssolid), and when the temperature increases, the entropy decreases (ΔS = qrev/T), so 1 mol of H₂(g) at 0ºC has the greatest entropy.

3) During a phase change, for a pure substance, the temperature remains constant, until all the substance changes for the other phase. The entropy is the measure of the disorganization of a system, so when the degrees of freedom and motion of the molecules increase, the system becomes more disorganized, so the entropy increases.

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NO2 Hope this helped!
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Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
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Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

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3 years ago
Which law is associated with inertia? *
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Newton first law of motion

Explanation:

This law by Newton states that an object will remain at rest, or a moving object will stay in motion in a straight line and with the same speed unless an external and unbalanced force acts on the object.

One application of this Newtonian law is in the launching of projectiles like ballistic missiles.  A ballistic missile parabolic path is affected by its inertia and the unbalanced forces acting on the object such as air resistance/ drag and gravity.

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Iron can be extracted from its ore with the used of blast furnace. The materials used for extraction of iron includes:

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--> haematite( iron ore)

--> limestone and

--> Hot air.

The iron ore is first roasted in air so that iron(III)oxide is produced. The iron(III)oxide is then mixed with coke and limestone and heated to a very high temperature. Hot air is introduced into it from the bottom of the furnace. The coke is oxidizes the the oxygen in the hot air blast to liberate carbondioxide.

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