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Troyanec [42]
2 years ago
12

Which of the following is a true statement? A. In the process of freezing, a liquid loses kinetic energy and turns into a solid.

B. In the process of freezing, a solid loses kinetic energy and turns into a liquid. C. In the process of freezing, a liquid gains kinetic energy and turns into a solid. D. In the process of freezing, a solid gains kinetic energy and turns into a liquid.
Chemistry
2 answers:
Firlakuza [10]2 years ago
7 0

The correct answer is A. In the process of freezing, a liquid loses kinetic energy and turns into a solid.

Explanation:

The kinetic energy in a substance is produced by the movement of the participles in it. Additionally, this depends on the state of the substance, for example, in gases substances move freely and therefore produce more kinetic energy than in liquids in which the movement is more restricted or in solids in which particles only vibrate. This means, in the process of freezing that implies a liquid turning into a solid the particles lose kinetic energy as their movement is limited in a solid state.

malfutka [58]2 years ago
4 0

its A. In the process of freezing, a liquid loses kinetic energy and turns into a solid. good luck ;)

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Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu
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Solution :

The balanced cell reaction will be,  

Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

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

E^0_{[Pb^{2+}/Pb]}=-0.13V

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E^0_{cell}=E^0_{cathode}-E^0_{anode}

E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}

E^0_{cell}=-0.13V-(0.34V)=-0.47V

Now we have to calculate the standard Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o_{cell}

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\Delta G^o = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole

Therefore, the standard Gibbs free energy is +91 kJ/mole

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