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kipiarov [429]
3 years ago
8

Help me the question number 1 , please !

Chemistry
2 answers:
Igoryamba3 years ago
7 0
The answer is 10x3.01x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x10x3.e+22=9.06e+44
maksim [4K]3 years ago
4 0
10•3.01•210•3.e+22=9.06e+44
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4 years ago
Which example is an endothermic process
Annette [7]

Answer:

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Explanation:

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4 0
3 years ago
The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

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

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

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

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

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I don’t see anything?
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