Answer : The 
for this reaction is, -88780 J/mole.
Solution :
The balanced cell reaction will be,
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 : 
Reduction : 
Now we have to calculate the Gibbs free energy.
Formula used :
where,
= Gibbs free energy = ?
n = number of electrons to balance the reaction = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = 0.46 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the for this reaction is, -88780 J/mole.
Answer:
There is an extra O2 molecule left over
Explanation:
The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.
Nitrogen pure
carbon is mixture
The random molecular movement from higher concentration to lower concentration
