The change in pH is calculated by:
pOH = Protein kinase B + log [NH4+]/ [NH3]
Protein kinase B of ammonia = 4.74
initial potential of oxygen hydroxide= 4.74 + log 0.100/0.100 = 4.74
pH = 14 - 4.74=9.26
moles NH4+ = moles NH3 = 0.100 L x 0.100 M = 0.0100
moles H+ added = 3.00 x 10^-3 L x 0.100 M=0.000300
NH3 + H+ = NH4+
moles NH3 = 0.0100 - 0.000300=0.00970
moles NH4+ = 0.0100 + 0.000300=0.0103
pOH = 4.74 + log 0.0103/ 0.00970= 4.77
oH = 14 - 4.77 = 9.23
the change is = 9.26 - 9.23 =0.03
Answer:- 21.4 grams of
are formed.
Solution:- The balanced equation is:

From this equation, lithium and nitrogen reacts in 6:1 mol ratio. Limiting reactant gives the theoretical yield of the product. We will calculate the grams of the product for the given grams of both the reactants and see which one of them gives the limited amount of the product. This limited amount of the product will be the theoretical yield.
The molar mass of Li is 6.94 gram per mol and for
It is 28.02 gram per mol. The molar mass of
is 34.83 gram per mol. The calculations for the grams of the product for given grams of both the reactants are shown below:

= 

= 
From above calculations, Li gives least amount of the product. So, 21.4 g of
are formed.
121.75 is really really really really really the right answer... For real
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.