___MgCl2(s) +___ MiCO3(s)
1 answer:
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Explanation:
The given reaction equation is as follows.
The value of = 59.1 kJ/mol
We know that ,
or,
= -(\frac{59.1 kJ/mol}{(8.314 \times 10^{-3} kJ/mol.K \times 310 K))}[/tex]
= -22.93
or,
=
Therefore,
=
=
Therefore, we can conclude that the value of in the intestinal tract is
.