There are ALOT because they would always come in and out and they will burst which creates more so techneclly there are infinate
The balanced equation for the ionization of the weak base pyridine,C5H5N in water, H2O
C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)
<h3>What is the balanced equation for the ionization?</h3>
Generally, Pyridine is characterized by a ring structure, in this characteristic ring structure N is sp2 hybridized, hence creating a lone pair present on N so s - character is more, as well as lone pair, is present.
Therefore, Considering The following functions of the equation:weak base pyridine,C5H5N in water, H2O
We write the balanced equation for the ionization as
C_5H_5N ( aq.) + H2O ( l) ---------> C5H5NH+ (aq.) + OH- (aq.)
Read more about Chemical Reaction
brainly.com/question/11231920

a)


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b)
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
Using the atomic mass given in the periodic table:




c)


Using the atomic mass given in the periodic table:




This result is an aproximation.