Answer:
a.
[tex]HBr + KOH → KBr + H _{2} O[/tex]
b.
[tex]1000 \: ml \: contains \: 0.685 \: moles \\ 55.4 \: ml \: contains \: ( \frac{55.4 \times 0.685}{1000} ) \\ = 0.038 \:moles \\ 1 \: mole \: of \: hydrobromic \: acid \: produces \: 1 \: mole \: of \: water \\ 0.038 \: moles \: produce \: (0.038 \times 1) \\ = 0.038 \: moles \\ 1 \: mole \: of \: water \: weighs \: 18 \: g \\ 0.038 \: moles \: weighs \: (0.038 \times 18) \: g \\ = 0.684 \: g[/tex]
c.
0.042 M
d.
1.4