If I have 15.40mL of 1.4M HBr(aq) and add 22.10mL KOH what is the molarity of
KOH being used?

The molarity of KOH is 0.98 M.

Explanation:

KOH + HBr →KBr + H₂O

As per the above reaction, equal moles of KBr and HBr reacts to form 1 mole of KBr and 1 mole of water. We have to find the molarity of KOH by using the law of Volumetric analysis using the equation as,

V1M1 = V2M2

Here V1 and M1 are the volume and molarity of HBr.

V2 and M2 are the volume and molarity of KOH .

We can find the molarity of KOH as,

Plugin the values as,

    =  0.98 M

So the molarity of KOH is 0.98 M.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-