Trial 12 1.Volume of Acid used ( mL) 2020 2.Molarity of acid used (M) 0.0160.016 3.Initial volume reading of base buret 018.6 4.Final volume reading of base used 18.636.9 5.Volume of base used (4-3, mL) 6.Molarity of base (1*2/5, M) 7.Average molarity of base

-- Volume of base used

--- Molarity of base

Explanation:

Given

-- Volume of acid used

--- Buret Initial reading

--- Buret Final reading

--- Molarity of the acid

Solving (a): Volume of base used (VB)

This is calculated by subtracting the initial reading from the final reading of the base buret.

i.e.

Solving (b): Molarity base (MB)

This is calculated using:

Make MB the subject

This gives:

Solving (c): There is no such thing as average molarity

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.