The first step is to write the <u>whole reactions</u>:
A)
B)
In A) reaction we will have an <u>acid-base reaction</u> (double replacement reaction). In this reaction, the <u>oxidation number</u> is the same for all atoms on both sides of the reaction.
In B) the <u>oxidation number</u> of Cu will change from zero to +2 and the <u>oxidation number</u> of Ag will change from +1 to zero. So, the Cu will be <u>oxidized </u>and the Ag will <u>reduce</u>. So, if we have a redox reaction for B) the charge must be equal on both sides.
The amount of HCl required for one experiment - 13.5 µl the volume in terms of L - 13.5 x 10⁻⁶ L the volume of HCl available - 0.250 L since one experiment uses up - 13.5 x 10⁻⁶ L then number of experiments - 0.250 L / 13.5 x 10⁻⁶ L = 1.8 x 10⁴ times the experiment can be carried out 18000 times