In standardizing the 0.1 m naoh in three trial runs suppose the concentrations of naoh obtained are

0.104M

Explanation:

1) The question is incomplete. This is the complete question:

In standardizing the 0.1M NaOH in three trial runs, suppose the concentrations of NaOH obtained are .108M, .102M, and .103M. Given this data, what concentration of NaOH should be used to calculate the moles of OH- used in the titration to analyze aspirin?

2) Since, 3 trials have bee run and each time it was obtained the concentration 3 different but similar concentrations for the NaOH solution, the best number to use is the average.

Precisely, there were run 3 trials to reduce the random errors of the measurements, and since all the 3 values make sense the best is to use the average of the three data.

3) Calculations:

Average = (0.108M + 0.102M + 0.103M) / 3 = 0.313M / 3 = 0.104M

4) It could also be considered that the number 0.108 is an outlier, but that can only be verified if you run more trials and so detect if 0.108M obeys to an  issue when the first trial was done. So, with more information you could tell whether you use 0.108M to calculate the average or should discard it.

No. an alpha particle is a pair of protons and a pair of neutrons, i.e. a helium

The number of medals awarded to the track team during the season.