Help I clearly shouldnt be doing a degree in forensics if I cant do this!!

 

A suspect in a shooting incident had his hands swabbed with 5% nitric acid. The swabs were then extracted into 20 cm3 of nitric acid. A portion of the sample solution was analysed by atomic emission spectroscopy and a response of 125 mV obtained. Next a 0.100 cm3 of a standard solution containing 40 µg of lead per cm3 was added to 10 cm 3 of the sample solution and the analysis repeated. A response of 157 mV was obtained from the "spiked" sample. Calculate the amount of lead recovered by the swabs in µg.

The additional 4000 µg of lead added to half of the original sample caused a raise of 32 mV. Considering you used only half of the sample, that means you started with 62.5 mV. This means; 4000 µg of lead is the equivalent of 94.5 mV (157 - 125/2). Dividing those gives the response of the detector: 4000/94.5 = 42.4 µg of lead/mV

 

As the sample gave a signal of 125 mV, this means there was 42.4 * 125 = 5291 µg of lead in the swabs.