In the minnesota department of health set a health risk limit for acetone in groundwater of 60.0 μg/l . suppose an analytical chemist receives a sample of groundwater with a measured volume of 79.0 ml.
calculate the maximum mass in micrograms of acetone which the chemist could measure in this sample and still certify that the groundwater from which it came met minnesota department of health standards. round your answer to significant digits.

m = 4.7 μg

Explanation:

Given data:

density of acetone = 60.0  μg/L

Volume = 79.0 mL

Mass = ?

Solution:

Formula:

d = m/v

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Now we will put the values on formula:

d = m/v

60.0  μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

So health risk limit for acetone = 4.7  μg

answer; /// i do believe that the correct answer is (a) boiling

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.