Chemistry, 12.06.2021 23:30 BWomack214

In the Minnesota Department of Health set a health risk limit for acetone in groundwater of . Suppose an analytical chemist receives a sample of groundwater with a measured volume of . Calculate the maximum mass in milligrams 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. Be sure your answer has the correct number of significant digits.

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Answer from: aedelfrance9250

The question is incomplete, the complete question is;

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.

0.0047mg

Explanation:

From the question, we have the following information;

density of acetone = 60.0 μg/L

Volume of water sample= 79.0 mL

Mass of acetone in water= ?

Solution:

Using:

d = m/v

d= density

m= mass

v=volume

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Substituting into the given formula:

d = m/v

60.0 μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

maximum mass in milligrams of acetone = 0.0047mg

Answer from: Quest

the electrical conductivity of asolution of an electrolyte is measured by determining the resistance of thesolution between two flat or cylindrical electrodes separated by a fixed distance.

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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In the Minnesota Department of Health set a health risk limit for acetone in groundwater of . Suppos...

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