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a_sh-v [17]
2 years ago
5

In the Minnesota Department of Health set a health risk limit for acetone in groundwater of 60.0 μg/L . Suppose an analytical ch

emist 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.
Chemistry
1 answer:
siniylev [52]2 years ago
3 0

Answer:

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

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A 2.0 L container of nitrogen gas had a pressure of 3.2 atm. What volume would be necessary to decrease the pressure to 1.0 atm
beks73 [17]

Answer:

6.4 L

Explanation:

When all other variables are held constant, you can use Boyle's Law to find the missing volume:

P₁V₁ = P₂V₂

In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the theoretical volume by plugging the given values into the equation and simplifying.

P₁ = 3.2 atm                      P₂ = 1.0 atm

V₁ = 2.0 L                          V₂ = ? L

P₁V₁ = P₂V₂                                                    <----- Boyle's Law

(3.2 atm)(2.0 L) = (1.0 atm)V₂                        <----- Insert values

6.4 = (1.0 atm)V₂                                           <----- Simplify left side

6.4 = V₂                                                        <----- Divide both sides by 1.0

6 0
2 years ago
The elements in yellow are referred to as ___________.
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Explanation:

hope this helps! :)

8 0
2 years ago
Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you
Alja [10]

<u>Answer:</u> The molality of the solution is 0.11 m

<u>Explanation:</u>

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (methanol) = 0.135 g

M_{solute} = Molar mass of solute (methanol) = 32 g/mol

W_{solvent} = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m

Hence, the molality of the solution is 0.11 m

6 0
3 years ago
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