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galina1969 [7]
3 years ago
6

Humans breathe about 500 mL of air per breath and take about 12 breaths per minute during normal activities. If a person is expo

sed to an atmosphere containing benzene at a concentration of 10 ppm (by volume), how many grams of benzene will be deposited in the lungs during an 8-hour shift if all the benzene that enters remains in the lungs
Chemistry
1 answer:
sasho [114]3 years ago
3 0

Answer:

mass of benzene in the lungs = 28.8 g of benzene

Explanation:

Volume of air breathed per hour = 500 mL * 12 * 60 = 360000 ml = 360 Litres of air

In 8 hours, volume of breathed air = 360 L * 8 = 2880 Litres of air

Concentration of benzene in air = 10 ppm

Note:  1 ppm = 1 mg per liter (mg/L)

10 ppm = 10 mg/L

Therefore, mass of benzene in the lungs in an 8-hour shift = 10 mg/L * 2880 L

mass of benzene in the lungs = 28800 mg

Converting to grams = 28800/1000

mass of benzene in the lungs = 28.8 g of benzene

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Answer:

Molar mass of vitamin K = 450.56\frac{g}{mol}[/tex]

Explanation:

The freezing point of camphor = 178.4 ⁰C

the Kf of camphor =  37.7°C/m

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the relation between freezing point depression and molality is

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we know that moles are related to mass and molar mass of a substance as:

moles=\frac{mass}{molarmass}

For vitamin K the mass is given = 0.802 grams

therefore molar mass = \frac{mass}{moles}=\frac{0.802}{0.00178}=450.56\frac{g}{mol}

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3 years ago
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