Answer:
T2 = 94.6 C
Explanation:
Use Clausius-Clayperyon equation.
ln P1/P2 = ∆Hvap/R (1/T2 - 1/T1) where R = 8.314 J/mol-K and T is in degrees K
P1 = 760 mmHg
P2 = 630 mmHg
T1 = 373 K
T2 = ?
∆Hvap = 40.7 kJ/mole
R = 0.008314 kJ/mole-K (NOTE: change R to units of kJ)
Plug in and solve for T2
ln 760 mmHg/630 mmHg = 40.7 kJ/mole (1/T2 - 1/373K)
T2 = 367.74 K = 94.6 C
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
706000 in scientific notation is 7.06x10^5
