Ok so O3 has a greater electronegativity and is taken into account first, -2*3=-6, so As has to equal 6/2=3, so As has a +3 oxidation number here
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
Answer:
for the given reaction is -2486.3 kJ
Explanation:
The given equation can be written as a combination of the following equation:
; 
; 
; 
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
