Answer:
kilograms of sulfur will be emitted from coal plants in two weeks
Step-by-step explanation:
Power produced by power plants in the United States :P
P=

1 Joule = Watts × Seconds
Watts = Joule/Second
P=

This means that
Joules of enrgy is produced in one second.
So, energy produced in 2 week:
1 week = 7 days
1 day = 24 hours
1 hour = 3600 seconds
2 week = 2 × 7 × 24 × 3600 s=1,209,600 s
Energy produced in 1,209,600 s: E

Coal has an energy content =

Mass of coal used while producing E amount of energy: m




Percentage of sulfur in coal = 4% by mass
Mass of sulfur produced from
in 2 weeks:
