Answer:
kilograms of sulfur will be emitted from coal plants in two weeks
Explanation:
Power produced by power plants in the United States :P
P=![2.6\times 10^{11} Watts](https://tex.z-dn.net/?f=2.6%5Ctimes%2010%5E%7B11%7D%20Watts%20)
1 Joule = Watts × Seconds
Watts = Joule/Second
P=![2.6\times 10^{11} J/s](https://tex.z-dn.net/?f=2.6%5Ctimes%2010%5E%7B11%7D%20J%2Fs)
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
![E=2.6\times 10^{11} J/s\times 1,209,600 s](https://tex.z-dn.net/?f=E%3D2.6%5Ctimes%2010%5E%7B11%7D%20J%2Fs%5Ctimes%201%2C209%2C600%20s)
Coal has an energy content =![2.4\times 10^7 J/Kg](https://tex.z-dn.net/?f=2.4%5Ctimes%2010%5E7%20J%2FKg)
Mass of coal used while producing E amount of energy: m
![m\times 2.4\times 10^7 J/Kg=E](https://tex.z-dn.net/?f=m%5Ctimes%202.4%5Ctimes%2010%5E7%20J%2FKg%3DE)
![m=\frac{E}{2.4\times 10^7 J/Kg}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7BE%7D%7B2.4%5Ctimes%2010%5E7%20J%2FKg%7D)
![m=\frac{2.6\times 10^{11} J/s\times 1,209,600 s}{2.4\times 10^7 J/Kg}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B2.6%5Ctimes%2010%5E%7B11%7D%20J%2Fs%5Ctimes%201%2C209%2C600%20s%7D%7B2.4%5Ctimes%2010%5E7%20J%2FKg%7D)
![m = 1.3104\times 10^{10} kg](https://tex.z-dn.net/?f=m%20%3D%201.3104%5Ctimes%2010%5E%7B10%7D%20kg)
Percentage of sulfur in coal = 4% by mass
Mass of sulfur produced from
in 2 weeks:
![\frac{4}{100}\times 1.3104\times 10^{10} kg=5.2416\times 10^8 kg](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B100%7D%5Ctimes%201.3104%5Ctimes%2010%5E%7B10%7D%20kg%3D5.2416%5Ctimes%2010%5E8%20kg)
If the.pressure exerted by a gas at [math]25^{\circ} \mathrm{C}[/math] in a volume of 0.044 L is 3.81 atm, how many moles of gas are present
Explanation:
that was awesome thanks for this lovley question
Answer:
I think the answer is Prokaryote and Eukaryote
Explanation: