That will be 0.78*10 raise to the power of -1 nanometers
Explanation:
The given data is as follows.
Volume of lake =
= 
Concentration of lake = 5.6 mg/l
Total amount of pollutant present in lake = 
=
mg
=
kg
Flow rate of river is 50 
Volume of water in 1 day = 
=
liter
Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are
or 
Flow rate of sewage = 
Volume of sewage water in 1 day =
liter
Concentration of sewage = 300 mg/L
Total amount of pollutants =
or 
Therefore, total concentration of lake after 1 day = 
= 6.8078 mg/l
= 0.2 per day
= 6.8078
Hence,
= 
=
= 1.234 mg/l
Hence, the remaining concentration = (6.8078 - 1.234) mg/l
= 5.6 mg/l
Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.
Gases are less dense and the molecules are farther apart which means that it can be compressed
Answer: genus
Explanation:
Amoebas do not form a single taxonomic group; instead, they are found in every major lineage of eukaryotic organisms. Amoeboid cells occur not only among the protozoa, but also in fungi, algae, and animals.
Potassium hydroxide (KOH) is formed when Potassium forms ionic bonds with OH-ions while Potassium Oxide (K2O) is formed when potassium forms ionic bonds with the Oxide (O2-) ions. i.e. This reaction is a neutralization reaction and occurs when an acid (HCl) reacts with a base (KOH).