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.
Could you show the characteristics?
<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
A non-polar covalent bond is defined as the bond which is formed between the atoms having no difference in electronegativity values. For Example: 
etc..
In this bond, the electrons are shared equally and 
value is equal to 0.
Hence, the correct answer is Option D.
Answer:
neutrons
Explanation:
isotopes have the same number of protons nut different number of neutrons. the difference in thr mumber of neutrons give them their uniques properties.
Answer:
a. 2,9x10⁻⁴ M HCl
Explanation:
A solution is considered acidic when its concentration of H⁺ is higher than 1x10⁻⁷. The higher concentration of H⁺ will be the most acidic solution.
a. 2,9x10⁻⁴ M HCl. In water, this solution dissolves as H⁺ and Cl⁻. That means concentration of H⁺ is 2,9x10⁻⁴ M.
b. 4,5x10⁻⁵M HNO₃. In the same way, concentration of H⁺ is 4,5x10⁻⁵M.
c. 1,0x10⁻⁷M NaCl. As this solution doesn't produce H⁺, the solution is not acidic
d. 1,5x10⁻²M KOH. This solution produce OH⁻. That means the solution is basic nor acidic.
Thus, the solution considered the most acidic is a. 2,9x10⁻⁴ M HCl, because has the higher concentration of H⁺.
I hope it helps!
