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.
Answer:
B) 7.7
Explanation:
For the reaction Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)
Kc = (CO₃²⁻) / (CrO₄²⁻)
and the Ksp given are
Ag₂CO₃ ⇒ 2 Ag⁺(aq) + CO₃²⁻(aq) Ksp₁ = (Ag⁺)²(CO₃²⁻)
Ag₂CrO₄ ⇒ 2 Ag⁺(aq)+ CrO₄²⁻(aq) Ksp₂ = (Ag⁺)²(CrO₄²⁻)
Where (...) indicate concentrations M
Notice if we divide the expressions for Ksp we get:
Ksp₁/Ksp₂ = (CO₃²⁻) / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7
which is the desired answer.
The state of matter with the slowest moving atoms is solid!
Answer:
14
Explanation:
Given parameters:
Number of protons = 6
Number of electrons = 6
Number of neutrons = 8
Unknown:
Mass number = ?
Solution:
The mass number of an atom is the sum of protons and neutrons in an atom;
Mass number = number of protons + number of neutrons
Mass number = 6 + 8 = 14