The path of energy flow from the sun to the humpback whale is as follows:
- Sun---> Plankton ---> Small fishes ---> Humpback whale.
<h3>What is energy?</h3>
Energy is the ability to do work.
The primary source of energy on the earth is the sun.
The energy from the sun is used by producers to produce food on which other organisms depend on.
The energy from the sun gets to the humpback whale through producers such as plankton.
The path of energy flow from the sun to the humpback whale is as follows:
- Sun---> Plankton ---> Small fishes ---> Humpback whale.
Learn more about energy flow at: brainly.com/question/21786633
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:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
Answer:
5.2 x 10⁻⁴ M.
Explanation:
- The relationship between gas pressure and the concentration of dissolved gas is given by Henry’s law:
<em>P = kC</em>
where P is the partial pressure of the gaseous solute above the solution.
k is a constant (Henry’s constant).
C is the concentration of the dissolved gas.
- At two different pressures, there is two different concentrations of dissolved gases and is expressed in a relation as:
<em>P₁C₂ = P₂C₁,</em>
P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.
P₂ = 0.76 atm, C₂ = ??? mol/L.
<em>∴ C₂ = (P₂C₁)/P₁ =</em> (0.76 atm)(6.8 x 10⁻⁴ mol/L)/(1.0 atm) = <em>5.168 x 10⁻⁴ mol/L ≅ 5.2 x 10⁻⁴ M.</em>
