Here are the examples: It includes e.g. sunlight, wind, biomass, rain, tides, waves and geothermal heat.
Answer:
0.133 mL
Explanation:
Given data
- Initial concentration (C₁): 15.0 M
- Initial volume (V₁): to be determined
- Final concentration (C₂): 0.001 M
- Final volume (V₂): 2.00 L
We can find the volume of the concentrated solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.001 M × 2.00 L / 15.0 M
V₁ = 1.33 × 10⁻⁴ L = 0.133 mL
Answer is: 0.330 ppm
Hope this helps
<h3><u>Answer;</u></h3>
Find the number of 1-foot cubes that fill the fish tank
<h3><u>Explanation;</u></h3>
Volume of a cuboid such as the fish tank is given by the product of length width and height;
Such that; Volume = length × width × height
Similarly, we can count the number of 1 foot cube that can fill the fish tank.
And since each cube has a volume of 1 cubic ft, then the number of cubes will be equivalent to the volume of the fish tank in cubic ft.
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
