Fresh water pollutants are substances which pollute fresh water and industrial waste, is most harmful fresh water pollutant to man and aquatic organisms.
<h3>What are pollutants?</h3>
Pollutants are substances which cause harm when they are present in the environment.
Pollutants include chemicals such as petroleum and material such as sewage.
The presence of pollutants in freshwater results in water pollution and make the water unfit for drinking purposes and also harms aquatic life in freshwaters.
Some freshwater pollutants include:
- Farming wastes
- Household pollutants
- Industrial wastes
- Erosion
- Oil and Gasoline
- heat
Of these pollutants, the most dangerous fresh water pollutant is industrial wastes as they kill aquatic organisms most due to the presence of harmful chemicals in them.
Therefore, fresh water pollutants such as industrial waste is most harmful to man and aquatic organisms.
Learn more about pollutants at: brainly.com/question/1235358
#SPJ1
Answer:
chlorine has higher ionization than carbon
Explanation:
Chlorine is only one row below carbon, but it is three columns to the right in this case the IP of chlorine would be predicted to be greater than the IP of carbon.
Answer:
1.60x10⁶ billions of g of CO₂
Explanation:
Let's calculate the production of CO₂ by a single human in a day. The molar mass of glucose is 180.156 g/mol and CO₂ is 44.01 g/mol. By the stoichiometry of the reaction:
1 mol of C₆H₁₂O₆ -------------------------- 6 moles of CO₂
Transforming for mass multiplying the number of moles by the molar mass:
180.156 g of C₆H₁₂O₆ ----------------- 264.06 g of CO₂
4.59x10² g ---------------- x
By a simple direct three rule:
180.156x = 121203.54
x = 672.77 g of CO₂ per day per human
So, in a year, 6.50 billion of human produce:
672.77 * 365 * 6.50 billion = 1.60x10⁶ billions of g of CO₂
Answer:
118.22 atm
Explanation:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
KP = 0.13 = 
Where p(SO₃) is the partial pressure of SO₃, p(SO₂) is the partial pressure of SO₂ and p(O₂) is the partial pressure of O₂.
- With 2.00 mol SO₂ and 2.00 mol O₂ if there was a 100% yield of SO₃, then 2 moles of SO₃ would be produced and 1.00 mol of O₂ would remain.
- With a 71.0% yield, there are only 2*0.71 = 1.42 mol SO₃, the moles of SO₂ that didn't react would be 2 - 1.42 = 0.58; and the moles of O₂ that didn't react would be 2 - 1.42/2 = 1.29.
The total number of moles is 1.42 + 0.58 + 1.29 = 3.29. With that value we can calculate the molar fraction (X) of each component:
The partial pressure of each gas is equal to the total pressure (PT) multiplied by the molar fraction of each component.
Rewriting KP and solving for PT:
