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
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I’m thinking it’s gold because lead isn’t with cooper meaning that if u switch lead with cooper it won’t work at all .
Answer:
a) Xbenzene = 0.283
b) Xtoluene = 0.717
Explanation:
At T = 20°C:
⇒ vapor pressure of benzene (P*b) = 75 torr
⇒ vapor pressure toluene (P*t) = 22 torr
Raoult's law:
∴ Pi: partial pressure of i
∴ Xi: mole fraction
∴ P*i: vapor pressure at T
a) solution: benzene (b) + toluene (t)
∴ Psln = 37 torr; at T=20°C
⇒ Psln = Pb + Pt
∴ Pb = (Xb)*(P*b)
∴ Pt = (Xt)*(P*t)
∴ Xb + Xt = 1
⇒ Psln = 37 torr = (Xb)(75 torr) + (1 - Xb)(22 torr)
⇒ 37 torr - 22 torr = (75 torr)Xb - (22 torr)Xb
⇒ 15 torr = 53 torrXb
⇒ Xb = 15 torr / 53 torr
⇒ Xb = 0.283
b) Xb + Xt = 1
⇒ Xt = 1 - Xb
⇒ Xt = 1 - 0.283
⇒ Xt = 0.717
Answer:
ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.
Explanation:
Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present
The reaction will be
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O
so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2
now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O
Hence answer is
molecules of CO2 formed = 6
Molecules of H2O formed = 8
molecules of C3H8 left = 1
molecules of O2 left = 0
