Answer:
the steady-state concentration of TCE in the treated water leaving the reactor is 5.88 mg/L
Explanation:
Given that;
Tank volume v = 3250 liters
wastewater flows into the tank Q = 200 L/min
TCE concentration Co= 25 mg/L
reactor decays TCE at a reaction rate K = 0.20 min-1
mass balance
we know that;
Accumulation = inflow - outflow ± generation
⇒dc/dt = QCo - Qc ± rc.V
now at a steady state; dc/dt = 0
so
0 = QCo - Qc + rcV
where rc = -kc
0 = QCo - Qc - kcV
Qc + kcV = QCo
c(Q + kV) = QCo
c = QCo / (Q + kV)
so we substitute
c = (200 × 25) / (200 + (0.2×3250))
c = 5000 / 850
c = 5.88 mg/L
Therefore, the steady-state concentration of TCE in the treated water leaving the reactor is 5.88 mg/L
Gold's natural state has a definite shape and a definite volume. What is gold's natural state(s)?
Answer:
If your asking what golds natural state of matter is it's solid.
Explanation:
Answer:
the answer is soild
Explanation:
i did it on edge :)
After going through a guided tutorial by selecting Run Grams Demonstration, you can create your own experiment by clicking the Run Experiment button at the end or by clicking the Overview tab and returning to the Experiment tab to select Run Experiment. There are nine reactions you can explore on your own. Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3). How much SO2 would you need to completely react with 6.00 g of O2 such that all reactants could be consumed
Answer: Thus 24.0 g of [tex]SO_2[/tex] would be needed.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} O_2=\frac{6.00g}{32g/mol}=0.1875moles[/tex]
[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(l)[/tex]
According to stoichiometry :
1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]
Thus 0.1875 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.1875=0.375moles[/tex] of [tex]SO_2[/tex]
Mass of [tex]SO_2=moles\times {\text {Molar mass}}=0.375moles\times 64g/mol=24.0g[/tex]
Thus 24.0 g of [tex]SO_2[/tex] would be needed to completely react with 6.00 g of [tex]O_2[/tex] such that all reactants could be consumed.
Early chemists, known at the time as alchemists, had a difficult time understanding the Law of Conservation of Mass. Using the burning of wood as an example, what property of matter made this law difficult for early scientists to understand?
The law of conservation of mass states that, for any isolated system, the mass can neither be created nor be destroyed
After burning a log, the remains that are left behind after it burns are lesser than what would initially seem. There is a difference in mass before the burning of the log after the burning of the log. This seems like a violation of the law of conservation of mass, which would have made it difficult for early chemists to understand it. But if we consider the surroundings as a system then we can see that the mass that is lost in burning is actually converted into smoke and energy or forming other forms of substances. Then considering the whole system we can see that the mass and energy are conserved and interrelated.learn more about conservation of mass and energy:
https://brainly.com/question/13971039
explain how liquid can enter the gas phase without reaching its boiling point
When lava cools outside the Earth, ____ igneous rocks are formed
Answer:The rock forms large crystals
Explanation: