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MariettaO [177]
3 years ago
8

Write an 250 words about Water Purification and Filtration

Chemistry
1 answer:
kotykmax [81]3 years ago
6 0

Answer:

Water purification is the process of removing undesirable chemicals, biological contaminants, suspended solids and gases from contaminated water. The goal is to produce water fit for a specific purpose. Most water is purified for human consumption (drinking water), but water purification may also be designed for a variety of other purposes, including meeting the requirements of medical, pharmacological, chemical and industrial applications. In general the methods used include physical processes such as filtration,sedimentation, and distillation, biological processes such as slow sand filters or biologically active carbon, chemical processes such asflocculation and chlorination and the use of electromagnetic radiation such as ultraviolet light.

Extreme lack or loss of water may lead to dehydration of the body and other health complications. For this reason, governments ensure that citizens have access to clean and safe water for domestic use. Clean water is essential in ensuring that no pathogens or impurities are ingested by people, either through direct drinking or through food.

To attain these standards of water, purification is important. Water purification involves physical and chemical processes, which are carried out stepwise to ensure the water is safe and free from any harm. This directional process essay synthesizes the steps, which have to be followed to achieve this task.

In essence, water purification denotes the process used to free water from impurities like bacteria and contaminants. Since the process is aimed at eliminating all the impurities present in the water, it is necessary to apply chemical and physical methods of separation in an orderly manner.

Explanation:

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5. Durante un estudio de la velocidad de la reacción A2(g) + 3B2(g)  2 AB3(g), se observa que en un recipiente cerrado que cont
weqwewe [10]

Answer:

a) Speed of the reaction = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time = 0.002083 mol/L.s

c) The rate of appearance of AB₃ = 0.004167 mol/L.s

Explanation:

English Translation

During a study of the reaction rate

A₂ (g) + 3B₂ (g) → 2 AB₃ (g),

it is observed that in a closed container containing a certain amount of A₂ and 0.75 mol / L of B₂, the concentration B₂ decreases to 0.5 mol / L in 40 seconds.

a) What is the speed of the reaction?

b) What is the rate of disappearance of A₂ during this period of time?

c) What is the rate of appearance of AB₃?

Solution

The rate of a chemical reaction is defined as the time rate at which a reactant is used up or the rate at which a product is formed.

It is the rate of change of the concentration of a reactant (rate of decrease of the concentration of the reactant) or a product (rate of increase in the concentration of the product) with time.

Mathematically, for a balanced reaction

aA → bB

Rate = -(1/a)(ΔA/Δt) = (1/b)(ΔB/Δt)

The minus sign attached to the change of the reactant's concentration indicates that the reactant's concentration decreases.

And the coefficients of each reactant and product in the balanced reaction normalize the rate of reaction for each of them

So, for our given reaction,

A₂ (g) + 3B₂ (g) → 2 AB₃ (g)

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

a) Speed of the reaction = Rate of the reaction

But we are given information on the change of concentration of B₂

Change in concentration of B₂ = ΔB₂ = 0.50 - 0.75 = -0.25 mol/L

Change in time = Δt = 40 - 0 = 40 s

(ΔB₂/Δt) = (-0.25/40) = -0.00625 mol/L.s

Rate of the reaction = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

b) The rate of disappearance of A₂ during this period of time

Recall

Rate = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

-(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt)

Rate of disappearance of A₂ = -(ΔA₂/Δt) = -(1/3)(ΔB₂/Δt) = (-1/3) × (-0.00625) = 0.002083 mol/L.s

c) The rate of appearance of AB₃

Recall

Rate = -(1/3)(ΔB₂/Δt) = (1/2)(ΔAB₃/Δt)

(1/2)(ΔAB₃/Δt) = -(1/3)(ΔB₂/Δt)

(ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt)

rate of appearance of AB₃ = (ΔAB₃/Δt) = -(2/3)(ΔB₂/Δt) = (-2/3) × (-0.00625) = 0.004167 mol/L.s

Hope this Helps!!!

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