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klio [65]
3 years ago
10

What are the three keys to keeping water clean? effective cleanup of oil and gasoline spills adding sediment to the water to red

uce sunlight absorption proper sewage treatment pouring pesticides and chemicals down drain pipes reduction of pollutants?
Chemistry
2 answers:
KengaRu [80]3 years ago
5 0
1. <span>Effective cleanup of oil and gasoline spills
2. </span><span>Proper sewage treatment
3. </span><span>Reduction of pollutants</span>
boyakko [2]3 years ago
4 0

Explanation:

Water is a natural resource and we should not waste it and try to keep rivers, ponds, lakes etc clean by not throwing any garbage, lather, human waste etc into it.

Three ways to keep the water clean are as follows.

  • Effective cleanup of oil and gasoline spills.
  • Proper sewage treatment.
  • Reduction of pollutants.
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Consider the reaction 5 Br− (aq) + BrO3− (aq) + 6 H+ (aq) → 3 Br2 (aq) + 3 H2O (l) The average rate of consumption of Br− is 1.8
kaheart [24]

Answer :  The average rate of consumption of H^+ during the same time interval is, 2.17 M/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)

The expression for rate of reaction :

\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}

\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}

\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}

\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}

As we are given:

\frac{d[Br^-]}{dt}=1.81M/s

Now we have to determine the average rate of consumption of H^+ during the same time interval.

As,

-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}

or,

-\frac{1}{6}\frac{d[H^+]}{dt}=-\frac{1}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\frac{d[Br^-]}{dt}

\frac{d[H^+]}{dt}=\frac{6}{5}\times (1.81M/s)

\frac{d[H^+]}{dt}=2.17M/s

Thus, the average rate of consumption of H^+ during the same time interval is, 2.17 M/s

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4 years ago
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