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S_A_V [24]
2 years ago
15

Choose a federal law from the list below.

Chemistry
1 answer:
goblinko [34]2 years ago
3 0

The principal law governing pollution of the nation’s surface waters is the Federal Water

Pollution Control Act, or Clean Water Act. Originally enacted in 1948, it was totally revised by

amendments in 1972 that gave the act its current dimensions. The 1972 legislation spelled out

ambitious programs for water quality improvement that have since been expanded and are still

being implemented by industries and municipalities.

This report presents a summary of the law, describing the statute without discussing its

implementation. Other CRS reports discuss implementation, including CRS Report R42883,

Water Quality Issues in the 113th Congress: An Overview, and numerous products cited in that

report.

The Clean Water Act consists of two major parts, one being the provisions which authorize

federal financial assistance for municipal sewage treatment plant construction. The other is the

regulatory requirements that apply to industrial and municipal dischargers. The act has been

termed a technology-forcing statute because of the rigorous demands placed on those who are

regulated by it to achieve higher and higher levels of pollution abatement under deadlines

specified in the law. Early on, emphasis was on controlling discharges of conventional pollutants

(e.g., suspended solids or bacteria that are biodegradable and occur naturally in the aquatic

environment), while control of toxic pollutant discharges has been a key focus of water quality

programs more recently.

Prior to 1987, programs were primarily directed at point source pollution, that is, wastes

discharged by industrial and municipal facilities from discrete sources such as pipes and outfalls.

Amendments to the law in that year authorized measures to address nonpoint source pollution

(runoff from farm lands, forests, construction sites, and urban areas), which is estimated to

represent more than 50% of the nation’s remaining water pollution problems. The act also

prohibits discharge of oil and hazardous substances into U.S. waters.

Under this act, federal jurisdiction is broad, particularly regarding establishment of national

standards or effluent limitations. Certain responsibilities are delegated to the states, and the act

embodies a philosophy of federal-state partnership in which the federal government sets the

agenda and standards for pollution abatement, while states carry out day-to-day activities of

implementation and enforcement.

To achieve its objectives, the act is based on the concept that all discharges into the nation’s

waters are unlawful, unless specifically authorized by a permit, which is the act’s principal

enforcement tool. The law has civil, criminal, and administrative enforcement provisions and also

permits citizen suit enforcement.

Financial assistance for constructing municipal sewage treatment plants and certain other types of

water quality improvements projects is authorized under Title VI. It authorizes grants to capitalize

State Water Pollution Control Revolving Funds, or loan programs. States contribute matching

funds, and under the revolving loan fund concept, monies used for wastewater treatment

construction are repaid to states, to be available for future construction in other communities.

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Answer:

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Explanation:

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nirvana33 [79]

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What is the maximum amount of water (in grams) that can be removed from 15ml of toluene by the addition?
Nataly [62]

Complete Question

Magnesium sulfate forms a hydrate with the formula MgSO_4. 7H_20. What is the maximum amount of water (in grams) that can be removed from 15 ml of toluene by the addition of 200 mg of anhydrous magnesium sulfate? The molar mass of MgSO_4 is 120.4 g/mol; H20 = 18 g/mol.

Answer:

The value  is  z =  0.2093 \  g of  H_2O

Explanation:

From the question we are told that

   The volume of toluene is  V = 15 mL

    The mass of  anhydrous magnesium sulfate is  m =  200m g  = 200 *10^{-3} \  g

   The formula of the hydrate is   MgSO_4. 7H_20

    The molar mass of   MgSO_4  is  z =120.4 \ g/mol

From the formula given we see that

  1 mole of  Mg SO_4 wil remove  7 moles of H_2O to for the given formula

Hence

  120.4 g (1 mole) will remove  7 moles (7 * 18 g = 126 g  ) of  H_2O to for the given formula

Therefore 1 g of  Mg SO_4  x g  of  H_2O  

So

     x  =  \frac{x]126 *  1}{ 120.4 }

=>     x  =  1.0465 \  g

From our calculation we obtained that

  1 g of Mg SO_4 will remove  x  =  1.0465 \  g  of  H_2O  

Then  

   200 *10^{-3} \  g of Mg SO_4 will remove z g of  x  =  1.0465 \  g  of  H_2O  

So

   z =  200 *10^{-3} *  1.0465

=>z =  200 *10^{-3} *  1.0465

=>z =  0.2093 \  g of  H_2O  

6 0
3 years ago
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