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1 year ago

The lowest concentration that can officially reported for any constituent in drinking water is known as the

Chemistry

1 answer:

allochka39001 [22]1 year ago

4 0

What is the legal limit?

The lowest concentration that can officially be reported for any constituent in drinking water is known as the "Legal Limit".

Contamination of water:

The level that safeguards human health and that water systems can attain with the finest technology is reflected in the regulatory limit for a contaminant.

<u><em>How does it occur:</em></u>

Plumbing components are the main way that lead and copper enters the drinking water. The health effects of lead and copper exposure can range from brain damage to stomach discomfort.
Nitrogen, bleach, salts, pesticides, metals, bacterial toxins, and human or animal medications are examples of chemical pollutants. Organisms in the water are biological pollutants. Other names for them include microorganisms and microbiological pollutants.

In order to control corrosion, the system must take a variety of additional measures if lead concentrations reach an action threshold of 15 ppb or copper concentrations exceed an action level of 1.3 ppm in more than 10% of measured customer taps.

Learn more about the contamination of water here,

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The following question appears on a quiz: ""You fill a tank with gas at 60Â°C to 100 kPa and seal it. You decrease the temperatu

dusya [7]

Answer: The final pressure will decrease ad the value is 85 kPa

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=100kPa\\T_1=60^0C=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K$

Putting values in above equation, we get:

$\frac{100kPa}{333K}=\frac{P_2}{283K}\\\\P_2=85kPa$

Hence, the final pressure will decrease ad the value is 85 kPa

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