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grigory [225]
1 year ago
11

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,

brainly.com/question/7910855

#SPJ4

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life will not be possible on earth without the water that covers much of its surface and the air that surrounds it earth water a
Vladimir [108]
Kdkskksidkdkfkfkfkfkfkkdkdkkdkdkkkf
5 0
2 years ago
A student attempted to identify an unknown compound by the method used in the experiment. She found that when she heated the sam
yanalaym [24]
Na2CO3 + 2Cl- ⇒ 2NaCl + CO3^-2 
<span>
1 mole of Na2CO3 = 106 g </span>
<span>2 moles of NaCl     = 2 x 58.4
                               = 116.8 g 
</span>Na2CO3 would increase by 116.8 / 106 = 1.10 to form 2NaCl.
<span>0.4862 g x 1.10 = 0.515 grams of NaCl.
</span>
K2CO3 + 2Cl- ⇒ 2KCl + CO3^-2 
<span>1 mole of K2CO3 = 138.2 g </span>
<span>2 moles of KCl = 149.1 </span>
<span>
K2CO3 would increase by </span>149.1 /138.2 = 1.079 <span>to form 2KCl
</span>
<span> 0.4862 x 1.079 = 0.5246 g</span>


3 0
3 years ago
Read 2 more answers
How many grams of butane were in 1. 000 atm of gas at room temperature?
dimaraw [331]

The mass in grams of butane at standard room temperature is 53.21 grams.

<h3>How can we determine the mass of an organic substance at room temperature?</h3>

The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:

PV = nRT

  • Pressure = 1.00 atm
  • Volume = 22.4 L
  • Rate = 0.0821 atm*L/mol*K
  • Temperature = 25° C = 298 k

1 × 22.4 L = n × (0.0821 atm*L/mol*K×  298 K)

n = 22.4/24.4658 moles

n = 0.91556 moles

Recall that:

  • number of moles = mass(in grams)/molar mass

mass of butane = 0.91556 moles × 58.12 g/mole

mass of butane = 53.21 grams

Learn more about calculating the mass of an organic substance here:

brainly.com/question/14686462

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3 0
2 years ago
The decomposition of NI3 to form N2 and I2 releases −290.0 kJ of energy. The reaction can be represented as 2NI3(s)→N2(g)+3I2(g)
EastWind [94]

Answer:

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

Explanation:

Mass of nitrogen triiodide = 20.0 g

Moles of nitrogen triiodide = \frac{20.0 g}{395 g/mol}=0.05063 mol

2NI_3(s)\rightarrow N_2(g)+3I_2(g), \Delta H_{rxn}=-290.0 kJ

According to reaction, 2 moles of nitrogen triiodide gives 290.0 kilo Joules of heat on decomposition ,then 0.05063 moles of nitrogen triiodide will give :

\frac{-290.0 kJ}{2}\times 0.05063=-7.34 kJ

-7.34 kilo Joules is the change in enthaply when 20.0 grams of nitrogen triiodide decomposes.

3 0
3 years ago
HELP PLEASE THE OTHER 'ANSWER' ISNT EVEN AN ANSWER!
hodyreva [135]

Answer:

most likely that (2) the replicated experiment was performed incorrectly.

Why, u ask? u dare question me:

1- The initial experiment invalidness cannot be proven.

2- <em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>s</u></em><em><u>e</u></em><em><u>c</u></em><em><u>o</u></em><em><u>n</u></em><em><u>d</u></em><em><u> </u></em><em><u>a</u></em><em><u>n</u></em><em><u>s</u></em><em><u>w</u></em><em><u>e</u></em><em><u>r</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>c</u></em><em><u>o</u></em><em><u>r</u></em><em><u>r</u></em><em><u>e</u></em><em><u>c</u></em><em><u>t</u></em>

3- Different labaratories does not effect the outcome, as long as the parameter and environment of the replicated experiment is the same as when the initial experiment was conducted.

4- Already knowing the data and errors would increase the precision of the replicated experiment.

5- Change in variables should still be in the objective (or purpose) of the experiment, thus, major difference in the outcome should not happen.

happy learning!

4 0
3 years ago
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