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Natalija [7]
3 years ago
12

The original list from the Stockholm Convention on Persistent Organic Pollutants (POPs) of 12 hazardous chemicals, called the __

__________________, which includes DDT and eight other chlorine-containing persistent pesticides, PCBs, dioxins, and furans.
Chemistry
1 answer:
o-na [289]3 years ago
5 0

Answer:

The dirty dozen.

Explanation:

This is mainly seen in the control of hazardous waste and its regulations through treaty in the control of persistent organic pollutant(POPs). This regulation is seen to occur in 12 widely used persistent organic pollutants; these are seen to have the ability to absorb and store tissues that make them fatty especially in humans and also animals found in the higher trophic levels or in the food web. That being said, it can be noted that the dirty dozen can move or even attain hundreds, or thousands of levels towards the environ and also in other cases, transporting them through air or water as the case may be.

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Answer:-  0.88

Solution:- Masses of silver and copper metals are given and we are asked to calculate the percentage of silver in the alloy.

mass percent of Ag = (\frac{mass of Ag}{mass of alloy})*100

Mass of Ag = 17.6 g

mass of Cu = 2.40 g

mass of alloy = 17.6 g + 2.40 g = 20.0 g

Let's plug in the values in the formula:

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What is the pOH of a 0.150 M solution of potassium nitrite? (Ka HNO2 = 4.5 x 10−4 )
yanalaym [24]

Answer:

11.9 is the pOH of a 0.150 M solution of potassium nitrite.

Explanation:

Solution :  Given,

Concentration (c) = 0.150 M

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The equilibrium reaction for dissociation of HNO_2 (weak acid) is,

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initially conc.         c                       0         0

At eqm.              c(1-\alpha)                c\alpha        c\alpha

First we have to calculate the concentration of value of dissociation constant (\alpha ).

Formula used :

k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}

Now put all the given values in this formula ,we get the value of dissociation constant (\alpha ).

4.5\times 10^{-4}=\frac{(0.150\alpha)(0.150\alpha)}{0.150(1-\alpha)}

4.5\times 10^{-4} - 4.5\times 10^{-4}\alpha =0.150\alpha ^2

0.150\alpha ^2+4.5\times 10^{-4}\alpha-4.5\times 10^{-4}=0

By solving the terms, we get

\alpha=0.0533

No we have to calculate the concentration of hydronium ion or hydrogen ion.

[H^+]=c\alpha=0.150\times 0.0533=0.007995 M

Now we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (0.007995 M)

pH=2.097\approx 2.1

pH + pOH = 14

pOH =14 -2.1 = 11.9

Therefore, the pOH of the solution is 11.9

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