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BigorU [14]
3 years ago
11

Gases and particles which are put into the air or emitted by various sources are called __________. photochemical smog emissions

carbon monoxide radon (geology)
Chemistry
1 answer:
Leto [7]3 years ago
7 0
I would say the answer is emissions. These are the particles that are not supposed to be present in air but due to the production of different substances from humans daily activities these substances go with the air we breath. Hope this helped.
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Viscosity is basically how thick a liquid is so pancake syrup would be the answer
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What has 6 valence electrons and 16 protons and ends withthe letter "r"
zmey [24]

Answer:

Sulphur

Explanation:

Sulphur(S) is the Sixteenth element of the periodic table and it has an electronic configuration of 2,8,6. This means it has six electrons in its outermost shell( valence electrons).

Sulphur has sixteen(16) protons which is equal to the atomic number which is also sixteen. Sulphur ends with the letter ‘r’. All the descriptions of the element are satisfied and it validates Sulphur.

4 0
3 years ago
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

Acid dissociation constant = k_a=4.5\times 10^{-4}

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

                           HNO_2+H_2O\rightleftharpoons NO_2^-+H_3O^+

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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