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emmasim [6.3K]
3 years ago
10

What are five different types of air pollutant each

Chemistry
1 answer:
vichka [17]3 years ago
7 0
5 types of air pollution would include: Carbon Monoxide, Lead, Nitrogen oxides, sulfur oxides. I only know those 4 actually… hope it helped
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Arrange the following elements in order of increasing electronegativity: <br><br> O, F, N, C, B
Vanyuwa [196]

Answer:

Electronegativity increases left to right across a row in the periodic table e.g. C < N < O < F<. b

3 0
3 years ago
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Which half-reaction correctly represents reduction? *
nasty-shy [4]

Answer:

C

Explanation:

Reduction can be seen through addition (gaining) of electrons, addition of Hydrogen or removal of Oxygen

Mn7+ is reduced to Mn4+ by the addition of 3 electrons.

3 0
3 years ago
Which of the following is an organic moleclue ?
Doss [256]

Answer:

water

Explanation:

6 0
3 years ago
Please help to me solve these ! :)
emmainna [20.7K]
I. The solubility of NaCl at 25 degrees C would be between the solubilities at 20 and 30 degrees C. A reasonable answer would be 36 grams/100 g water
ii. From the table, it’s clear that the salts are more soluble at higher temperatures, indicating that an increase in temperature increases solubility.
iii. At 50 degrees C, a saturated ammonium chloride solution will have 50.6 grams of salt per 100 g water. At 20 degrees C, the solution can hold only 37.3 grams of salt per 100 g water. Thus, 13.3 grams of salt will precipitate per 100 grams of water.
4 0
4 years ago
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The rate constant of the first-order reaction A→3B is 0.291 s−1. The concentration of A at t=5 seconds is 0.081 mol/L. What was
zimovet [89]

Answer : The initial concentration of A is, 0.347 mol/L

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 0.291s^{-1}

t = time passed by the sample  = 5 s

a = initial amount of the reactant  = ?

a - x = amount left after decay process = 0.081 mol/L

Now put all the given values in above equation, we get

5=\frac{2.303}{0.291}\log\frac{a}{0.081}

a=0.347mol/L

Therefore, the initial concentration of A is, 0.347 mol/L

7 0
3 years ago
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