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victus00 [196]
2 years ago
8

7 Coal was produced in the earth over millions of years. Which of the following is NOT a step in

Chemistry
1 answer:
svp [43]2 years ago
7 0

Answer:

D. Time and pressure changes dead plant material to peat.

Explanation:

There are four steps in the production of coal: peat, lignite, bituminous and anthracite.

Peat: in this stage dead plants are oxidized to water and carbon dioxide and buried under sediments. The partial decomposition of plant matter due to the absence of oxygen is called peat. There is no factor of time and pressure that changes dead plant material to peat.

Lignite: Peat is subjected to heat, pressure, and time to form lignite.

Bituminous: More pressure in lignite removes all the traces of plant matter and form “soft coal”, bituminous coal.

Anthracite: It is the last stage, in which hard coal forms with the combined pressure and high temperature.

Hence, the correct answer is "D. Time and pressure changes dead plant material to peat."

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Find the percent ionization of a 0.337 m hf solution. the ka for hf is 3.5 x 10-4. 1.1 % 1.2 x 10-2 % 3.2 % 3.5 x 10-2 % 4.7 %
Ugo [173]

To determine the percent ionization of the acid given, we make use of the acid equilibrium constant (Ka) given. It is the ration of the equilibrium concentrations of the dissociated ions and the acid. The dissociation reaction of the HF acid would be as follows:<span>

HF = H+ + F-

The acid equilibrum constant would be expressed as follows:

Ka = [H+][F-] / [HF] = 3.5 x 10-4

To determine the equilibrium concentrations we use the ICE table,
         HF             H+              F-
I      0.337           0                 0
C      -x              +x               +x
---------------------------------------------
E    0.337-x        x                   x 

3.5 x 10-4 = [H+][F-] / [HF] 
3.5 x 10-4 = [x][x] / [0.337-x] </span>

Solving for x,

x = 0.01069 = [H+] = [F-]

percent ionization = 0.01069 / 0.337 x 100  = 3.17%

8 0
3 years ago
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ad-work [718]

Answer:

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7 0
3 years ago
What is the correct equilibrium constant
Elza [17]

Answer:[co2][CF4]

[COF2]^2

Explanation:

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Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. When samples of these were decomposed the sulfur dioxide produce
Zinaida [17]

Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

Explanation : Given,

Mass of oxygen in sulfur dioxide = 3.49 g

Mass of sulfur in sulfur dioxide = 3.50 g

Mass of oxygen in sulfur trioxide = 9.00 g

Mass of sulfur in sulfur trioxide = 6.00 g

Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.

Mass of oxygen per gram of sulfur for sulfur dioxide = \frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}

Mass of oxygen per gram of sulfur for sulfur dioxide = \frac{3.49}{3.50}=0.997g

and,

Mass of oxygen per gram of sulfur for sulfur trioxide = \frac{\text{Mass of oxygen}}{\text{Mass of sulfur}}

Mass of oxygen per gram of sulfur for sulfur trioxide = \frac{9.00}{6.00}=1.5g

Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.

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3 years ago
Is it always easy to classify research as applied or pure? Explain.
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No..since research can take different format

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