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Tcecarenko [31]
3 years ago
8

In old times, before there were train signals, people who worked on the railroad would often put their ear down to the train tra

ck to listen for an oncoming train.
Why would this result in hearing the train further off than just listening for the train through the air?

Question options:

People did not do this because sound travels faster through less dense material, like air, rather than more dense materiel, like train tracks.


Sound travels slower, but louder through solid material because it is more dense, so the train could be heard better this way.


Sound travels slower in air because there are often trees and other items blocking the sound.


Sound waves travel faster in more dense material, and the more dense the material the louder the sound is.
Chemistry
2 answers:
Roman55 [17]3 years ago
6 0
The fourth option is correct because sound waves travel faster through denser compounds.
julsineya [31]3 years ago
5 0

The answer to your question is D. Sound waves travel faster in more dense material, and the more dense the material the louder the sound is.

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Is the element CH4 a metal or non metal
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What is the enthalpy for 2H2S(g) + 3O2(g) -&gt; 2H2O(I) + 2SO2(g). (∆H)​
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2H2S (g) + 3O2 (g) → 2H2O (l) + 2SO2 (g)

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The equation for another reaction used in industry isCO(g) + H₂O(g) <img src="https://tex.z-dn.net/?f=%5Crightleftharpoons" id="
Sloan [31]

Answer:

(i) CO = 0.4 mol; H₂O = 1.6 mol; Kc = 4

(ii) CO = 0.67 mol; H₂O = 0.67 mol; CO₂ = 1.33 mol; H₂ = 1.33 mol

Explanation:

(i) For the equation given let's make a table of the concentrations for equilibrium (the volume is constant, so, we can do it with moles number)

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol    3.2 mol      0          0              <em>Initial</em>

-x              -x                +x        +x            <em>Reacts</em> (stoichiometry is 1: 1: 1: 1)

2.0-x       3.2-x            x           x             <em>Equilibrium</em>

In the equilibrum, the moles number of hydrogen and carbon dioxide are 1.6 mol, so x = 1.6 mol

The amounts of CO and H₂O are:

CO = 2.0 - 1.6 = 0.4 mol

H₂O = 3.2 - 1.6 = 1.6 mol

The constant of the equilibrium is the multiplications of the concentrations of products divided by the multiplication of the concentration of the reactants (all the concentrations elevated to the coefficient). So:

Kc = (1.6x1.6)/(0.4x1.6)

Kc = 1.6/0.4

Kc = 4

(ii) Kc must remais constant (it only changes with the temperature), so let's construct a new table of equilibrium:

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol  2.0 mol      0          0                 <em>Initial</em>

-x              -x             +x         +x               <em>Reacts</em> (stoichiometry is 1: 1: 1: 1)

2.0-x        2.0-x         x           x                <em>Equilibrium</em>

Kc = (x*x)/((2.0-x)*(2.0-x))

4 = x²/(4 - 4x + x²)

16 - 16x + 4x² = x²

3x² - 16x + 16 = 0

Using Baskhara's equation:

Δ =(-16)² - 4x3x16

Δ = 256 - 192

Δ = 64

x = (-(-16) +/- √64)/(2*3)

x' = (16 + 8)/6 = 4

x'' = (16 - 8)/6 = 1.33

x must be small than 2.0, so x = 1.33 mol, which is the amount of hydrogen and carbon dioxide at equilibrium. The both reactants has 2.0 - 1.33 = 0.67 mol at equilibrium.

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