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3 years ago

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Air is made of nitrogen, carbon dioxide, and small amounts of other gases. What is the pressure of O₂, total is 1 atm if PN₂ = 5

93.4 mm Hg, PCO₂ = 3.0 mm Hg, and Pothers = 7.1 mm Hg? *
0.3 atm
0.2 atm
0.4 atm
0.8 atm

1 answer:

Stolb23 [73]3 years ago

6 0

Answer:

The pressure of O₂ is 0.8 atm.

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

PT = PA + PB

This relationship is due to the assumption that there are no attractive forces between the gases.

In this case:

PT=Pnitrogen + Pcarbon dioxide + Pother gases

Being:

Pnitrogen: 593.4 mmHg
Pcarbon dioxide: 3 mmHg
Pother gases: 7.1 mmHg

and replacing:

PT= 593.4 mmHg + 3 mmHg + 7.1 mmHg

you get:

PT= 603.5 mmHg

Being 760 mmHg= 1 atm, you get:

PT= 603.5 mmHg= 0.8 atm

<u><em>The pressure of O₂ is 0.8 atm.</em></u>

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Read 2 more answers

Hcl and nh3 react to form a white solid, nh4cl. if cotton plugs saturated with aqueous solutions of each are placed at the ends

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24.4 cm.

<h3>Explanation</h3>

HCl and NH₃ reacts to form NH₄Cl immediately after coming into contact. Where NH₄Cl is found is the place the two gases ran into each other. To figure out where the two gases came into contact, you'll need to know how fast they move relative to each other.

The speed of a HCl or NH₃ molecule depends on its <em>kinetic energy</em>.

$E_\text{k} = 1/2 \; m \cdot v^{2}$

Where

$E_\text{k}$ is the <em>kinetic energy</em> of the molecule,
$m$ its mass, and
$v^{2}$ the square of its speed.

Besides, the <em>kinetic theory</em> <em>of gases</em> suggests that for an ideal gas,

$E_\text{k} \propto T$

where $\text{T}$ its temperature in degrees kelvins. The two quantities are directly proportional to each other. In other words, the <em>average kinetic energy</em> of molecules shall be the same for <em>any ideal gas </em>at the same<em> temperature</em>. So is the case for HCl and NH₃

$E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3)$

$m(\text{HCl}) \cdot v^{2}(\text{HCl}) = E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

Where

$m(\text{HCl})$ , $v(\text{HCl})$ , and $E_\text{k}(\text{NH_3})$ the mass, speed, and kinetic energy of an HCl molecule;
$m(\text{NH}_3)$ , $v(\text{NH}_3)$ , and $E_\text{k}(\text{NH}_3)$ the mass, speed, and kinetic energy of a NH₃ molecule.

The ratio between the mass of an HCl molecule and a NH₃ molecule equals to the ratio between their <em>molar mass</em>. HCl has a molar mass of 35.45; NH₃ has a molar mass of 17.03. As a result, $m(\text{HCl}) = 36.45 / 17.03 \; m(\text{NH}_3)$ . Therefore:

$36.45 /17.03\; m(\text{NH}_3) \cdot v^{2}(\text{HCl}) = m(\text{HCl}) \cdot v^{2}(\text{HCl}) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)$

$36.45 /17.03\; v^{2}(\text{HCl}) = v^{2}(\text{NH}_3)$

$\sqrt{36.45 /17.03}\; v(\text{HCl}) = v(\text{NH}_3)$

The <em>average </em>speed NH₃ molecules would be $\sqrt{36.45/17.03} \approx 1.463$ <em>if</em> the <em>average </em>speed of HCl molecules $v(\text{HCl})$ is 1.

$\text{Time before the two gases meet} = \frac{\text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}$

$\text{Distance from the HCl end} = v(\text{HCl}) \times \text{Time before the two gases meet}\\\phantom{\text{Distance from the HCl end}} = v(\text{HCl}) \times \frac{ \text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}\\\phantom{\text{Distance from the HCl end}} = \frac{v(\text{HCl})}{v(\text{HCl}) + v(\text{NH}_3)} \times \text{Length of the Tube}\\\phantom{\text{Distance from the HCl end}} = \frac{1}{1 + 1.463} \times 60.0\; \text{cm} \\\phantom{\text{Distance from the HCl end}} = 24.4 \; \text{cm}$

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3 years ago