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2 years ago

Where would you expect snow or rain to fall on a mountain?

Chemistry

2 answers:

Paraphin [41]2 years ago

3 0

<h2>Answer:</h2>

<u>The windward side</u>

<h2>Explanation:</h2>

There are two sides of the mountain i.e windward side and leeward side. The windward side is that side of a mountain subject to the prevailing wind, and is thus the wetter side. When air reaches the mountain, it rises because the mountains are in the way. As the air rises, it cools, and because cool air can carry less moisture than warm air, there is usually snow or rainfall. So windward side cause rain or snow.

Yuliya22 [10]2 years ago

3 0

Answer: The correct answer is windward sides.

Explanation:

Sides of mountain:

Windward side: Area or side where high rainfall occurs. Wind prevailing loose their moisture on this side which leads to high rainfall on this side.
Leeward side : Area or side where low rainfall occurs.

Rain shadow is the dry area on the leeward sides of the mountainous area. As windward side blocks the moisture which prevents the rainfall on the Leeward sides.This effect is known as Rain shadow effect.

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Which element is the first element in group 13 of the periodic table of elements?

joja [24]

Answer:

Boron

Explanation:

8 0

3 years ago

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A 5L sample of CO2 at 540 torr and 25 degrees celsius expands to 15L at 32 degrees celsius. What is the pressure of the new cont

saw5 [17]

By use of combined gas law
P2= T2P1V1/V2
v1=5L
P1=540 torr
T1=25+273=298k
V2=15 L
T2=32+273=305 k

P2 is therefore=( 305k x 540 torr x5 L) /( 15L x 298)= 184.23 torr

6 0

3 years ago

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

IgorLugansk [536]

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.

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

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