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3 years ago

Define and compare evaporation and boiling.

2 answers:

3 0

Evaporation is the process of conversion of a liquid into vapor at a temperature below its boiling point. It is a surface phenomenon. Boiling is the process by which a liquid turns into a vapor when it is heated at its boiling point. It occurs only at a particular temperature - boiling point.

kkurt [141]3 years ago

3 0

Answer: So the difference between the two is, 1, boiling can occurs everywhere, depending on the heat source. Whereas evaporation only takes place at surface of the liquid.

2, boiling requires certain temperature at certain pressure, and continuous heating source, while evaporation works at any temperature and pressure.

3, evaporation is a slow and soft process, while boiling is violent and fast.

4. Wind and surface area affect evaporation. Wind and surface area do not affect boiling

Explanation: To begin with, i’ll talk about this two phenomena first.

When the liquid get heated, tiny vapor bubbles come out around the condensation nucleus in liquid, which could be air in water or ashes. Vapor has its pressure at certain temperature, which is called Saturated Vapor Pressure. When the vapor pressure is lower than the pressure that vapor bubbles received from out side, such as the air pressure (and the liquid pressure considering the depth of liquid, however, in most cases, the liquid pressure is much too low), the bubbles stay there and waiting. As long as the Saturated Vapor Pressure in vapor bubbles reach the same as air pressure, the vapor bubbles will unstoppablely expand and move up, and that’s how boiling works.

But for the evaporation, it occurs on the surface of liquid. The molecules of liquid move at the surface of liquid. Most of the molecules are trapped by intermolecular forces, whereas a few molecules escaped to the air as vapor. Meanwhile, some vapor molecules move back to the liquid surface and are trapped by the liquid again. So if the escaping rate is higher than the tracpping rate, this is how evaporation works. if you blow a liquid, the vapor on that area is gone, and the evaporation speeds up.

Sorry if this was a bit long, wanted to include as much detail as possible! Hope I could help! :)

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Farmer brown is planting crops in his feild s. He wants to prevent the topsoil from being blown away by the wind or washed away

Ratling [72]

Hey!

I would say the answer is A. Plow the soil many times to lessen the chance of the topsoil being blown away.

Hope this helps !

4 0

3 years ago

Read 2 more answers

What volume of air is needed to burn an entire 55-L (approximately 15-gal) tank of gasoline? Assume that the gasoline is pure oc

White raven [17]

Answer:

The volume of air required is 527,686.25L.

Explanation:

When the question says "burn", it refers to a combustion reaction, where a substance (in this case octane) reacts with oxygen to produce carbon dioxide and water.

Step 1: Write a balanced equation

Considering it is a combustion reaction, the balanced equation is:

C₈H₁₈ + 12.5 O₂ ⇄ 8 CO₂ + 9 H₂O

In this step, we start balancing elements that are present only in one compound on each side of the equation, namely, carbon and hydrogen.

To finish, it is important to count that there are the same number of atoms on both sides of the equation. In this case there are 8 atoms of Carbon, 18 atoms of Hydrogen and 25 atoms of Oxygen, so it is balanced.

Step 2: Find out the mass of C₈H₁₈

Since the balanced equation gives us information about the mass of C₈H₁₈ involved in the reaction, we need to find out how many grams we have.

The info we have is:

55 L of gasoline (assuming gasoline to be pure octane).
The density of octane is 0,70 g/cm³

Density relates mass and volume, so we can find out how many grams are represented by 55 L. Since the units used are different, first we need to convert liters into cm³. We use the conversión factor 1 L = 1000 cm³.

$55L.\frac{1000cm^{3} }{1L} =55000cm^{3}$

Since density = mass/volume, we can solve for mass:

$mass = density.volume=0.70\frac{g}{cm^{3} } .55,000cm^{3} =38,500g$

Step 3: Establish the theoretical relationship between the mass of octane and the moles of oxygen.

This relationship comes from the balanced equation.

For octane:

molar mass of C₈H₁₈ = molar mass of C . 8 + molar mass of H . 18 =

12 g/mol . 8 + 1g/mol . 18 = 114 g/mol

According to the balanced equation reacts 1 mol of octane, which means 114 grams of it.

For oxygen:

According to the balanced equation, 12.5 moles of oxygen react.

Then, the relationship is 114 g octane : 12.5 moles of oxygen

Step 4: Use the theorethical relationship to find the moles of oxygen that reacted

We use the mass of octane found in step 2 and apply the proper conversión factor.

$38,500g (octane) . \frac{12.5mol (oxygen)}{114g (octane)} = 4,221.49mol(oxygen)$

Step 5: Find out the volume of oxygen.

We know that 1 mol of any gas at room temperature occupies about 25 L. Then,

$4,221.49mol(oxygen).\frac{25L(oxygen)}{1mol(oxygen)} =105,537.25 L (oxygen)$

Step 6: Look the volume of air that contains such amount of oxygen

Given oxygen represents 20% of air, we can use that relationship to find the volume of air.

$105,537.25L(oxygen).\frac{100L(air)}{20L(oxygen)} =527,686.25L (air)$

4 0

4 years ago

Where are the most massive elements found on the periodic table?

iragen [17]

By atomic radius, the bottom left

5 0

3 years ago

Read 2 more answers

For an electron that interacts with visible light, which visible photon would give the most accurate (lowest error) measurement

Rom4ik [11]

Answer:

violet

Explanation:

When an electron interacts with visible light, energy is transferred from the photon to the electron.

The most accurate (lowest error) measurement of momentum (or speed) of an electron is obtained using a photon of high energy and low wavelength.

Violet light which lies at the high energy end of spectrum is most suitable.

4 0

3 years ago

Buffer preparation. You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acid plus acet

Hoochie [10]

Answer:

0.182 moles of acetic acid are needed, this means 10.93 g.

0.318 moles of sodium acetate are needed, this means 26.08 g.

Explanation:

The Henderson–Hasselbalch (H-H) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:

pH = pka + $log\frac{[A^{-} ]}{[HA]}$

In this case, [A⁻] is the concentration of sodium acetate, and [HA] is the concentration of acetic acid. The pka is a value that can be looked up in literature: 4.76.

From the problem we know that

[A⁻] + [HA] = 250 mM = 0.250 M eq. 1

We use the H-H equation, using the data we know, to describe [A⁻] in terms of [HA]:

5.0 = 4.76 + $log\frac{[A^{-} ]}{[HA]}$

$0.24=log\frac{[A^{-} ]}{[HA]}\\\\10^{0.24}=\frac{[A^{-} ]}{[HA]}\\ 1.74 [HA] = [A^{-}]$ eq.2

Now we replace the value of [A⁻] in eq. 1, to calculate [HA]:

1.74 [HA] + [HA] = 0.250 M

[HA] = 0.091 M

Then we calculate [A⁻]:

[A⁻] + 0.091 M = 0.250 M

[A⁻] = 0.159 M

Using the volume, we can calculate the moles of each substance:

moles of acetic acid = 0.091 M * 2 L = 0.182 moles
moles of sodium acetate = 0.159 M * 2 L = 0.318 moles

Using the molecular weight, we can calculate the grams of each substance:

grams of acetic acid = 0.182 mol * 60.05 g/mol = 10.93 g
grams of sodium acetate = 0.318 mol * 82.03 g/mol = 26.08 g

8 0

3 years ago