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4 years ago

How do you evaporation and boiling differ?

Chemistry

2 answers:

alekssr [168]4 years ago

6 0

evaporation- the surface of the liquid is heated slowly and eventually becomes a gas

boiling- the entire liquid is heated and the entire liquid becomes a gas

maksim [4K]4 years ago

4 0

Evaporation is called a surface phenomenon because water molecules present on the surface of liquid are bonded weakly as compared to inner molecules and when temperature increase hydrogen bonding between molecules break .Due to this water molecules tend to evaporation so that's why evaporation called the surface phenomenon.

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Compared to a normal body cell a normal egg cell contains

lisov135 [29]

Sex cells only contain one chromosome from each pair. When an egg cell andsperm<span> cell join together, the fertilised egg cell contains 23 pairs of chromosomes. One chromosome in each pair comes from the mother, the other from the father.

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5 0

3 years ago

Read 2 more answers

What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 11.0 L of water to produce an antifreeze solution wi

shusha [124]

Answer:

The volume of ethylene glycol, we must add is 8.2 L

Explanation:

Let's apply the formula for the colligative property of Depression of freezing point, to solve this

ΔT = Kf . m . i

Where ΔT = Fussion T° in pure solvent - Fussion T° in solution

Kf = Cryoscopic constant (equal to 1.86 °C kg/mol for the freezing point of water)

m = molality (mol of solute in 1kg of solvent)

i = The number of ions dissolved in solution. As it is a non-electrolytic compound, the i values 1 (Van't Hoff factor)

0°C - (-25°C) = 1.86 kg°C/mol . m

25°C / 1.86 m/kg°C = m

13.4 = mol/kg

As water density is 1 g/ml, let's convert firstly 11L in mL

11L .1000 = 11000mL

In conclusion, we have 11000 g of water.

density = mass / volume

So, if we have 13.4 moles in 1 kg of water, how many moles of ethylene glycol do we have, in 11000 g of water.

11000 g = 11kg

1 kg _____ 13.4 moles of ethylene glycol

11 kg _____ (11 . 13.4)/1 = 147.4 moles

Molar mass of ethylene glycol = 62.07 g/m

Moles . molar mass = mass

147.4 m . 62.07g/m = 9149.1 g

Now we can apply density of ethylene glycol to find out the volume

Density ethylene glycol = ethylene glycol mass / ethylene glycol volume

1.11 g/ml = 9149.1 g / ethylene glycol volume

ethylene glycol volume = 9149.1 g/ 1.11 g/ml

ethylene glycol volume = 8242 mL

8242 mL = 8.2 L

8 0

4 years ago

Throughout the reflection, make sure you have a copy of the Student Guide and your data tables. Use the drop-

Inessa05 [86]

Answer:

1) mass and type of material

2) type of material

3) temperature

Explanation:

8 0

2 years ago

Choose the correct statement regarding the behavior of water.

AfilCa [17]

Answer:

d. The energy required to evaporate 1 kg of liquid water equals the energy released when 1 kg of water vapor condenses into liquid.

Explanation:

Hello,

Since we're considering the same amount of water, the vapor phase has a higher energy content than the liquid phase, thus, for the specified amount of water particles (those contained in the given 1 kg) the energy MUST be same when taking them either to a gaseous phase or to a liquid phase, the only difference is the sign which is negative from gaseous to liquid (heat withdrawal) and positive from liquid to gaseous (heat adding).

Best regards.

5 0

3 years ago

In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?

Bezzdna [24]

It would be the same amount. So, 45 ml of NaOH is required to be added to the 45 ml of HCI to neutralize the acid fully. Here is a brief calculation:

Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH)
With the values put in: 0.35 x 45 = 0.35 x V(NaOH)
= 45 ml.
There is 45 ml of V(NaOH)

Let me know if you need anything else. :)

- Dotz

6 0

3 years ago