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2 years ago

water droplets tend to have a spherical, round shape, Explain ths fact in terms of the polar nature of water molecules

1 answer:

6 0

What does it happen in smallest scale?
You have to look for "DUPRE' EQUATION" which treat about 
"TWO-PHASE INTERFACIAL FORCEs". 
At boundary surface between two aggregation phases, you may attribute upper energy level to molecules standing at the boundary zone. So, it has to define INTERFACIAL FORCEs WHICH ACT TO REMODEL SHAPE AND EXTENSION OF MEETING PHASEs. 
In your case, Water's Droplet is a liquid phase surrounded by Air (e.g. gas phase) and Interfacial Forces act to dominate Boundary-Surface. 
Mathematically, smallest possible Surface comes to SPHERICAL SHAPE.

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Write the number 345 in scientific notation, with the same number of significant figures:

uranmaximum [27]

Answer:

3.45 x 10^2

Explanation:

345= 3 SF

3.45 x 10^2= 3 SF

8 0

3 years ago

How much energy is lost to condense 300. grams of steam at 100.C?

True [87]

Energy lost to condense = 803.4 kJ

<h3>Further explanation</h3>

Condensation of steam through 2 stages:

1. phase change(steam to water)

2. cool down(100 to 0 C)

1. phase change(condensation)

Lv==latent heat of vaporization for water=2260 J/g

$\tt Q=300\times 2260=678000~J$

2. cool down

c=specific heat for water=4.18 J/g C

$\tt Q=300\times 4.18\times (100-0)=125400$

Total heat =

$\tt 678000+125400=803400~J$

3 0

2 years ago

A 6.0 Liter solution is made with 2 moles of solute. What is the molarity of the solution?

kodGreya [7K]

The formula for molarity is: mol/L. so it would be 2.0 mol/ 6.0 L.

the answer is: .33 M

hope this helps!

3 0

3 years ago

How much heat energy is needed to heat 300g of water from 10 degrees Celsius to 50 degrees Celsius

elixir [45]

Answer:

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

Explanation:

Step 1: Data given

mass of water = 300 grams

initial temperature = 10°C

final temperature = 50°C

Temperature rise = 50 °C - 10 °C = 40 °C

Specific heat capacity of water = 4.184 J/g °C

Step 2: Calculate the heat

Q = m*c*ΔT

Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)

Q = 50208 Joule = 50.2 kJ

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

8 0

2 years ago

Which of the following changes can alter the form of a substance but does not necessarily change it to an other substance?

Goshia [24]

Answer:

The physical changes can alter the form of a substance but does not necessarily change it to an other substance

Explanation:

The physical properties of matter is one that does not depend on composition.It just change the form of matter into another form with changing into composition.

Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others.

The three states of matter are: solid, liquid, and gas. The melting point and boiling point are related to changes of the state of matter. All matter may exist in any of three physical states of matter.

Example:

When water is boiled then it change into vapors. Now the form of water is change but its composition is same like water
When water freezes into ice, now phase are different i.e water is liquid while ice is solid but both have same composition.

8 0

2 years ago