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mash [69]
3 years ago
8

As an object falls freely near the earths surface , the loss in gravitational potential energy of the object is equal to its wha

t
Chemistry
2 answers:
Romashka [77]3 years ago
6 0

Answer:

Therefore the loss (transfer) of this potential energy is equal to the increase in kinetic energy

Explanation:

First remember that energy is not destroyed, it only transpires

There are different types of energy, an example of different types of energy is the heat energy (the one that is transferred from an object that is hotter to another, or also by friction), the gravitational potential energy (increases when the height of an object increases with respect to the zero point) another example would be the kinetic energy (increases with increasing the speed of an object)

When object "loses" energy is really transforming into another type of energy

While an object falls and loses its potential energy (it loses height) it increases its speed (it falls faster and faster), as its speed increases the kinetic energy increases

Therefore the loss (transfer) of this potential energy is equal to the increase in kinetic energy

Nitella [24]3 years ago
4 0
Increase in kinetic energy as well as energy loss to the surroundings in the form of heat ( negligible)
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Name gaseous element in period 2 group 16?
Sholpan [36]

Answer:

mark me as a brainlist answer

Explanation:

Oxygen

3 0
3 years ago
Pressure is the force exerted by a substance on its container. Of the three phases of matter which one would exert the highest p
Luden [163]

Answer:

Gaseous phase exert more pressure on container.

Explanation:

Pressure:

It is the force exerted by the substance on its container.

Mathematical expression:

P = F/A

P = pressure

F = force

A = area

We know that there are phases of matter solid liquid and gas.

Gases don't have definite volume and shape and take the shape and volume of container in which it present.  Molecule of gases randomly move everywhere and occupy all available space.  The molecules of gas randomly move collide with each other and also the wall of container their kinetic energy also increased because of this collision and pressure is produced

While in case of liquid molecules are packed and they can not move freely thus did not exert pressure like gaseous molecules.

In case of solids molecules are tightly packed and form more compact structure. They can not move thus did not exert pressure to the wall of container.

Properties of gases:

Molecule of gases randomly move everywhere and occupy all available space.

Gases don't have definite volume and shape and take the shape and volume of container in which it present.

Their densities are very low as compared to the liquid and solids.

Gas molecules are at long distance from each other therefore by applying pressure gases can be compressed.

The very weak inter molecular forces are present between gas molecules.

Properties of Liquid:

Liquid have definite volume but don,t have definite shape.

Their densities are high as compared to the gases but low as compared to the solids.

In liquid, molecules are close to each other and have greater inter molecular forces as compared to the gas molecules.

Properties of solids:

Solids have definite volume and shape.

In solids molecules are tightly pack and very close to each other.

Their melting and boiling point are every high.

The densities of solids are also very high as compared to the liquid and gas.

There are very strong inter molecular forces are present between solid molecules.

6 0
3 years ago
A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
Romashka [77]

<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

W_{solvent} = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

8 0
3 years ago
A 2.500g sample of compound containing only Carbon and Hydrogen is found containing 2.002g of Carbon. Determine the empirical fo
Dima020 [189]

The empirical formula : CH₃

<h3>Further explanation</h3>

Given

2.5 g sample

2.002 g Carbon

Required

The empirical formula

Solution

Mass of Hydrogen :

= 2.5 - 2.002

= 0.498

Mol ratio C : H :

C : 2.002/12 = 0.167

H : 0.498/1 = 0.498

Divide by 0.167 :

C : H = 1 : 3

7 0
2 years ago
In the compound CO2, how many lone pairs are on the central atom?
Cloud [144]
There are 4 lone pairs of electrons present in the carbon dioxide molecule 
4 0
3 years ago
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