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3 years ago

How do molecules in a solid differ from those in a liquid or gas

1 answer:

5 0

Molecules in solids are packed a lot closer together than molecules in liquids and gases, so there is far less space between solid particles. Because of this, solid particles cannot move around quickly and freely like liquids and gases, instead they only vibrate so are fairly 'locked' into the same place, so solids have fixed volume and space.

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Which factor can increase the potential energy of an object?

OLEGan [10]

Answer:

Increase in height from the ground.

Explanation:

Potential energy =mass×acceleration due to gravity×height.

P.E= mgh

4 0

3 years ago

5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time

11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}$

$\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142$

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen = $d_{O_2}=\frac{1 m}{t}$

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane= $d_{CH_4}=\frac{y }{t}$

$\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142$

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0

3 years ago

Why pie bond are not perticipate in hybridaization

DedPeter [7]

The first bond between two atoms is always a sigma bond and the other bonds are always pi bonds and a hybridized orbital cannot be involved in a pi bond. Thus we need to leave one electron (in case of Carbon double bond) to let the Carbon have the second bond as a pi bond.

3 0

3 years ago

When hydrated copper(II) sulfate crystals are heated, anhydrous copper(II) sulfate forms. A mass of 12.5 g of hydrated copper(II

pychu [463]

The formula of hydrated copper(II) sulfate is CuSO4.10H2O

<h3>What is the formula of the hydrated copper (ii) sulfate salt?</h3>

The formula of the hydrated copper (ii) sulfate is determined as follows:

Mass of hydrated salt = 12.5 g

Mass of anhydrous salt = 8.0 g

Mass of water = 12.5 - 8 = 4.5 g

mole ratio of water and anhydrous salt is;

4.5/18 : 8.0/159.5

0.562 : 0.05

10 : 1

Water of crystallization (n) = 10.

Therefore, the formula of hydrated copper(II) sulfate is CuSO4.10H2O

Learn more about water of crystallization at: brainly.com/question/26146814

#SPJ1

7 0

2 years ago

What factors affect freezing-point depression of a solution? (select all that apply)

ANTONII [103]

Answer:

A and F i think

Explanation:

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all lower than that of the pure solvent and is directly proportional to the molality of the solute

8 0

3 years ago