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3 years ago

Which statement describes the water gas particles in the air above the cup compared with the water liquid

Chemistry

1 answer:

Law Incorporation [45]3 years ago

8 0

Answer:

The particles move faster and are far apart

Explanation:

A substance may exist in three states of matter; solid, liquid and gas.

In the solid state, there is very strong intermolecular forces between the particles of the substance. They can only vibrate or rotate about their mean positions but can not translate.

In the liquid state, the particles of the substance have a greater degree of freedom than in the solid. The magnitude of intermolecular forces is lower than in solids, the molecules can move at low speeds.

In a gas, the molecules are separated from each other with negligible intermolecular interaction hence they move at very high speed.

Therefore, for the water gas particles in the air above the cup; the particles move faster and are far apart.

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A human lung at maximum capacity has a volume of 3.0 liters. If the partial pressure of oxygen in the air is 21.1 kilopascals an

Juliette [100K]

V = maximum capacity of human lung = 3 liter = 3 x 0.001 m³ = 0.003 m³ (Since 1 liter = 0.001 m³)

P = pressure of oxygen = 21.1 kilo pascal = 21.1 x 1000 = 21100 Pa (since 1 kilo = 1000)

T = temperature of air = 295 K

n = number of moles of oxygen

Using the ideal gas equation

PV = n RT

inserting the above values in the equation

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Calculate the volume(in L), at STP, of carbon dioxide produced by the complete combustion of 6.75 lbs of charcoal.

gtnhenbr [62]

535.8 L volume(in L), at STP, of carbon dioxide produced by the complete combustion of 6.75 lbs of charcoal.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Given:

$C(s) + O_2(g)$ ⇒ $CO_2(g)$

1 mole of $CO_2(g)$ is formed from 1 mole of charcoal, that means 1 mole of $CO_2(g)$ is formed from 12 g of charcoal.

1 lbs = 453.592 grams

6.75 lbs = 6.75 x 453.592 =3061.746g

So, $CO_2(g)$ formed from 3061.746g of charcoal is:

= $\frac{1}{12}$ x 3061.746g

=255.14 mol.

Now, at STP, the volume of 1 mol of gas =22.4 L

Volume of 255.14 mol of $CO_2(g)$ = 255.14 x 22.4 L

Volume of 255.14 mol of $CO_2(g)$ = 535.8 L

Hence, 535.8 L volume(in L), at STP, of carbon dioxide produced by the complete combustion of 6.75 lbs of charcoal.

Learn more about the ideal gas here:

brainly.com/question/27691721

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2 years ago

The Bohr model of the atom suggest that the protons and neutrons are found in the nucleus and the electrons orbit the nucleus mu

kondaur [170]

Answer:

The Bohr model suggested that electrons orbited the nucleus in circular paths where as the modern model views the atom to consist of positively charged nucleus surrounded by electrons.

Explanation:

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I think it is
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