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2 years ago

Which of these best describes the particle motion taking place as CO2 gas is exposed to freezing temperatures?

Chemistry

2 answers:

nirvana33 [79]2 years ago

6 0

The options in the question are as follows

a. The motion of particles becomes random.

b. The motion of the particles is unchanged.

c. The particles move with more force.

d. The particles decrease in speed.

Answer:

The particles decrease in speed.

Explanation:

In a gas at room temperature, the particles of the gas move at a very high speed and in a random manner. A gas has no definite volume and it is also assumed that there is minimal interaction between gas particles. Recall that temperature is a measure of the average kinetic energy of the particles of a system. As the temperature is decreased and heat is withdrawn as the CO2 approaches freezing temperature, the speed and kinetic energy of the particles decreases and significant interaction between particles is observed. Hence the CO2 particles slow down.

mote1985 [20]2 years ago

5 0

CO2 is carbon dioxide which is most famous for being in gas form so i would figure if it was exposed to freezing temperatures it would turn into a liquid then maybe a solid

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2 years ago

The pressure of 5.00 L of gas increases from 1.50 atm to 1240 mmHg. What is the final volume of the gas, assuming constant tempe

Vsevolod [243]

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2 years ago

Read 2 more answers

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Step 1

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Step 2

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Step 3

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Step 4

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Step 5

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

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