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A sample of gas at 62.0°C occupies a volume of 4.80 L at a pressure of 2.90 atm. How many moles of gas are contained in the sample?

Answers

Answer 1
PV = nRT —> n = PV/RT

P = 2.90 atm
V = 4.80 L
R = 0.08206 L atm / mol K
T = 62.0 + 273 = 335 K (make sure you convert from celsius to kelvin)

n = (2.90 • 4.80) / (0.08206 • 335) = 0.506 moles of gas
Answer 2

A sample of gas at 62.0°C occupies a volume of 4.80 L at a pressure of 2.90 atm. 0.506 moles of gas are contained in the sample.

What is mole?

Inside the World System of Units, the mole is indeed the unit of substance amount (SI). How many fundamental entities of a particular substance are present in an object and sample is determined by the quantity of that material. There are precisely 6.022×10²³ elementary entities in a mole.

An elementary entity can be an atoms, a molecule, an ion, and ion pair, or a fundamental particle like an electron depending upon what the substance is. For instance, although having differing volumes and masses, 10 moles liquid water as well as ten moles of mercury both contain the same quantity of material, as well as the mercury includes exactly one atom with each molecule of water.

PV = nRT

n = PV/RT

P = 2.90 atm

V = 4.80 L

R = 0.08206 L atm / mol K

T = 62.0 + 273 = 335 K

n = (2.90 •4.80) / (0.08206 • 335)

   = 0.506 moles of gas

Therefore, 0.506 moles of gas are contained in the sample.

To know more about mole, here:

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Answers

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C9H20+14O2→9CO2+10H2O a. How many molecules of nonane will burn in a tank that has 770 mL of oxygen at STP? b. If the container were heated to 78O C before the combustion, what would the pressure be?

Answers

Explanation:

C9H2O + 14O2 → 9CO2 + 10H2O

a. How many molecules of nonane will burn in a tank that has 770 mL of oxygen at STP?

From the reaction;

1 mol of nonane reacts with 14 mol of Oxygen

At STP, 1 mol = 22.4 L

Converting 770 ml (0.77 L) to mol;

1 = 22.4

x = 0.77

x = 0.77 / 22.4 = 0.034375 mol

Back to the equation;

1 mol = 14 mol

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b. If the container were heated to 78O C before the combustion, what would the pressure be?

Initial Pressure P1 = ?

Initial Temperature T1 = 780 +273 = 1053 K

Final Temperature (STP) T2 = 273 K

Final Pressure (STP) P2 = 1 atm

Using gay lusaac's law;

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1 point

1. A balloon has 1.55 L of air in it exerting a pressure of 105.2 kPa. At

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Answers

Answer:

[tex]V_2=1.34L[/tex]

Explanation:

Hello there!

In this case, since this problem is describing how pressure changes as a function of volume and vice versa, it is possible recall the Boyle's law as shown below:

[tex]P_1V_1=P_2V_2[/tex]

Whereas we are asked to compute the volume when the change is pressure is performed (V₂); thus, we proceed as follows:

[tex]V_2=\frac{P_1V_1}{P_2}\\\\V_2=\frac{1.55L*105.2kPa}{122.0kPa}\\\\V_2=1.34L[/tex]

Best regards!

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