Answer:
1.11 atm
(P, T)
2.83 atm
0.740 g
0.179 g/L, 0.158 g/L
N₂
Explanation:
<em>One mole of an ideal gas is sealed in a 22.4-L container at a pressure of 1 atm and a temperature of 273 K. The temperature is then increased to 304 K, but the container does not expand. What will the new pressure be?</em>
Assuming ideal behavior, we can calculate the new pressure (P₂) using Gay-Lussac's law.

<em>The most appropriate formula for solving this problem includes only which variables?</em>
Gay-Lussac's law includes pressure (P) and absolute temperature (T).
<em>Q2) A sample of nitrogen gas in a 1.69-L container exerts a pressure of 1.37 atm at 17 °C. What is the pressure if the volume of the container is maintained constant and the temperature is raised to 327 °C?
</em>
Initially the system is at 17°C (290 K) and the temperature is raised to 327°C (600 K). We can calculate the new pressure using Gay-Lussac's law.

<em>Q3) A gas mixture with a total pressure of 770 mmHg contains each of the following gases at the indicated partial pressures: 120 mmHg CO₂, 227mmHg Ar, and 190 mmHg O₂. The mixture also contains helium gas
.</em>
What mass of helium gas is present in a 14.0-L sample of this mixture at 282 K?
First, we have to calculate the pressure of Helium. We know that the total pressure is the sum of partial pressures.
Ptotal = pCO₂ + pAr + pO₂ + pHe
pHe = Ptotal - pCO₂ - pAr - pO₂
pHe = 770mmHg - 120mmHg - 227mmHg - 190mmHg=233mmHg
We can calculate the moles of Helium using the ideal gas equation.

The molar mass of He is 4.00g/mol.

<em>Calculate the density of oxygen, O₂, under each of the following conditions:
</em>
- <em>1.00 atm and 35.0 ∘C
</em>
<em>
Express your answers numerically in grams per liter. Enter the density at STP first and separate your answers by a comma.</em>
<em />
STP stands for Standard Temperature and Pressure. The standard temperature is 273 K and the standard pressure is 1 atm.
We can calculate the density using the following expression:

<em>At 1.00 atm and 35.0 °C (308 K)</em>

<em>To identify a diatomic gas (X₂), a researcher carried out the following experiment: She weighed an empty 4.1-L bulb, then filled it with the gas at 2.00 atm and 24.0 ∘C and weighed it again. The difference in mass was 9.5 g . Identify the gas. Express your answer as a chemical formula.</em>
We will look for the molar mass of the compound using the ideal gas equation.

If the molar mass of X₂ is 28 g/mol, the molar mass of X is 14 g/mol. Then, X is nitrogen and X₂ is N₂.