If 4.73 g Ar are added to 3.22 atm He in a 2.00 L cylinder at 27.0 °C. what is the total pressure of the resulting gaseous mixture?

4.67 atm

Explanation:

Step 1: Given data

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 27.0 °C + 273.15 = 300.2 K

Step 3: Calculate the moles of Ar

The molar mass of Ar is 39.94 g/mol.

4.73 g × 1 mol/39.94 g = 0.118 mol

Step 4: Calculate the partial pressure of Ar

We will use the ideal gas equation.

pAr × V = nAr × R × T

pAr = nAr × R × T / V

pAr = 0.118 mol × (0.0821 atm.L/mol.K) × 300.2 K / 2.00 L

pAr = 1.45 atm

Step 5: Determine the total pressure of the gaseous mixture

The total pressure of the gaseous mixture is equal to the sum of the partial pressures of the individual gases.

P = pAr + pHe = 1.45 atm + 3.22 atm = 4.67 atm

Hello!

A 25.5 liter balloon holding 3.5 moles of carbon dioxide leaks. If we are able to determine that 1.9 moles of carbon dioxide escaped before the container could be sealed, what is the new volume of the container?​

V1 (initial volume) = 25.5 L

n1 (initial number of moles) = 3.5 mol

V2 (final volume) = ? (in L )

Note: there was a leak in the number of moles, so we have:

n2 (final number of moles) = 3.5 mol - 1.9 mol = 1.6 mol

By Avogadro's Law it is known that the volume is directly proportional to the number of gas particles, that is, the larger the number of moles of gas, the greater its volume, on which we have the following relation:

We apply the data to the formula, we have:

multiply the means by the extremes

The new volume of the container is approximately 11.66 liters

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