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

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

the answer is barium and copper bromide