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

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answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.