An experiment calls for 50 ml of a 0.50M aqueous solution of a sodium bicarnonate. Work with your group to describe the steps you would use in order to make up such a solution making sure you have none left over

We have to weight out 2.1 grams of NaHCO3 and dissolve it in 50 mL water.

Explanation:

Step 1: Data given

Volume = 50 mL = 0.050 L

Molarity sodium bicarbonate (NaHCO3) = 0.50 M

Molar mass NaHCO3 = 84.0 g/mol

Step 2: Calculate moles NaHCO3

Moles NaHCO3 = molarity NaHCO3 * volume NaHCO3

Moles NaHCO3 = 0.50 M * 0.050 L

Moles NaHCO3 = 0.025 moles

Step 3: Calculate mass NaHCO3

Mass NaHCO3 = moles NaHCO3 * molar mass NaHCO3

Mass NaHCO3 = 0.025 moles * 84.0 g/mol

Mass NaHCO3 = 2.1 grams NaHCO3

We have to weight out 2.1 grams of NaHCO3 and dissolve it in 50 mL water. This gives us a 0.50 M aqueous solution of a sodium bicarnonate

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)