Answer:
5mL of 6 M HCl is needed to make 100 mL of 0.3 M of HCl.
Explanation:
Using the dilution formula:
M1V1 = M2V2
M1 = 0.30 M
V1 = 100 mL
M2 = 6.0 M
V2 = unknown
Re-arranging the formula by making V2 the subject of the equation:
V2 = M1V1 / M2
V2 = 0.30 * 100 / 6
V2 = 5 mL
The volume of 6 M of HCl needed to make 0.3 M at 100 mL is 5 mL.
A 2% saltwater is a dilute solution in which the solvent is water and the solute is salt. The solvent is the component of the solution which comprise a large percentage while the solute comprise the smaller part. When alcohol is added, the solute still is salt and the solvent is the mixture of water and alcohol.
Answer:
0.50603kg when you add given masses
Answer:
The molar mass of the unknown gas is 100.4 g/mol
Explanation:
Step 1: Data given
Molar mass of argon = 39.95 g/mol
After filling with argon the flask gained 3.221 grams
After filling with an unknown gas, the flask gained 8.107 grams
Step 2: Calculate the molar mass of the unknown gas
The gas with the higher molar mass will have the higher density.
Ar - 3.224 g; molar mass = 39.95 g/mol
X = 8.102 g; molar mass = ??
Molar mass of the unknown gas = 8.102g X *(39.95 g/mol / 3.224 g) = 100.4 g/mol
The molar mass of the unknown gas is 100.4 g/mol
