Answer:
111.15 g are required to prepare 500 ml of a 3 M solution
Explanation:
In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.
Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of
(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.
Answer:
The answer to your question is P2 = 9075000 atm
Explanation:
Data
Pressure 1 = P1 = 5 atm
Volume 1 = V1 = 363 ml
Pressure 2 = P2 = ?
Volume 2 = 0.0002 ml
Process
To solve this problem use Boyle's law
P1V1 = P2V2
-Solve for P2
P2 = P1V1/V2
-Substitution
P2 = (5 x 363) / 0.0002
-Simplification
P2 = 1815 / 0.0002
-Result
P2 = 9075000 atm
Answer:it is wrong answer
Explanation:estro man
Answer:
The number before any molecular formula applies to the entire formula. So here you have five molecules of water with two hydrogen atoms and one oxygen atom per molecule. Thus you have ten hydrogen atoms and five oxygen atoms in total.
Answer:
0.50 M
Explanation:
Given data
- Mass of sodium sulfate (solute): 7.1 g
- Volume of solution: 100 mL
Step 1: Calculate the moles of the solute
The molar mass of sodium sulfate is 142.04 g/mol. The moles corresponding to 7.1 grams of sodium sulfate are:
Step 2: Convert the volume of solution to liters
We will use the relation 1 L = 1000 mL.
Step 3: Calculate the molarity of the solution
