part 1 : the final volume : 1.404 L
part 2 : the initial concentration : 4.06 M
<h3>Further explanation
</h3>
Dilution is the process of adding a solvent to get a more dilute solution.
The moles(n) before and after dilution are the same.
Can be formulated :
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
part 1 :
M₁=44.8%
V₁=0.73 L
M₂=23.3%
part 2 :
V₁=739 ml=0.739 L
V₂=1.5 L
M₂=2
If the object has an irregular shape, the volume can be measured using a displacement can. The displacement can is filled with water above a narrow spout and allowed to drain until the water is level with the tap. As the irregular object is lowered into the displacement can, the water level rises.
Answer:
75 g
Explanation:
1 half life = 600/2 = 300g
2 half life = 300/2 = 150g
3 half life = 150/2 = 75g
Hi!
To make 500 mL of a 1,500 M solution of NaCl you'll require
43,83 g
To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:
Have a nice day!
Answer:
10.5g
Explanation:
First, let us calculate the number of mole of NaHCO3 present in the solution. This is illustrated below:
Volume = 250mL = 250/1000 = 0.25L
Molarity = 0.5M
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.5 x 0.25
Mole = 0.125 mole
Now, we shall be converting 0.125 mole of NaHCO3 to grams to obtain the desired result. This can be achieved by doing the following:
Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol
Number of mole of NaHCO3 = 0.125 mole
Mass of NaHCO3 =?
Mass = number of mole x molar Mass
Mass of NaHCO3 = 0.125 x 84
Mass of NaHCO3 = 10.5g
Therefore, 10.5g of NaHCO3 is needed.
