When concentration is expressed in molarity, this is equivalent to the number of moles of the solute per liter of solution. We are given with the amount of volume which is 239 mL or 0.239 L. However, there is no known information of the amount of solute. So, I can't give an exact answer. For sample purposes, let's just assume that there is 1 mole of KCl in the solution. The molarity would be:
Molarity = 1 moles/0.239 L = 4.184 M
Answer:
1.90 L
Explanation:
Using Boyle's law
Given ,
V₁ = 595 L
V₂ = ?
P₁ = 1.00 atm
P₂ = 55.0 atm
Using above equation as:
<u>The volume would be 1.90 L.</u>
We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows,
M1V1 = M2V2
where M1 is the concentration
of the stock solution, V1 is the volume of the stock solution, M2 is the
concentration of the new solution and V2 is its volume.
M1V1 = M2V2
0.266 M x V1 = 0.075 M x 150 mL
V1 = 42.29 mL
Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.
That looks like cells of a multicellular organism, so B.
Answer:
46.40 g.
Explanation:
- It is a stichiometric problem.
- The balanced equation of the reaction: 4K + O₂ → 2K₂O.
- It is clear that 4.0 moles of K reacts with 1.0 mole of oxygen produces 2.0 moles of K₂O.
- We should convert the mass of K (38.5 g) into moles using the relation:
<em>n = mass / molar mass,</em>
n = (38.5 g) / (39.098 g/mol) = 0.985 mole.
<em>Using cross multiplication:</em>
4.0 moles of K produces → 2.0 moles of K₂O, from the stichiometry.
0.985 mole of K produces → ??? moles of K₂O.
∴ The number of moles of K₂O produced = (0.985 mole) (2.0 mole) / (4.0 mole) = 0.4925 mole ≅ 0.5 mole.
- Now, we can get the mass of K₂O:
∴ mass = n x molar mass = (0.5 mole) (94.2 g/mol) = 46.40 g.
