Answer:
The volume of solution in liters required to make a 0.250 M solution from 3.52 moles of solute is 14.08 liters of solution
Explanation:
The question relates to the definition of the concentration of a solution which is the number of moles per liter (1 liter = 1 dm³) of solution
Therefore we have;
The concentration of the intended solution = 0.250 M
Therefore, the number of moles per liter of the required resolution = 0.250 moles
Therefore, the concentration of the required solution = 0.250 moles/liter
The volume in liters of the required solution that will have 3.52 moles of the solute is given as follows;
The required volume of solution = The number of moles of the solute/(The concentration of the solution)
∴ The required volume of solution = 3.52 moles/(0.250 moles/liter) = 14.08 liters
The required volume of solution to make a 0.250 M solution from 3.52 moles of solute = 14.08 liters.
Therefore the number of liters required to make a 0.250 M solution from 3.52 moles of solute = 14.08 liters.
I'm not to sure but try soy
According to Boyle's law, volume is inversely proportional to pressure. thus P=k/V
Therefore PV=k
P1V1=P2V2
In the question above,
P1=3.67atm
P2=1.94atm
V1=2.22L
V2=?
Thus substituting for the values in the gas equation;
3.67atm*2.22L=1.94atm*V2
V2=3.67atm*2.22L/1.94atm
=4.21L
Answer:
23.2 g of Al will be left over when the reaction is complete
Explanation:
2Al + 3S → Al₂S₃
1 mol of Al = 26.98 g
1 mol of S = 32.06 g
Mole = Mass / Molar mass
63.8 g/ 26.98 g/m = 2.36 mole of Al
72.3 g / 32.06 g/m = 2.25 mole of S
2 mole of Aluminun react with 3 mole of sulfur
2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S
As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.
3 mole of S react with 2 mole of Al
2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole
I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.
2.36 mole of Al - 1.50 mole of Al = 0.86 mole
This is the quantity of Al without reaction.
Molar mass . mole = Mass → 26.98 g/m . 0.86 m = 23.2 g
(2.32g/cm³) x (1kg/1000g)x(1 000 000 cm³/1m³) = 2320 kg/m³
1 ml= 1 cm³
