3I₂ + 2Al → 2AlI₃
m(I₂)=3M(I₂)m(Al)/{2M(Al)}
m(I₂)=3*253.8*20.4/{2*27.0}=287.64 g
Answer:
Final concentration of NaOH = 0.75 M
Explanation:
For
:-
Given mass = 90.0 g
Molar mass of NaOH = 39.997 g/mol
The formula for the calculation of moles is shown below:
Thus,

Molarity is defined as the number of moles present in one liter of the solution. It is basically the ratio of the moles of the solute to the liters of the solution.
The expression for the molarity, according to its definition is shown below as:
Where, Volume must be in Liter.
It is denoted by M.
Given, Volume = 3.00 L
So,
<u>Final concentration of NaOH = 0.75 M</u>
For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:
Kd = c </span>× α²
α = √(Kd/c) × 100%
Kd = 6.0×10⁻⁷
c(HA) = 0.1M
α = √(6.0×10⁻⁷/0.1) × 100% = 0.23%
So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>
Answer:
Mass = 0.158 g
Explanation:
Formula used,
P V = n R T
Or,
n = P V / R T
Putting values,
n = 0.948 atm . 0.025 L / 0.0821 L.atm.K⁻¹.mol⁻¹ . 291.45
n = 0.00099 mol
Note: we have changed pressure from mmHg to atm, volume from mL to L and temperature from C to K)
Also,
Mass = n . Molecular Mass
Mass = 0.00099 mol × 159.808 g/mol
Mass = 0.158 g