Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)
v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol
n(Na₂S)=n{Cd(NO₃)₂}=cv
m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv
m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
Answer:
11482 ppt of Li
Explanation:
The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:
<em>Moles Mg²⁺:</em>
0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA
<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>
0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium
That means mass of lithium is (Molar mass Li=6.941g/mol):
4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:
0.00287g * (1000000μg / g) = 2870μg of Li
As ppt is μg of solute / Liter of solution, ppt of the solution is:
2870μg of Li / 0.250L =
<h3>11482 ppt of Li</h3>
Pressure of argon = 546.8 kPa
Conversion factor: 1 atm = 101.325 kPa
Pressure of argon = 546.8 kPa x 1 atm/101.325 kPa = 5.4 atm
Moles of argon = 15.82
Volume of argon = 75.0 L
According to Ideal gas law,
PV = nRT
where P is the pressure, V is the volume , n is the number of moles, R is the universal gas constant, and T is the temperature
T = PV/nR = (5.4 atm x 75.0 L) / (15.82 x 0.0821 L.atm.mol⁻¹K⁻¹)
T = 311.82 K
Hence the temperature of the canister is 311.82 K.
<h2>
Hello!</h2>
The answer is:
<h2>
Why?</h2>
Since there is not information about the solute but only its mass, we need to assume that we are calculating the molar concentration of a solution or molarity. So, need to use the following formula:
Now, we know that the mass of the solute is equal 3.5 moles and the volume is equal to 1500 mL or 1.5L
Then, substituting into the equation, we have:
Have a nice day!
