Answer:
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Explanation:
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Answer:
0.0693M Fe
Explanation:
It is possible to quantify Fe in a sample using Mn as internal standard using response factor formula:
F = A(analyte)×C(std) / A(std)×C(analyte) <em>(1)</em>
Where A is area of analyte and std, and C is concentration.
Replacing with first values:
F = 1.05×2.00mg/mL / 1.00×2.50mg/mL
<em>F = 0.84</em>
In the unknown solution, concentration of Mn is:
13.5mg/mL × (1.00mL/6.00mL) = <em>2.25 mg Mn/mL</em>
Replacing in (1) with absorbances values and F value:
0.84 = 0.185×2.25mg/mL / 0.128×C(analyte)
C(analyte) = <em>3.87 mg Fe / mL</em>
As molarity is moles of solute (Fe) per liter of solution:
= <em>0.0693M Fe</em>
Answer:
The solution will not form a precipitate.
Explanation:
The Ksp of PbI₂ is:
PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)
Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>
When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:
[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>
[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>
<em />
Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>
If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.
Replacing:
Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹
As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.
The hydrogen bonds that form between water molecules account for some of the essential and unique properties of water. The attraction created by hydrogen bonds keepswater liquid over a wider range of temperature than is found for any other molecule its size.
Hope this helped!
