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>.
<span>The </span>octet rule<span> is a chemical </span>rule<span> of thumb that reflects observation that atoms of main-group elements tend to combine in such a way that each atom has eight electrons in its valence shell, giving it the same electronic configuration as a noble gas.
Hope this helps</span>
For an element whose third shell contains six electrons, the appropriate electron configuration is; 1s2 2s2 2p6 3s2 3p4.
The electron configuration shows the distribution of electrons in the shells of an atom and in orbitals.
We have been told that the six electrons are found in the third shell. This shell has n=3 and the configuration of this shell must ns2 np4.
The only electron configuration that meets this standard is 1s2 2s2 2p6 3s2 3p4.
Learn more: brainly.com/question/18704022
Answer:
<h2>6.64 moles</h2>
Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
We have the final answer as
<h3>6.64 moles</h3>
Hope this helps you
