A solution containing lead(II) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0630 M in Pb(NO3)2 and 0.0103 M in NaBr. What is the value of Q for the insoluble product? Express the reaction quotient to three significant figures.

Q = 6.68x10⁻⁶

Explanation:

The initial solutions are Pb(NO₃)₂ and NaBr. As sodium and nitrates salts are soluble, the insoluble product must be the formed from the other ions, PbBr₂.

The soluble product equilibrium is written as:

PbBr₂(s) ⇄ Pb²⁺(aq) + 2Br⁻(aq)

Where Q is defined as:

Q = [Pb²⁺] [Br⁻]²

As:

[Pb²⁺] = 0.0630M

[Br⁻] = 0.0103M

Q = [Pb²⁺] [Br⁻]²

Q = [0.0630M] [0.0103M]²

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"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."