0.0200 m fe3+ is initially mixed with 1.00 m oxalate ion, c2o42-, and they react according to the equation: fe3+(aq) + 3 c2o42-(aq) ⇄ [fe(c2o4)3]3-(aq) kc = 1.67 × 1020 what is the concentration of fe3+(aq) when equilibrium is reached? 1.67 × 1020 m 8.35 × 10-19 m 6.9a × 1021 m 1.44 × 10-22 m 0.980 am 0.940 m 0.0100 atm

Answer : The concentration of at equilibrium is 0 M.

Solution :  Given,

Concentration of = 0.0200 M

Concentration of = 1.00 M

The given equilibrium reaction is,

Initially conc.       0.02         1.00                   0

At eqm.             (0.02-x)    (1.00-3x)                x

The expression of will be,

By solving the term, we get:

Concentration of at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of at equilibrium is 0 M.

your parents i think this could be it

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.