An x would represent the gained electrons
A . Would represent the valence electrons
You would just draw [ ] around the diagram
And the charge should be placed outside the brackets
Answer:
The Lewis dot diagram is supposed to have dots on each side. What's incorrect is that there isn't a dot on the bottom, only the left and right side and the top. What's correct about this is that there are 5 outer valence electrons, and they correctly put 5 dots, even though they're in the wrong place.
Explanation:
Answer:
d. the conjugate base of the weak acid
Explanation:
The strong base (BOH) is completely dissociated in water:
BOH → B⁺ + OH⁻
The resulting conjugate acid (OH⁻) is a weak acid, so it remains in solution as OH⁻ ions.
By other hand, the weak acid (HA) is only slightly dissociated in water:
HA ⇄ H⁺ + A⁻
The resulting conjugate base (A⁻) is a weak base. Thus, it reacts with H⁺ ions from water to form HA, increasing the concentration of OH⁻ ions in the solution.
Therefore, the resulting solution will have a pH > 7 (basic).
Answer: The mass percentage of
is 5.86%
Explanation:
To calculate the mass percentage of
in the sample it is necessary to know the mass of the solute (
in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).
To calculate the mass of the solute, we must take the mass of the
precipitate. We can establish a relation between the mass of
and
using the stoichiometry of the compounds:

Since for every mole of Tl in
there are two moles of Tl in
, we have:

Using the molar mass of
we have:

Finally, we can use the mass percentage formula:

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