if we did not use an excess of the BaCl2 solution it would decrease the mass percentage of sulfate in the unknown sample.
The net precipitation equation would be.
Ba2+(aq) + SO42-(aq) → BaSO4(s)
If BaCl2 (Ba2+) is not taken in excess then the precipitation would not be completed as some of the sulfate ions would still be remaining in the solution. This would decrease the mass percentage of sulfate in the unknown sample.
If some tiny pieces of filter paper still remained mixed with the precipitate(BaSO4) then the mass of sulfate would increase and it gives a high mass percentage of the sulfate.
mass percentage of sulfate = (mass of sulfate/mass of sample)*100
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The answer would be letter A - bromine. Bromine <span>(B<span>r2</span>)</span>
has only london dispersion forces and are weak, but it's large electron
cloud allows for transient polarization creating dipole moments that
strengthen their interaction.
In an ionic bond :
=》B. one atom accepts electrons from another.
in this bond an atom ( <em><u>metallic</u></em> ) loses its electrons and another atom ( <em><u>non- metallic</u></em> ) accepts the electrons, and since there isn't the equal positive and negative charges in the atoms, they forms <em><u>cations</u></em> ( +ve charge ) and <em><u>anions </u></em>( -ve charge )
and get stacked or <em><u>attracted</u></em> to each other by strong <em><u>electrostatic force</u></em>.
