The reaction between the lead iodide, PbI2 and sodium carbonate, Na2CO3, will form the chemical compounds, sodium iodide and lead carbonate as shown below.
PbI2 + Na2CO3 --> NaI + PbCO3
The white precipitate formed in the reaction is PbCO3.
Answer:
C₆H₅COOH Benzoic Acid
Explanation:
Here we are using the extraction method to separate benzoic acid from naphtalene in the ether solution by reacting it with the base sodium bicarbonate to produce the salt sodium benzoate ( naphtalene remains in the organic layer ) .
The sodium benzoate is completely soluble in the aqueous layer being a salt, and reacts with hydrochloric acid ( again an acid base reaction ) which precipitates the benzoic acid since the it is insoluble in water hence separating it.
The 2 parts or components that make up a solution would be the solute and the solvent.
There is 1 unpaired electrons
The innermost shell has 2 electrons.
<span>The next shell has 8 electrons. </span>
<span>The next shell has 18 electrons. </span>
<span>The next shell has 12 electrons </span>
<span>The most outer shell has just 1 single electron (no pairs)
</span>have a good day :)